$\displaystyle(u-t)\,D_{1R}^{0235}(t,u,Q^{2})\biggr{]}=\frac{1}{2ut}\,\biggl{[}(Q^{2}-u)\,C_{0}(u,Q^{2})$
$\displaystyle-$
$\displaystyle(Q^{2}-t)\,C_{0}(t,Q^{2})+u\,C_{0}(u)-t\,C_{0}(t)+ut\,D_{0}(t,u,Q^{2})\biggr{]}\,.$

Relations:

$\displaystyle\left\\{\begin{array}[]{lccl}D_{1P/1T}^{0125}(s,u,Q^{2})&=&-&D_{1P/1T}^{0126}(s,u,Q^{2})\,,\\\\[2.84526pt]
D_{1R}^{0125}(s,u,Q^{2})&=&+&D_{1R}^{0126}(s,u,Q^{2})\,,\\\\[2.84526pt]
D_{1P}^{0125/0126}(s,u,Q^{2})&=&\pm&D_{1R}^{0125/0126}(s,u,Q^{2})\mp\dfrac{C_{0}(s,Q^{2})-C_{0}(u,Q^{2})}{s}\,,\\\\[2.84526pt]
D_{1P}^{0235}(t,u,Q^{2})&=&-&D_{1R}^{0235}(t,u,Q^{2})+\dfrac{C_{0}(t,Q^{2})-C_{0}(u)}{s}\,\,.\end{array}\right.$

###  B.3 Tensor box integrals

Here we list scalar functions occurring in expansion of tensor box integrals.

Integral $D_{2}^{\mu\nu;0125}(p_{1},p_{2},k_{1})$

$\displaystyle D_{2PP}^{0125}(s,u,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{P_{\mu}P_{\nu}}{P^{4}}\,D_{2}^{\mu\nu;0125}(p_{1},p_{2},k_{1})=\frac{1}{2s}\,\biggl{[}C_{1P}^{025}(s,u,Q^{2})-C_{1P}^{015}(s,u,Q^{2})\biggr{]}\,,$
$\displaystyle D_{2PR}^{0125}(s,u,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{P_{\mu}R_{\nu}}{P^{2}R^{2}}\,D_{2}^{\mu\nu;0125}(p_{1},p_{2},k_{1})=\frac{1}{2s}\,\biggl{[}C_{1R}^{025}(s,u,Q^{2})-C_{1R}^{015}(s,u,Q^{2})\biggr{]}\,,$
$\displaystyle D_{2PT}^{0125}(s,u,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{P_{\mu}T_{\nu}}{P^{2}T^{2}}\,D_{2}^{\mu\nu;0125}(p_{1},p_{2},k_{1})=\frac{1}{2s}\,\biggl{[}C_{1T}^{025}(u,Q^{2})-C_{1T}^{015}(u)\biggr{]}\,,$
$\displaystyle D_{2RR}^{0125}(s,u,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{R_{\mu}R_{\nu}}{R^{4}}\,D_{2}^{\mu\nu;0125}(p_{1},p_{2},k_{1})=\frac{1}{2s}\,\biggl{[}-C_{1R}^{025}(s,u,Q^{2})-C_{1R}^{015}(s,u,Q^{2})$
$\displaystyle+$ $\displaystyle
2C_{1R}^{078}(s,u,Q^{2})-C_{0}(s,Q^{2})\biggr{]}\,,$ $\displaystyle
D_{2RT}^{0125}(s,u,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{R_{\mu}T_{\nu}}{R^{2}T^{2}}\,D_{2}^{\mu\nu;0125}(p_{1},p_{2},k_{1})=\frac{1}{2s}\,\biggl{[}-C_{1T}^{025}(u,Q^{2})-C_{1T}^{015}(u)$
$\displaystyle+$ $\displaystyle 2C_{1T}^{078}(s,Q^{2})\biggr{]}\,,$
$\displaystyle D_{2TT}^{0125}(s,u,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{T_{\mu}T_{\nu}}{T^{4}}\,D_{2}^{\mu\nu;0125}(p_{1},p_{2},k_{1})=\frac{s}{2ut}\,\biggl{[}\frac{u}{s}\,C_{1T}^{015}(u)-\frac{s+t}{s}\,C_{1T}^{025}(u,Q^{2})$
$\displaystyle-$
$\displaystyle\frac{u-t}{s}\,C_{1T}^{078}(s,Q^{2})-u\,D_{1T}^{0125}(s,u,Q^{2})\biggr{]}\,,$
$\displaystyle D_{2g}^{0125}(s,u,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{g_{\mu\nu}^{\perp}}{D-3}\,D_{2}^{\mu\nu;0125}(p_{1},p_{2},k_{1})=\frac{1}{D-3}\,\biggl{[}C_{0}(s,Q^{2})-s\,D_{2PP}^{0125}(s,u,Q^{2})$
(433) $\displaystyle+$ $\displaystyle
s\,D_{2RR}^{0125}(s,u,Q^{2})+\frac{ut}{s}\,D_{2TT}^{0125}(s,u,Q^{2})\biggr{]}\,.$

Integral $D_{2}^{\mu\nu;0126}(p_{1},p_{2},k_{1})$

$\displaystyle D_{2PP}^{0126}(s,t,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{P_{\mu}P_{\nu}}{P^{4}}\,D_{2}^{\mu\nu;0126}(p_{1},p_{2},k_{1})=\frac{1}{2s}\,\biggl{[}C_{1P}^{026}(s,t,Q^{2})-C_{1P}^{016}(s,t,Q^{2})\biggr{]}$
$\displaystyle=$ $\displaystyle+D_{2PP}^{0125}(s,t,Q^{2})\,,$ $\displaystyle
D_{2PR}^{0126}(s,t,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{P_{\mu}R_{\nu}}{P^{2}R^{2}}\,D_{2}^{\mu\nu;0126}(p_{1},p_{2},k_{1})=\frac{1}{2s}\,\biggl{[}C_{1R}^{026}(s,t,Q^{2})-C_{1R}^{016}(s,t,Q^{2})\biggr{]}$
$\displaystyle=$ $\displaystyle-D_{2PP}^{0125}(s,t,Q^{2})\,,$ $\displaystyle
D_{2PT}^{0126}(s,t,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{P_{\mu}T_{\nu}}{P^{2}T^{2}}\,D_{2}^{\mu\nu;0126}(p_{1},p_{2},k_{1})=\frac{1}{2s}\,\biggl{[}C_{1T}^{026}(t)-C_{1T}^{016}(t,Q^{2})\biggr{]}$
$\displaystyle=$ $\displaystyle+D_{2PT}^{0125}(s,t,Q^{2})\,,$ $\displaystyle
D_{2RR}^{0126}(s,t,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{R_{\mu}R_{\nu}}{R^{4}}\,D_{2}^{\mu\nu;0126}(p_{1},p_{2},k_{1})=\frac{1}{2s}\,\biggl{[}-C_{1R}^{026}(s,t,Q^{2})-C_{1R}^{016}(s,t,Q^{2})$
$\displaystyle+$ $\displaystyle
2C_{1R}^{079}(s,t,Q^{2})-C_{0}(s,Q^{2})\biggr{]}=+D_{2RR}^{0125}(s,t,Q^{2})\,,$
$\displaystyle D_{2RT}^{0126}(s,t,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{R_{\mu}T_{\nu}}{R^{2}T^{2}}\,D_{2}^{\mu\nu;0126}(p_{1},p_{2},k_{1})=\frac{1}{2s}\,\biggl{[}-C_{1T}^{026}(t,Q^{2})-C_{1T}^{016}(t)$
$\displaystyle+$ $\displaystyle
2C_{1T}^{079}(s,Q^{2})\biggr{]}=-D_{2RT}^{0125}(s,t,Q^{2})\,,$ $\displaystyle
D_{2TT}^{0126}(s,t,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{T_{\mu}T_{\nu}}{T^{4}}\,D_{2}^{\mu\nu;0126}(p_{1},p_{2},k_{1})=\frac{s}{2ut}\,\biggl{[}-\frac{t}{s}\,C_{1T}^{026}(t)+\frac{s+u}{s}\,C_{1T}^{016}(t,Q^{2})$
$\displaystyle-$
$\displaystyle\frac{u-t}{s}\,C_{1T}^{079}(s,Q^{2})+t\,D_{1T}^{0126}(s,t,Q^{2})\biggr{]}=+D_{2TT}^{0125}(s,t,Q^{2})\,,$
$\displaystyle D_{2g}^{0126}(s,t,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{g_{\mu\nu}^{\perp}}{D-3}\,D_{2}^{\mu\nu;0126}(p_{1},p_{2},k_{1})$
(434) $\displaystyle=$
$\displaystyle\frac{1}{D-3}\,\biggl{[}C_{0}(s,Q^{2})-s\,D_{2PP}^{0126}(s,t,Q^{2})+s\,D_{2RR}^{0126}(s,t,Q^{2})$
$\displaystyle+$
$\displaystyle\frac{ut}{s}\,D_{2TT}^{0126}(s,t,Q^{2})\biggr{]}=+D_{2g}^{0125}(s,t,Q^{2})\,.$

Integral $D_{2}^{\mu\nu;0235}(p_{1},p_{2},k_{1})$

$\displaystyle D_{2PP}^{0235}(t,u,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{P_{\mu}P_{\nu}}{P^{4}}\,D_{2}^{\mu\nu;0235}(p_{1},p_{2},k_{1})=\frac{1}{2s}\,\biggl{[}\frac{C_{0}(t,Q^{2})}{2}-C_{1P}^{046}(s,t,Q^{2})$
$\displaystyle+$ $\displaystyle
C_{1P}^{023}(s,t,Q^{2})-C_{1P}^{025}(s,u,Q^{2})-C_{1P}^{035}(s,u,Q^{2})-uD_{1P}^{0235}(t,u,Q^{2})\biggr{]}\,,$
$\displaystyle D_{2PR}^{0235}(t,u,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{P_{\mu}R_{\nu}}{P^{2}R^{2}}\,D_{2}^{\mu\nu;0235}(p_{1},p_{2},k_{1})=\frac{1}{2s}\,\biggl{[}-\frac{C_{0}(t,Q^{2})}{2}-C_{1R}^{046}(s,t,Q^{2})$
$\displaystyle+$ $\displaystyle
C_{1R}^{023}(s,t,Q^{2})-C_{1R}^{025}(s,u,Q^{2})-C_{1R}^{035}(s,u,Q^{2})-uD_{1R}^{0235}(t,u,Q^{2})\biggr{]}\,,$
$\displaystyle D_{2PT}^{0235}(t,u,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{P_{\mu}T_{\nu}}{P^{2}T^{2}}\,D_{2}^{\mu\nu;0235}(p_{1},p_{2},k_{1})=\frac{1}{2s}\,\biggl{[}-C_{1T}^{046}(t,Q^{2})+C_{1T}^{023}(t)$
$\displaystyle-$ $\displaystyle
C_{1T}^{025}(u,Q^{2})-C_{1T}^{035}(u)-uD_{1T}^{0235}(t,u,Q^{2})\biggr{]}\,,$
$\displaystyle D_{2RR}^{0235}(t,u,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{R_{\mu}R_{\nu}}{R^{4}}\,D_{2}^{\mu\nu;0235}(p_{1},p_{2},k_{1})=\frac{1}{2s}\,\biggl{[}-\frac{C_{0}(t,Q^{2})}{2}-C_{1R}^{046}(s,t,Q^{2})$
$\displaystyle-$ $\displaystyle
C_{1R}^{023}(s,t,Q^{2})+C_{1R}^{025}(s,u,Q^{2})-C_{1R}^{035}(s,u,Q^{2})+uD_{1R}^{0235}(t,u,Q^{2})\biggr{]}\,,$
$\displaystyle D_{2RT}^{0235}(t,u,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{R_{\mu}T_{\nu}}{R^{2}T^{2}}\,D_{2}^{\mu\nu;0235}(p_{1},p_{2},k_{1})=\frac{1}{2s}\,\biggl{[}-C_{1T}^{046}(t,Q^{2})-C_{1T}^{023}(t)+C_{1T}^{025}(u,Q^{2})$
$\displaystyle-$ $\displaystyle
C_{1T}^{035}(u)+uD_{1T}^{0235}(t,u,Q^{2})\biggr{]}\,,$ $\displaystyle
D_{2TT}^{0235}(t,u,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{T_{\mu}T_{\nu}}{T^{4}}\,D_{2}^{\mu\nu;0235}(p_{1},p_{2},k_{1})=\frac{s}{2ut}\,\biggl{[}\frac{s+u}{s}\,C_{1T}^{046}(t,Q^{2})+\frac{s+t}{s}\,C_{1T}^{025}(u,Q^{2})$
$\displaystyle+$
$\displaystyle\frac{u}{s}\,C_{1T}^{035}(u)-\frac{t}{s}\,C_{1T}^{023}(t)+\frac{ut}{s}D_{1T}^{0235}(t,u,Q^{2})\biggr{]}\,,$
$\displaystyle D_{2g}^{0235}(t,u,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{g_{\mu\nu}^{\perp}}{D-3}\,D_{2}^{\mu\nu;0235}(p_{1},p_{2},k_{1})=\frac{1}{D-3}\,\biggl{[}C_{0}^{046}(t,Q^{2})-s\,D_{2PP}^{0235}(t,u,Q^{2})$
(435) $\displaystyle+$ $\displaystyle
s\,D_{2RR}^{0235}(t,u,Q^{2})+\frac{ut}{s}\,D_{2TT}^{0235}(t,u,Q^{2})\biggr{]}\,.$

Relations:

$\displaystyle\left\\{\begin{array}[]{lccl}D_{i}^{0125}(s,u,Q^{2})&=&+&D_{i}^{0126}(s,u,Q^{2})\,,\quad
i=2g,2PP,2RR,2TT,2PT\\\\[2.84526pt]
D_{i}^{0125}(s,u,Q^{2})&=&-&D_{i}^{0126}(s,u,Q^{2})\,,\quad i=2PR,2TR\\\
\end{array}\right.$

Functions occurring in the expansion of tensor rank-3 box integrals (54)-(56)
read:

Integral $D_{3}^{\mu\nu\alpha}(s,u,Q^{2})$

$\displaystyle D_{3PPP}^{0125}(s,u,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{P_{\mu}P_{\nu}P_{\alpha}}{P^{6}}\,D_{3}^{\mu\nu\alpha}(s,u,Q^{2})=\frac{1}{2s}\,\biggl{[}C_{2PP}^{025}(s,u,Q^{2})-C_{2PP}^{015}(s,u,Q^{2})\biggr{]}\,,$
$\displaystyle D_{3PPR}^{0125}(s,u,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{P_{\mu}P_{\nu}R_{\alpha}}{P^{4}R^{2}}\,D_{3}^{\mu\nu\alpha}(s,u,Q^{2})=\frac{1}{2s}\,\biggl{[}C_{2PR}^{025}(s,u,Q^{2})-C_{2PR}^{015}(s,u,Q^{2})\biggr{]}\,,$
$\displaystyle D_{3PPT}^{0125}(s,u,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{P_{\mu}P_{\nu}T_{\alpha}}{P^{4}T^{2}}\,D_{3}^{\mu\nu\alpha}(s,u,Q^{2})=\frac{1}{2s}\,\biggl{[}C_{2PT}^{025}(s,u,Q^{2})-C_{2PT}^{015}(s,u,Q^{2})\biggr{]}\,,$
$\displaystyle D_{3RRP}^{0125}(s,u,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{P_{\mu}R_{\nu}R_{\alpha}}{R^{4}P^{2}}\,D_{3}^{\mu\nu\alpha}(s,u,Q^{2})=\frac{1}{2s}\,\biggl{[}C_{2RR}^{025}(s,u,Q^{2})-C_{2RR}^{015}(s,u,Q^{2})\biggr{]}\,,$
$\displaystyle D_{3TTP}^{0125}(s,u,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{P_{\mu}T_{\nu}T_{\alpha}}{T^{4}P^{2}}\,D_{3}^{\mu\nu\alpha}(s,u,Q^{2})=\frac{1}{2s}\,\biggl{[}C_{2TT}^{025}(s,u,Q^{2})-C_{2TT}^{015}(s,u,Q^{2})\biggr{]}\,,$
$\displaystyle D_{3PRT}^{0125}(s,u,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{P_{\mu}R_{\nu}T_{\alpha}}{P^{2}R^{2}T^{2}}\,D_{3}^{\mu\nu\alpha}(s,u,Q^{2})=\frac{1}{2s}\,\biggl{[}C_{2RT}^{025}(s,u,Q^{2})-C_{2RT}^{015}(s,u,Q^{2})\biggr{]}\,,$
$\displaystyle D_{3RRR}^{0125}(s,u,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{R_{\mu}R_{\nu}R_{\alpha}}{R^{6}}\,D_{3}^{\mu\nu\alpha}(s,u,Q^{2})$
$\displaystyle=$
$\displaystyle\frac{1}{2s}\,\biggl{[}-C_{2RR}^{025}(s,u,Q^{2})-C_{2RR}^{015}(s,u,Q^{2})+\frac{C_{0}^{078}(s,Q^{2})}{2}$
$\displaystyle-$ $\displaystyle
2C_{1R}^{078}(s,u,Q^{2})+2C_{2RR}^{078}(s,u,Q^{2})\biggr{]}\,,$ $\displaystyle
D_{3RRT}^{0125}(s,u,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{R_{\mu}R_{\nu}T_{\alpha}}{R^{4}T^{2}}\,D_{3}^{\mu\nu\alpha}(s,u,Q^{2})$
$\displaystyle=$
$\displaystyle\frac{1}{2s}\,\biggl{[}-C_{2RT}^{025}(s,u,Q^{2})-C_{2RT}^{015}(s,u,Q^{2})$
$\displaystyle+$ $\displaystyle
2C_{2RT}^{078}(s,u,Q^{2})-C_{1T}^{078}(s,Q^{2})\biggr{]}\,,$ $\displaystyle
D_{3TTR}^{0125}(s,u,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{R_{\mu}T_{\nu}T_{\alpha}}{T^{4}R^{2}}\,D_{3}^{\mu\nu\alpha}(s,u,Q^{2})$
$\displaystyle=$
$\displaystyle\frac{1}{2s}\,\biggl{[}-C_{2TT}^{025}(s,u,Q^{2})-C_{2TT}^{015}(s,u,Q^{2})+2C_{2TT}^{078}(s,u,Q^{2})\biggr{]}\,,$
$\displaystyle D_{3TTT}^{0125}(s,u,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{T_{\mu}T_{\nu}T_{\alpha}}{T^{6}}\,D_{3}^{\mu\nu\alpha}(s,u,Q^{2})$
$\displaystyle=$
$\displaystyle\frac{s}{2ut}\,\biggl{[}-uD_{2TT}^{0125}(s,u,Q^{2})+C_{2TT}^{012}(s)+\frac{u}{s}\,C_{2TT}^{015}(s,u,Q^{2})$
$\displaystyle-$
$\displaystyle\frac{s+t}{s}\,C_{2TT}^{025}(s,u,Q^{2})-\frac{u-t}{s}\,C_{2TT}^{078}(s,u,Q^{2})\biggr{]}\,,$
$\displaystyle D_{3gP}^{0125}(s,u,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{g_{\mu\nu}^{\perp}P_{\alpha}}{(D-3)P^{2}}\,D_{3}^{\mu\nu\alpha}(s,u,Q^{2})$
$\displaystyle=$
$\displaystyle\frac{1}{D-3}\,\biggl{[}\frac{C_{0}(s,Q^{2})}{2}+C_{1P}^{078}(s,u,Q^{2})+s\,D_{3P}^{0125}(s,u,Q^{2})\biggr{]}\,,$
$\displaystyle D_{3gR}^{0125}(s,u,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{g_{\mu\nu}^{\perp}R_{\alpha}}{(D-3)R^{2}}\,D_{3}^{\mu\nu\alpha}(s,u,Q^{2})$
$\displaystyle=$
$\displaystyle\frac{1}{D-3}\,\biggl{[}-\frac{C_{0}(s,Q^{2})}{2}+C_{1R}^{078}(s,u,Q^{2})+s\,D_{3R}^{0125}(s,u,Q^{2})\biggr{]}\,,$
$\displaystyle D_{3gT}^{0125}(s,u,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{g_{\mu\nu}^{\perp}T_{\alpha}}{(D-3)T^{2}}\,D_{3}^{\mu\nu\alpha}(s,u,Q^{2})$
(437) $\displaystyle=$
$\displaystyle\frac{1}{D-3}\,\biggl{[}C_{1T}^{078}(s,Q^{2})+s\,D_{3T}^{0125}(s,u,Q^{2})\biggr{]}\,.$

Here and in the following

$\displaystyle D_{3L}^{0ijk}(s,u,Q^{2})$ $\displaystyle=$ $\displaystyle-
D_{3LPP}^{0ijk}(s,u,Q^{2})+D_{3LRR}^{0ijk}(s,u,Q^{2})+\frac{ut}{s^{2}}D_{3LTT}^{0ijk}(s,u,Q^{2})\,,$
$\displaystyle L$ $\displaystyle=$ $\displaystyle P,R,T\,.$ (438)

Integral $D_{3}^{\mu\nu\alpha}(s,t,Q^{2})$

$\displaystyle D_{3PPP}^{0126}(s,t,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{P_{\mu}P_{\nu}P_{\alpha}}{P^{6}}\,D_{3}^{\mu\nu\alpha}(s,t,Q^{2})=\frac{1}{2s}\,\biggl{[}C_{2PP}^{026}(s,t,Q^{2})-C_{2PP}^{016}(s,t,Q^{2})\biggr{]}\,,$
$\displaystyle D_{3PPR}^{0126}(s,t,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{P_{\mu}P_{\nu}R_{\alpha}}{P^{4}R^{2}}\,D_{3}^{\mu\nu\alpha}(s,t,Q^{2})=\frac{1}{2s}\,\biggl{[}C_{2PR}^{026}(s,t,Q^{2})-C_{2PR}^{016}(s,t,Q^{2})\biggr{]}\,,$
$\displaystyle D_{3PPT}^{0126}(s,t,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{P_{\mu}P_{\nu}T_{\alpha}}{P^{4}T^{2}}\,D_{3}^{\mu\nu\alpha}(s,t,Q^{2})=\frac{1}{2s}\,\biggl{[}C_{2PT}^{026}(s,t,Q^{2})-C_{2PT}^{016}(s,t,Q^{2})\biggr{]}\,,$
$\displaystyle D_{3RRP}^{0126}(s,t,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{P_{\mu}R_{\nu}R_{\alpha}}{R^{4}P^{2}}\,D_{3}^{\mu\nu\alpha}(s,t,Q^{2})=\frac{1}{2s}\,\biggl{[}C_{2RR}^{026}(s,t,Q^{2})-C_{2RR}^{016}(s,t,Q^{2})\biggr{]}\,,$
$\displaystyle D_{3TTP}^{0126}(s,t,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{P_{\mu}T_{\nu}T_{\alpha}}{T^{4}P^{2}}\,D_{3}^{\mu\nu\alpha}(s,t,Q^{2})=\frac{1}{2s}\,\biggl{[}C_{2TT}^{026}(s,t,Q^{2})-C_{2TT}^{016}(s,t,Q^{2})\biggr{]}\,,$
$\displaystyle D_{3PRT}^{0126}(s,t,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{P_{\mu}R_{\nu}T_{\alpha}}{P^{2}R^{2}T^{2}}\,D_{3}^{\mu\nu\alpha}(s,z,Q^{2})=\frac{1}{2s}\,\biggl{[}C_{2RT}^{026}(s,t,Q^{2})-C_{2RT}^{016}(s,t,Q^{2})\biggr{]}\,,$
$\displaystyle D_{3RRR}^{0126}(s,t,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{R_{\mu}R_{\nu}R_{\alpha}}{R^{6}}\,D_{3}^{\mu\nu\alpha}(s,t,Q^{2})=\frac{1}{2s}\,\biggl{[}-C_{2RR}^{026}(s,t,Q^{2})-C_{2RR}^{016}(s,t,Q^{2})$
$\displaystyle+$
$\displaystyle\frac{C_{0}(s,Q^{2})}{2}-2C_{1R}^{079}(s,t,Q^{2})+2C_{2RR}^{079}(s,t,Q^{2})\biggr{]}\,,$
$\displaystyle D_{3RRT}^{0126}(s,t,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{R_{\mu}R_{\nu}T_{\alpha}}{R^{4}T^{2}}\,D_{3}^{\mu\nu\alpha}(s,t,Q^{2})$
$\displaystyle=$
$\displaystyle\frac{1}{2s}\,\biggl{[}-C_{2RT}^{026}(s,t,Q^{2})-C_{2RT}^{016}(s,t,Q^{2})$
$\displaystyle+$ $\displaystyle
2C_{2RT}^{079}(s,t,Q^{2})-C_{1T}^{079}(t,Q^{2})\biggr{]}\,,$ $\displaystyle
D_{3TTR}^{0126}(s,t,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{R_{\mu}T_{\nu}T_{\alpha}}{T^{4}R^{2}}\,D_{3}^{\mu\nu\alpha}(s,t,Q^{2})$
$\displaystyle=$
$\displaystyle\frac{1}{2s}\,\biggl{[}-C_{2TT}^{026}(s,t,Q^{2})-C_{2TT}^{016}(s,t,Q^{2})+2C_{2TT}^{079}(s,t,Q^{2})\biggr{]}\,,$
$\displaystyle D_{3TTT}^{0126}(s,t,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{T_{\mu}T_{\nu}T_{\alpha}}{T^{6}}\,D_{3}^{\mu\nu\alpha}(s,t,Q^{2})$
$\displaystyle=$
$\displaystyle\frac{s}{2ut}\,\biggl{[}t\,D_{2TT}^{0126}(s,t,Q^{2})-C_{2TT}^{012}(s)-\frac{t}{s}\,C_{2TT}^{026}(s,t,Q^{2})$
$\displaystyle+$
$\displaystyle\frac{s+u}{s}\,C_{2TT}^{016}(s,t,Q^{2})+\frac{t-u}{s}\,C_{2TT}^{079}(s,t,Q^{2})\biggr{]}\,,$
$\displaystyle D_{3gP}^{0126}(s,t,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{g_{\mu\nu}^{\perp}P_{\alpha}}{(D-3)P^{2}}\,D_{3}^{\mu\nu\alpha}(s,t,Q^{2})$
$\displaystyle=$
$\displaystyle\frac{1}{D-3}\,\biggl{[}\frac{C_{0}(s,Q^{2})}{2}+C_{1P}^{079}(s,t,Q^{2})+s\,D_{3P}^{0126}(s,t,Q^{2})\biggr{]}\,,$

$\displaystyle D_{3gR}^{0126}(s,t,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{g_{\mu\nu}^{\perp}R_{\alpha}}{(D-3)R^{2}}\,D_{3}^{\mu\nu\alpha}(s,t,Q^{2})$
$\displaystyle=$
$\displaystyle\frac{1}{D-3}\,\biggl{[}-\frac{C_{0}(s,Q^{2})}{2}+C_{1R}^{079}(s,t,Q^{2})+s\,D_{3R}^{0126}(s,t,Q^{2})\biggr{]}\,,$
$\displaystyle D_{3gT}^{0126}(s,t,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{g_{\mu\nu}^{\perp}T_{\alpha}}{(D-3)T^{2}}\,D_{3}^{\mu\nu\alpha}(s,t,Q^{2})$
(439) $\displaystyle=$
$\displaystyle\frac{1}{D-3}\,\biggl{[}C_{1T}^{079}(t,Q^{2})+s\,D_{3T}^{0126}(s,t,Q^{2})\biggr{]}\,.$

Relations:

$\displaystyle\left\\{\begin{array}[]{l}D_{i}^{0125}(s,u,Q^{2})=+D_{i}^{0126}(s,u,Q^{2})\,,\\\
i=3gR,3PPR,3TTR,3PRT,3RRR\\\ \\\
D_{i}^{0125}(s,u,Q^{2})=-D_{i}^{0126}(s,u,Q^{2})\,,\\\
i=3gP,3gT,3PPP,3TTT,3PPT,3TTP,3RRP,3RRT\\\ \end{array}\right.$ (444)

Integral $D_{3}^{\mu\nu\alpha}(t,u,Q^{2})$

$\displaystyle D_{3PPP}^{0235}(t,u,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{P_{\mu}P_{\nu}P_{\alpha}}{P^{6}}\,D_{3}^{\mu\nu\alpha}(t,u,Q^{2})$
$\displaystyle=$
$\displaystyle\frac{1}{2s}\,\biggl{[}C_{2PP}^{023}(s,t,Q^{2})-C_{2PP}^{025}(s,u,Q^{2})-C_{2PP}^{035}(s,u,Q^{2})$
$\displaystyle+$
$\displaystyle\frac{C_{0}(t,Q^{2})}{4}-C_{1P}^{046}(s,t,Q^{2})+C_{2PP}^{046}(s,t,Q^{2})-uD_{2PP}^{0235}(u,t,Q^{2})\biggr{]}\,,$
$\displaystyle D_{3PPR}^{0235}(t,u,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{P_{\mu}P_{\nu}R_{\alpha}}{P^{4}R^{2}}\,D_{3}^{\mu\nu\alpha}(t,u,Q^{2})$
$\displaystyle=$
$\displaystyle\frac{1}{2s}\,\biggl{[}C_{2PR}^{023}(s,t,Q^{2})-C_{2PR}^{025}(s,u,Q^{2})-C_{2PR}^{035}(s,t,Q^{2})$
$\displaystyle-$
$\displaystyle\frac{C_{0}(t,Q^{2})}{4}+\frac{C_{1P}^{046}(s,t,Q^{2})}{2}-\frac{C_{1R}^{046}(s,t,Q^{2})}{2}+C_{2PR}^{046}(s,t,Q^{2})$
$\displaystyle-$ $\displaystyle uD_{2PR}^{0235}(t,u,Q^{2})\biggr{]}\,,$
$\displaystyle D_{3PPT}^{0235}(t,u,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{P_{\mu}P_{\nu}T_{\alpha}}{P^{4}T^{2}}\,D_{3}^{\mu\nu\alpha}(t,u,Q^{2})$
$\displaystyle=$
$\displaystyle\frac{1}{2s}\,\biggl{[}C_{2PT}^{023}(s,t,Q^{2})-C_{2PT}^{025}(s,u,Q^{2})-C_{2PT}^{035}(s,u,Q^{2})$
$\displaystyle-$
$\displaystyle\frac{C_{1T}^{046}(s,t,Q^{2})}{2}+C_{2PT}^{046}(s,t,Q^{2})-uD_{2PT}^{0235}(t,u,Q^{2})\biggr{]}\,,$
$\displaystyle D_{3RRP}^{0235}(t,u,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{P_{\mu}R_{\nu}R_{\alpha}}{R^{4}P^{2}}\,D_{3}^{\mu\nu\alpha}(t,u,Q^{2})$
$\displaystyle=$
$\displaystyle\frac{1}{2s}\,\biggl{[}C_{2RR}^{023}(s,t,Q^{2})-C_{2RR}^{025}(s,u,Q^{2})-C_{2RR}^{035}(s,u,Q^{2})$
$\displaystyle+$
$\displaystyle\frac{C_{0}(t,Q^{2})}{4}+C_{1R}^{046}(s,t,Q^{2})+C_{2RR}^{046}(s,t,Q^{2})-uD_{2RR}^{0235}(t,u,Q^{2})\biggr{]}\,,$
$\displaystyle D_{3TTP}^{0235}(t,u,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{P_{\mu}T_{\nu}T_{\alpha}}{T^{4}P^{2}}\,D_{3}^{\mu\nu\alpha}(t,u,Q^{2})$
$\displaystyle=$
$\displaystyle\frac{1}{2s}\,\biggl{[}C_{2TT}^{023}(s,t,Q^{2})-C_{2TT}^{025}(s,u,Q^{2})-C_{2TT}^{035}(s,u,Q^{2})$
$\displaystyle+$ $\displaystyle
C_{2TT}^{046}(s,t,Q^{2})-uD_{2TT}^{0235}(t,u,Q^{2})\biggr{]}\,,$
$\displaystyle D_{3PRT}^{0235}(t,u,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{P_{\mu}R_{\nu}T_{\alpha}}{P^{2}R^{2}T^{2}}\,D_{3}^{\mu\nu\alpha}(t,u,Q^{2})$
$\displaystyle=$
$\displaystyle\frac{1}{2s}\,\biggl{[}C_{2RT}^{023}(s,t,Q^{2})-C_{2RT}^{025}(s,u,Q^{2})-C_{2RT}^{035}(s,u,Q^{2})$
$\displaystyle+$
$\displaystyle\frac{C_{1T}^{046}(t,Q^{2})}{2}+C_{2RT}^{046}(s,t,Q^{2})-uD_{2RT}^{0235}(t,u,Q^{2})\biggr{]}\,,$
$\displaystyle D_{3RRR}^{0235}(t,u,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{R_{\mu}R_{\nu}R_{\alpha}}{R^{6}}\,D_{3}^{\mu\nu\alpha}(t,u,Q^{2})$
$\displaystyle=$
$\displaystyle\frac{1}{2s}\,\biggl{[}-C_{2RR}^{023}(s,t,Q^{2})+C_{2RR}^{025}(s,u,Q^{2})-C_{2RR}^{035}(s,u,Q^{2})$
$\displaystyle+$
$\displaystyle\frac{C_{0}^{046}(t,Q^{2})}{4}+C_{1R}^{046}(s,t,Q^{2})+C_{2RR}^{046}(s,t,Q^{2})+uD_{2RR}^{0235}(t,u,Q^{2})\biggr{]}\,,$
$\displaystyle D_{3RRT}^{0235}(t,u,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{R_{\mu}R_{\nu}T_{\alpha}}{R^{4}T^{2}}\,D_{3}^{\mu\nu\alpha}(t,u,Q^{2})$
$\displaystyle=$
$\displaystyle\frac{1}{2s}\,\biggl{[}-C_{2RT}^{023}(s,t,Q^{2})+C_{2RT}^{025}(s,u,Q^{2})-C_{2RT}^{035}(s,u,Q^{2})$
$\displaystyle+$
$\displaystyle\frac{C_{1T}^{046}(t,Q^{2})}{2}++C_{2RT}^{046}(s,t,Q^{2})+uD_{2RT}^{0235}(u,t,Q^{2})\biggr{]}\,,$
$\displaystyle D_{3TTR}^{0235}(t,u,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{R_{\mu}T_{\nu}T_{\alpha}}{T^{4}R^{2}}\,D_{3}^{\mu\nu\alpha}(t,u,Q^{2})$
$\displaystyle=$
$\displaystyle\frac{1}{2s}\,\,\biggl{[}-C_{2TT}^{023}(s,t,Q^{2})+C_{2TT}^{025}(s,u,Q^{2})-C_{2TT}^{035}(s,u,Q^{2})$
$\displaystyle+$ $\displaystyle
C_{2TT}^{046}(s,t,Q^{2})+uD_{2TT}^{0235}(u,t,Q^{2})\biggr{]}\,,$
$\displaystyle D_{3TTT}^{0235}(t,u,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{T_{\mu}T_{\nu}T_{\alpha}}{T^{6}}\,D_{3}^{\mu\nu\alpha}(t,u,Q^{2})$
$\displaystyle=$
$\displaystyle\frac{1}{2ut}\,\biggl{[}-tC_{2TT}^{023}(s,t,Q^{2})+(s+t)\,C_{2TT}^{025}(s,u,Q^{2})+u\,C_{2TT}^{035}(s,u,Q^{2})$
$\displaystyle-$
$\displaystyle(s+u)\,C_{2TT}^{046}(s,t,Q^{2})+D_{2TT}^{0235}(u,t,Q^{2})\biggr{]}\,,$
$\displaystyle D_{3gP}^{0235}(t,u,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{g_{\mu\nu}^{\perp}P_{\alpha}}{(D-3)P^{2}}\,D_{3}^{\mu\nu\alpha}(t,u,Q^{2})$
$\displaystyle=$
$\displaystyle\frac{1}{D-3}\,\biggl{[}\frac{C_{0}(t,Q^{2})}{2}-C_{1P}^{046}(s,t,Q^{2})+s\,D_{3P}^{0235}(u,t,Q^{2})\biggr{]}\,,$
$\displaystyle D_{3gR}^{0235}(t,u,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{g_{\mu\nu}^{\perp}R_{\alpha}}{(D-3)R^{2}}\,D_{3}^{\mu\nu\alpha}(s,t,Q^{2})$
$\displaystyle=$
$\displaystyle\frac{1}{D-3}\,\biggl{[}-\frac{C_{0}(t,Q^{2})}{2}+C_{1R}^{046}(s,t,Q^{2})+s\,D_{3R}^{0235}(u,t,Q^{2})\biggr{]}\,,$
$\displaystyle D_{3gT}^{0235}(t,u,Q^{2})$ $\displaystyle=$
$\displaystyle\frac{g_{\mu\nu}^{\perp}T_{\alpha}}{(D-3)T^{2}}\,D_{3}^{\mu\nu\alpha}(s,t,Q^{2})$
(445) $\displaystyle=$
$\displaystyle\frac{1}{D-3}\,\biggl{[}-C_{1T}^{046}(t,Q^{2})+s\,D_{3T}^{0235}(u,t,Q^{2})\biggr{]}\,.$

##  Appendix C Spherical integration in dimensional regularization

The key tool to handle PSIs in dimensional regularization is a generalization
of spherical integration to arbitrary dimension. In our consideration we
follow handbook Sterman:1994ce . For any integer $n$-dimension we can
introduce spherical coordinates. For $n=2$ dimension we have

$\displaystyle\vec{x\,}^{(2)}=r_{2}\vec{u\,}^{(2)}(\theta_{1})\quad\text{with}\quad\vec{u\,}^{(2)}(\theta_{1})=(\sin\theta_{1},\,\,\cos\theta_{1})\,,$
(446)

with module of vector $0\leq r_{2}<\infty$ and angle $0\leq\theta_{1}\leq
2\pi$. Here and in the following, we denote the $n$-dimensional length of a
vector as $r_{n}$ and $\vec{u\,}^{(n)}(\theta_{1},\ldots,\theta_{n-1})$ is a
unit vector depending on $n-1$ angular coordinates $\theta_{i}$.

Suppose that we know how to express the unit vector
$\vec{u\,}^{(n-1)}(\theta_{1},\ldots,\theta_{n-2})$ in terms of $n-2$ angular
variables. Then we can write $\vec{x\,}^{(n)}$ in cylindrical coordinates

$\displaystyle\vec{x\,}^{(n)}=(r_{n-1}\vec{u\,}^{(n-1)}(\theta_{1},\ldots,\theta_{n-2}),x_{n})\,.$
(447)

Changing variables to spherical ones by the relations

$\displaystyle r_{n-1}=r_{n}\sin\theta_{n-1}\,,\qquad
x_{n}=r_{n}\cos\theta_{n-1}\,,$ (448)

where $0\leq r_{n}\leq\infty$ and $0\leq\theta_{n-1}<\pi$ gives

$\displaystyle\vec{x\,}^{(n)}$
$\displaystyle=r_{n}\,(\sin\theta_{n-1}\vec{u\,}^{(n-1)}(\theta_{1},\ldots,\theta_{n-2}),\cos\theta_{n-1})$
$\displaystyle=r_{n}\,\vec{u\,}^{(n)}(\theta_{1},\ldots,\theta_{n-1})\,.$
(449)

With this procedure we have constructed spherical coordinates for arbitrary
integer dimension $n\geq 2$.

Now we are facing the issue of how to evaluate integrals in $n$ dimensions
without specifying $n$. This will enable us to formally extend the resulting
formula to non-integer $n$. The central idea to do so is to first examine on
which set of angular variables the integrand depends. For our purposes, it
suffices to consider scalar integrands where the integration variable
$\vec{x\,}^{(n)}$ only appears in scalar products of the form
$\vec{x\,}^{(n)}\cdot\vec{y\,}^{(n)}$ with a fixed $n$-dimensional vector of
the specific form $\vec{y\,}^{(n)}=(\vec{0\,}_{n-2},y_{2},y_{1})$ or as an
absolute square $\vec{x\,}^{(n)}\cdot\vec{x\,}^{(n)}=r_{n}^{2}$. We can expand
$\vec{x\,}^{(n)}$ with our recursion in terms of angular variables, obtaining

$\displaystyle\vec{x\,}^{(n)}$
$\displaystyle=r_{n}\vec{u\,}^{(n)}(\theta_{1},\ldots,\theta_{n-1})$
$\displaystyle=r_{n}\,(\sin\theta_{n-1}\vec{u}^{(n-1)}(\theta_{1},\ldots,\theta_{n-2}),\cos\theta_{n-1})$
$\displaystyle=r_{n}\,(\sin\theta_{n-1}\sin\theta_{n-2}\vec{u}^{(n-2)}(\theta_{1},\ldots,\theta_{n-3}),\sin\theta_{n-1}\cos\theta_{n-2},\cos\theta_{n-1})\,.$
(450)

Therefore,

$\displaystyle\vec{x\,}^{(n)}\cdot\vec{y\,}^{(n)}=r_{n}\,(y_{2}\sin\theta_{n-1}\cos\theta_{n-2}+y_{1}\cos\theta_{n-1})\,.$
(451)

This implies that we can integrate over all angular variables except for
$\theta_{n-2}$ and $\theta_{n-1}$ as we have integrands of the form
$f(r_{n},\theta_{n-1},\theta_{n-2})$. The next step is to express the
$\int\mathrm{d}^{n}x$ integral in terms of our spherical coordinates with
explicit $\theta_{n-2}$ and $\theta_{n-1}$ integrations.

The integral in spherical coordinates in $n$-dimensions has the form

$\displaystyle\int\mathrm{d}^{n}x=\int_{0}^{\infty}\mathrm{d}r_{n}r_{n}^{n-1}\int\mathrm{d}\Omega_{n-1}\,.$
(452)

Constructing $\int\mathrm{d}\Omega_{n-1}$ iteratively using the spherical
coordinates defined above yields

$\displaystyle\int\mathrm{d}\Omega_{n-1}=\int_{0}^{2\pi}\mathrm{d}\theta_{1}\,\prod_{k=2}^{n-1}\int_{0}^{\pi}\mathrm{d}\theta_{k}\,\sin^{k-1}\theta_{k}\,.$
(453)

The second part is

$\displaystyle\int_{0}^{\pi}\mathrm{d}\theta\,\sin^{k}\theta=B\Big{(}\frac{1}{2},\frac{k+1}{2}\Big{)}\,,$
(454)

and integrating out all angles yields

$\displaystyle\Omega_{n}=\int\mathrm{d}\Omega_{n}=2\pi\prod_{k=1}^{n-1}\,B\Big{(}\frac{1}{2},\frac{k+1}{2}\Big{)}=\frac{2\pi^{\frac{n+1}{2}}}{\Gamma\left(\frac{n+1}{2}\right)}\,.$
(455)

Formula (455) allows for continuation to arbitrary complex $n$ (except for
poles). This observation takes us a step closer to formally evaluate integrals
in non-integer dimensions.

The idea for calculating the integrals of interest $\int\mathrm{d}\Omega_{n}$
with $n$ non-integer is to split it in two parts. One part is the integral
over $k$ angles, where $k$ is a fixed integer. In our case this part contains
the integrand $f(r_{n},\theta_{n-1},\theta_{n-2})$ with $k=2$. The remaining
part is an $n-k$ dimensional spherically symmetric integral, which can be also
evaluated for non-integer $n-k$ using the identity (455). The splitting we
need is

$\displaystyle\int\mathrm{d}\Omega_{n-1}$
$\displaystyle=\int_{0}^{\pi}\mathrm{d}\theta_{n-1}\sin^{n-2}\theta_{n-1}\int\mathrm{d}\Omega_{n-2}$
$\displaystyle=\int_{0}^{\pi}\mathrm{d}\theta_{n-1}\sin^{n-2}\theta_{n-1}\int_{0}^{\pi}\mathrm{d}\theta_{n-3}\sin^{n-2}\theta_{n-2}\int\mathrm{d}\Omega_{n-3}\,.$
(456)

Thus we obtain for our integral

$\displaystyle\int\mathrm{d}^{n}x\,f(r_{n},\theta_{n-1},\theta_{n-2})=\int_{0}^{\infty}\mathrm{d}r_{n}\,r_{n}^{n-1}\int\mathrm{d}\Omega_{n-1}\,f(r_{n},\theta_{n-1},\theta_{n-2})$
(457) $\displaystyle=$
$\displaystyle\Omega_{n-3}\int_{0}^{\infty}\mathrm{d}r_{n}\,r_{n}^{n-1}\int_{0}^{\pi}\mathrm{d}\theta_{n-1}\sin^{n-2}\theta_{n-1}\int_{0}^{\pi}\mathrm{d}\theta_{n-2}\sin^{n-3}\theta_{n-2}\,f(r_{n},\theta_{n-1},\theta_{n-2})\,.$

Commonly the angles are relabeled $\theta_{n-1}\rightarrow\theta_{1}$ and
$\theta_{n-2}\rightarrow\theta_{2}$ at this stage.

##  Appendix D Partial fraction decomposition of linear propagators

A common phrase to read about PSI is that it involves a tedious partial
fraction decomposition. Algorithms of how to perform them are known. However,
to the best of the authors knowledge, the partial fraction decomposition was
never viewed in a wider context. Especially the question under which
circumstances one can reduce the number of angular integrals to be calculated
often lacks an explanation. In this section we want to discuss a simple
geometric representation of the occurring propagators, which is
straightforwardly connected with Minkowski momentum space. We will see that
any partial fractioning has its root in just one basic formula (466). This
formula can then be extended to not only a general picture of partial fraction
decompositions, but also to a nice and useful formula for the splitting of
double-massive propagators to single-massive ones used in Sec. 3.2.6 to
calculate the double massive integral in $D$ dimensions.

To set the starting stage, we consider Euclidean vectors
$\mathbf{v_{i}}\in\mathbb{R}^{D-1}$ and define there scaled linear propagator

$\displaystyle\Delta_{\mathbf{k}}\left(\mathbf{v_{i}}\right)=\frac{1}{1-\mathbf{v_{i}}\cdot\mathbf{k}}\,,$
(458)

where $\mathbf{k}$ is the PSI integration momentum choosen for convenience as
an arbitrary unit vector. How $\Delta_{\mathbf{k}}\left(\mathbf{v_{i}}\right)$
is connected with physical propagators will be clarified in Sec. D.5. For
convenience in our consideration we introduce the shorten notation for the
product of $N$ propagators:

$\displaystyle\Delta_{\mathbf{k}}\left(\mathbf{v_{1}},\mathbf{v_{2}},\cdots,\mathbf{v_{N}}\right)\equiv\Delta_{\mathbf{k}}\left(\mathbf{v_{1}}\right)\Delta_{\mathbf{k}}\left(\mathbf{v_{2}}\right)\cdots\Delta_{\mathbf{k}}\left(\mathbf{v_{N}}\right)\,.$
(459)

###  D.1 Two-point splitting lemma

We start by proving the so-called two-point splitting lemma allowing to
factorize two variables in two-propagator product
$\Delta_{\mathbf{k}}\left(\mathbf{v_{1}},\mathbf{v_{2}}\right)$ [see
definition in Eq. 459] by introducing two-propagator products of the vectors
$v_{1}$ and $v_{2}$ with another (third) arbitrary Euclidean vector $v_{3}$.
In particular, we are interested in deriving a decomposition of the form

$\displaystyle\Delta_{\mathbf{k}}\left(\mathbf{v_{1}},\mathbf{v_{2}}\right)=\lambda\,\Delta_{\mathbf{k}}\left(\mathbf{v_{1}},\mathbf{v_{3}}\right)+\mu\,\Delta_{\mathbf{k}}\left(\mathbf{v_{2}},\mathbf{v_{3}}\right)\,,$
(460)

where $\lambda$ and $\mu$ are two real parameters. Such identity will be
useful for manipulations of products of propagators containing
$\mathbf{v_{1}}$ and $\mathbf{v_{2}}$ in PSIs. Starting from r.h.s. of Eq.
(460), we arrive at

$\displaystyle\lambda\,\Delta_{\mathbf{k}}\left(\mathbf{v_{1}},\mathbf{v_{3}}\right)+\mu\,\Delta_{\mathbf{k}}\left(\mathbf{v_{2}},\mathbf{v_{3}}\right)$
(461) $\displaystyle=$
$\displaystyle\Big{(}\lambda\,\Delta^{-1}_{\mathbf{k}}\left(\mathbf{v_{2}}\right)+\mu\,\Delta^{-1}_{\mathbf{k}}\left(\mathbf{v_{1}}\right)\Big{)}\Delta_{\mathbf{k}}\left(\mathbf{v_{1}},\mathbf{v_{2}},\mathbf{v_{3}}\right)$
$\displaystyle=$
$\displaystyle\Big{[}\Bigl{(}\lambda\,\Delta^{-1}_{\mathbf{k}}\left(\mathbf{v_{2}}\right)+\mu\,\Delta^{-1}_{\mathbf{k}}\left(\mathbf{v_{1}}\right)\Big{)}\,\Delta_{\mathbf{k}}\left(\mathbf{v_{3}}\right)\Big{]}\,\Delta_{\mathbf{k}}\left(\mathbf{v_{1}},\mathbf{v_{2}}\right)\,,$

Comparing the latter with l.h.s.of Eq. (460) we conclude that the condition

$\displaystyle\Big{(}\lambda\,\Delta^{-1}_{\mathbf{k}}\left(\mathbf{v_{2}}\right)+\mu\,\Delta^{-1}_{\mathbf{k}}\left(\mathbf{v_{1}}\right)\Big{)}\,\Delta_{\mathbf{k}}\left(\mathbf{v_{3}}\right)\equiv
1\,,$ (462)

must hold. It gives a system of two equations on two arbitrary real parameters
$\lambda$ and $\mu$ and vector $\mathbf{v_{3}}$:

$\displaystyle\left\\{\begin{array}[]{c}\lambda+\mu=1\\\
\mu\mathbf{v_{1}}+\lambda\mathbf{v_{2}}=\mathbf{v_{3}}\,,\end{array}\right.$
(465)

From Eq. (465) it follows that the parameter $\lambda$ can be chosen
arbitrarily, while the second parameter $\mu$ and vector $\mathbf{v_{3}}$ are
then fixed as $\mu=1-\lambda$ and
$\mathbf{v_{3}}=(1-\lambda)\mathbf{v_{1}}+\lambda\mathbf{v_{2}}$.

Therefore, we are now in the position to formulate the Two-Point Splitting
Lemma:

$\displaystyle\Delta_{\mathbf{k}}\left(\mathbf{v_{1}},\mathbf{v_{2}}\right)=\lambda\,\Delta_{\mathbf{k}}\left(\mathbf{v_{1}},\mathbf{v_{3}}\right)+(1-\lambda)\,\Delta_{\mathbf{k}}\left(\mathbf{v_{2}},\mathbf{v_{3}}\right)\,,$
(466)

where $\mathbf{v_{3}}=(1-\lambda)\,\mathbf{v_{1}}+\lambda\,\mathbf{v_{2}}$.

Geometrical interpretation of the two-point splitting lemma is that the heads
of the vector $\mathbf{v_{3}}$ defined as
$\mathbf{v_{3}}=(1-\lambda)\,\mathbf{v_{1}}+\lambda\,\mathbf{v_{2}}$ with
$\lambda$ being arbitrary real parameter represents the line passing through
the heads of $\mathbf{v_{1}}$ and $\mathbf{v_{2}}$. In other words, it is a
definition of a line by two linear independent vectors. All three vectors
$\mathbf{v_{1}}$, $\mathbf{v_{2}}$, and $\mathbf{v_{3}}$ obviously lie in the
two-dimensional plane spanned by $\mathbf{v_{1}}$ and $\mathbf{v_{2}}$. This
is depicted in Fig. 2.

Figure 2: A graphical representation of the two-point splitting lemma. The
two-dimensional $(xy)$ plane is spanned by two arbitrary linearly independent
vectors $\mathbf{v_{1}}$ and $\mathbf{v_{2}}$. The head of the third vector
$\mathbf{v_{3}}$ (we show one typical example) lies in the straight line
connecting the heads of $\mathbf{v_{1}}$ and $\mathbf{v_{2}}$. The circle
represents all vectors with $\mathbf{v}\cdot\mathbf{v}=1$.

For $j,l\in\mathbb{N}$ the two-point splitting lemma can be generalized to

$\displaystyle\Delta^{j}_{\mathbf{k}}\left(\mathbf{v_{1}}\right)\Delta^{l}_{\mathbf{k}}\left(\mathbf{v_{2}}\right)$
$\displaystyle=\sum_{n=0}^{j-1}\binom{l-1+n}{l-1}\lambda^{l}(1-\lambda)^{n}\Delta^{j-n}_{\mathbf{k}}\left(\mathbf{v_{1}}\right)\Delta^{l+n}_{\mathbf{k}}\left(\mathbf{v_{3}}\right)$
$\displaystyle+\sum_{n=0}^{l-1}\binom{j-1+n}{j-1}\lambda^{n}(1-\lambda)^{j}\Delta^{l-n}_{\mathbf{k}}\left(\mathbf{v_{2}}\right)\Delta^{j+n}_{\mathbf{k}}\left(\mathbf{v_{3}}\right)\,.$
(467)

To prove Eq. (467) we employ the binomial summation formula

$\displaystyle\sum_{n=0}^{N}\binom{j-1+n}{j-1}=\binom{j+N+1}{j}\,,$ (468)

and use mathematical induction.

###  D.2 Two-mass splitting and two-point partial fractioning

The two-mass splitting now becomes a simple corollary of the two-point
splitting lemma. We just have to choose $\mathbf{v_{3}}$ such that
$\mathbf{v_{3}}\cdot\mathbf{v_{3}}=1$. To achieve this we have to take
$\mathbf{v_{3}}$ as the intersection of the line
$(1-\lambda)\,\mathbf{v_{1}}+\lambda\,\mathbf{v_{2}}$ with the unit circle.
The condition

$\displaystyle\left|(1-\lambda)\,\mathbf{v_{1}}+\lambda\,\mathbf{v_{2}}\right|^{2}\overset{!}{=}$
$\displaystyle 1\,,$ (469)

leads to the quadratic equation (see Fig. 3)

$\displaystyle\lambda^{2}\,(\mathbf{v_{2}}-\mathbf{v_{1}})^{2}+2\,\lambda\,\mathbf{v_{1}}\cdot(\mathbf{v_{2}}-\mathbf{v_{1}})+\mathbf{v_{1}}\cdot\mathbf{v_{1}}-1\overset{!}{=}0\,.$
(470)

Defining the scaled Mandelstam variables

$\displaystyle v_{11}=1-\mathbf{v_{1}}\cdot\mathbf{v_{1}}\,,\quad
v_{12}=1-\mathbf{v_{1}}\cdot\mathbf{v_{2}}\,,\quad
v_{22}=1-\mathbf{v_{2}}\cdot\mathbf{v_{2}}\,,$ (471)

we write down Eq. (470) as

$\displaystyle\lambda^{2}\,(2v_{12}-v_{11}-v_{22})+2\lambda\,(v_{11}-v_{12})-v_{11}=0\,.$
(472)

The solution of the latter equation is

$\displaystyle\lambda_{\pm}=\frac{v_{12}-v_{11}\pm\sqrt{v_{12}^{2}-v_{11}v_{22}}}{2v_{12}-v_{11}-v_{22}}\,.$
(473)

Figure 3: Graphical representation of double-to-single-massive splitting.
$\mathbf{v_{3}^{\pm}}$ is chosen as one of the intersections of the unit
circle with the line through $\mathbf{v_{1}}$ and $\mathbf{v_{2}}$.

The massless vectors associated with $\mathbf{v_{1}}$ and $\mathbf{v_{2}}$
thus are

$\displaystyle\mathbf{v_{3}^{\pm}}=\frac{v_{12}-v_{22}\pm\sqrt{v_{12}^{2}-v_{11}v_{22}}}{2v_{12}-v_{11}-v_{22}}\,\mathbf{v_{1}}+\frac{v_{12}-v_{11}\mp\sqrt{v_{12}^{2}-v_{11}v_{22}}}{2v_{12}-v_{11}-v_{22}}\,\mathbf{v_{2}}\,.$
(474)

Using the $\lambda$ from Eq. (473) and the associated vector from (474) we
have the Double-to-Single-Massive Splitting (or simply Two-mass splitting)

$\displaystyle\Delta_{\mathbf{k}}\left(\mathbf{v_{1}},\mathbf{v_{2}}\right)=\lambda_{\pm}\,\Delta_{\mathbf{k}}\left(\mathbf{v_{1}},\mathbf{v_{3}^{\pm}}\right)+(1-\lambda_{\pm})\Delta_{\mathbf{k}}\left(\mathbf{v_{2}},\mathbf{v_{3}^{\pm}}\right)\,.$
(475)

It splits the double-massive propagator on the left into two single-massive
propagators on the right. This proves very useful for the calculation of the
double-massive integral with integer coefficients.

The two mass-splitting becomes even more powerful in combination with the
generalized form of the two point splitting lemma. For $j,l\in\mathbb{N}$ it
holds

$\displaystyle\Delta^{j}_{\mathbf{k}}\left(\mathbf{v_{1}}\right)\Delta^{l}_{\mathbf{k}}\left(\mathbf{v_{2}}\right)$
$\displaystyle=\sum_{n=0}^{j-1}\binom{l-1+n}{l-1}\lambda_{\pm}^{l}(1-\lambda_{\pm})^{n}\Delta^{j-n}_{\mathbf{k}}\left(\mathbf{v_{1}}\right)\Delta^{l+n}_{\mathbf{k}}\left(\mathbf{v^{\pm}_{3}}\right)$
$\displaystyle+\sum_{n=0}^{l-1}\binom{j-1+n}{j-1}\lambda_{\pm}^{n}(1-\lambda_{\pm})^{j}\Delta^{l-n}_{\mathbf{k}}\left(\mathbf{v_{2}}\right)\Delta^{j+n}_{\mathbf{k}}\left(\mathbf{v^{\pm}_{3}}\right)\,.$
(476)

Another direct consequence of the two-point splitting lemma is the partial
fractioning of the propagator of two different linear dependent vectors. If
$\lambda_{1}\mathbf{v_{1}}+\lambda_{2}\mathbf{v_{2}}=\mathbf{0}$ with
$\lambda_{1}+\lambda_{2}\neq 0$, then the line connecting $\mathbf{v_{1}}$ and
$\mathbf{v_{2}}$ passes through the origin. Thus $\mathbf{0}$ is associated
with them. As $\Delta_{\mathbf{k}}\left(\mathbf{\mathbf{0}}\right)=1$ the two-
point splitting lemma can be used to perform a reduction of the number of
propagators. Thus we have the two-point partial fractioning (see Fig. 4)

$\displaystyle\Delta_{\mathbf{k}}\left(\mathbf{v_{1}},\mathbf{v_{2}}\right)=\frac{1}{\lambda_{1}+\lambda_{2}}\left[\lambda_{2}\,\Delta_{\mathbf{k}}\left(\mathbf{v_{1}}\right)+\lambda_{1}\,\Delta_{\mathbf{k}}\left(\mathbf{v_{2}}\right)\right]\,,$
(477)

with two constraints
$\lambda_{1}\mathbf{v_{1}}+\lambda_{2}\mathbf{v_{2}}=\mathbf{0}$ and
$\lambda_{1}+\lambda_{2}\neq 0$. The most important special case of the two
point partial fractioning is $\mathbf{v_{2}}=-\mathbf{v_{1}}$. Then one can
write Eq. (477) as

$\displaystyle\Delta_{\mathbf{k}}\left(\mathbf{v},\mathbf{-v}\right)=\frac{1}{2}\left[\Delta_{\mathbf{k}}\left(\mathbf{v}\right)+\Delta_{\mathbf{k}}\left(\mathbf{-v}\right)\right]\,.$
(478)

Figure 4: Graphical illustration of two-point partial fractioning. The origin
lies on the line $\overline{\mathbf{v_{1}}\mathbf{v_{2}}}$, we can use
$\Delta_{\mathbf{k}}\left(\mathbf{0}\right)$ to split
$\Delta_{\mathbf{k}}\left(\mathbf{v_{1}}\right)$ and
$\Delta_{\mathbf{k}}\left(\mathbf{v_{2}}\right)$.

###  D.3 Three-point partial fractioning and general denominator reduction

Eq. (477) gives a way to reduce the propagator of two linear dependent vectors
to just one. Together with the two-point splitting lemma (8) we will
geometrically construct the formula for the reduction of three propagators
with linear dependent vectors to the sum of two propagators. The derived
formula will be at the heart of the partial fraction decomposition for the
NNLO real correction of the Drell-Yan angular distribution (which we plan to
investigate in the near future in the context of small $Q_{T}$ resummation,
compare Ref. Boer:2006eq  and Berger:2007jw ).

Let us start with three linear dependent and pairwise linear independent
vectors $\mathbf{v_{1}}$, $\mathbf{v_{2}}$ and $\mathbf{v_{3}}$ satisfying

$\displaystyle\lambda_{1}\mathbf{v_{1}}+\lambda_{2}\mathbf{v_{2}}+\lambda_{3}\mathbf{v_{3}}=\mathbf{0}\,,$
(479)

with $\lambda_{1}+\lambda_{2}+\lambda_{3}\neq 0$. We want to derive an
expression of the form

$\displaystyle\Delta_{\mathbf{k}}\left(\mathbf{v_{1}},\mathbf{v_{2}},\mathbf{v_{3}}\right)=a_{1}\Delta_{\mathbf{k}}\left(\mathbf{v_{1}},\mathbf{v_{2}}\right)+a_{2}\Delta_{\mathbf{k}}\left(\mathbf{v_{1}},\mathbf{v_{3}}\right)+a_{3}\Delta_{\mathbf{k}}\left(\mathbf{v_{2}},\mathbf{v_{3}}\right)\,,$
(480)

with parameters $a_{1,2,3}\equiv
a_{1,2,3}(\lambda_{1},\lambda_{2},\lambda_{3})$ to be determined. In our
graphical representation $\mathbf{v_{1}}$, $\mathbf{v_{2}}$ and
$\mathbf{v_{3}}$ form a triangle. We proceed in three steps depicted in Fig.
5.

Figure 5: Graphical illustration of three-point partial fractioning.

First, we find the intersection of $\overline{\mathbf{v_{1}}\mathbf{v_{2}}}$
and $\overline{\mathbf{0}\mathbf{v_{3}}}$ and name it
$\mathbf{v_{\text{int}}}$555In the case, where these lines are parallel such
that no intersection exists, change the roles of $\mathbf{v_{1}}$,
$\mathbf{v_{2}}$, and $\mathbf{v_{3}}$ appropriately..
$\mathbf{v_{\text{int}}}$ is associated with $\mathbf{v_{1}}$ by the two-point
splitting lemma and therefore there exists a $\lambda$ such that

$\displaystyle\Delta_{\mathbf{k}}\left(\mathbf{v_{1}},\mathbf{v_{2}}\right)=(1-\lambda)\Delta_{\mathbf{k}}\left(\mathbf{v_{1}},\mathbf{v_{\text{int}}}\right)+\lambda\Delta_{\mathbf{k}}\left(\mathbf{v_{2}},\mathbf{v_{\text{int}}}\right)\,.$
(481)

Therefore, upon multiplication of
$\Delta_{\mathbf{k}}\left(\mathbf{v_{1}},\mathbf{v_{2}}\right)$ with
$\Delta_{\mathbf{k}}\left(\mathbf{v_{3}}\right)$ it follows

$\displaystyle\Delta_{\mathbf{k}}\left(\mathbf{v_{1}},\mathbf{v_{2}},\mathbf{v_{3}}\right)=(1-\lambda)\Delta_{\mathbf{k}}\left(\mathbf{v_{1}},\mathbf{v_{\text{int}}},\mathbf{v_{3}}\right)+\lambda\Delta_{\mathbf{k}}\left(\mathbf{v_{2}},\mathbf{v_{\text{int}}},\mathbf{v_{3}}\right)\,.$
(482)

Secondly, we observe that $\mathbf{v_{\text{int}}}$ is linear dependent on
$\mathbf{v_{3}}$, thus $\mathbf{v_{\text{int}}}=\mu\mathbf{v_{3}}$ for some
$\mu$. Hence, according to Eq. (477) we can use the two-point partial
fractioning

$\displaystyle\Delta_{\mathbf{k}}\left(\mathbf{v_{\text{int}}},\mathbf{v_{3}}\right)=\frac{1}{1-\mu}\Delta_{\mathbf{k}}\left(\mathbf{v_{3}}\right)-\frac{\mu}{1-\mu}\Delta_{\mathbf{k}}\left(\mathbf{v_{\text{int}}}\right).$
(483)

Thus,

$\displaystyle\Delta_{\mathbf{k}}\left(\mathbf{v_{1}},\mathbf{v_{2}},\mathbf{v_{3}}\right)$
$\displaystyle=\frac{1-\lambda}{1-\mu}\Delta_{\mathbf{k}}\left(\mathbf{v_{1}},\mathbf{v_{3}}\right)-\frac{(1-\lambda)\,\mu}{1-\mu}\Delta_{\mathbf{k}}\left(\mathbf{v_{1}},\mathbf{v_{\text{int}}}\right)$
$\displaystyle+\frac{\lambda}{1-\mu}\Delta_{\mathbf{k}}\left(\mathbf{v_{2}},\mathbf{v_{3}}\right)-\frac{\lambda\,\mu}{1-\mu}\Delta_{\mathbf{k}}\left(\mathbf{v_{2}},\mathbf{v_{\text{int}}}\right)\,,$
(484)

with each of the summands corresponding to one of the four triangles with the
corners $(\mathbf{0},\mathbf{v_{1}},\mathbf{v_{3}})$,
$(\mathbf{0},\mathbf{v_{1}},\mathbf{v_{\text{int}}})$,
$(\mathbf{0},\mathbf{v_{2}},\mathbf{v_{3}})$,
$(\mathbf{0},\mathbf{v_{2}},\mathbf{v_{\text{int}}})$.

In a third step we get rid of $\mathbf{v_{\text{int}}}$ by combining the
triangles $(\mathbf{0},\mathbf{v_{1}},\mathbf{v_{\text{int}}})$ and
$(\mathbf{0},\mathbf{v_{2}},\mathbf{v_{\text{int}}})$ with two-point splitting
in reverse via Eq (466) such that

$\displaystyle\Delta_{\mathbf{k}}\left(\mathbf{v_{1}},\mathbf{v_{2}},\mathbf{v_{3}}\right)=-\frac{\mu}{1-\mu}\Delta_{\mathbf{k}}\left(\mathbf{v_{1}},\mathbf{v_{2}}\right)+\frac{1-\lambda}{1-\mu}\Delta_{\mathbf{k}}\left(\mathbf{v_{1}},\mathbf{v_{3}}\right)+\frac{\lambda}{1-\mu}\Delta_{\mathbf{k}}\left(\mathbf{v_{2}},\mathbf{v_{3}}\right)\,.$
(485)

Solving the intersection equation

$\displaystyle\mu\mathbf{v_{3}}\overset{!}{=}\lambda\mathbf{v_{1}}+(1-\lambda)\mathbf{v_{2}}\,,$
(486)

with boundary condition (479) yields for $\lambda$ and $\mu$

$\displaystyle\lambda=\frac{\lambda_{1}}{\lambda_{1}+\lambda_{2}}\quad\text{and}\quad\mu=-\frac{\lambda_{3}}{\lambda_{1}+\lambda_{2}}\,.$
(487)

Plugging solution (487) into Eq. (485) we have the three-point partial
fractioning

$\displaystyle\Delta_{\mathbf{k}}\left(\mathbf{v_{1}},\mathbf{v_{2}},\mathbf{v_{3}}\right)=\frac{1}{\lambda_{1}+\lambda_{2}+\lambda_{3}}\left[\lambda_{3}\Delta_{\mathbf{k}}\left(\mathbf{v_{1}},\mathbf{v_{2}}\right)+\lambda_{2}\Delta_{\mathbf{k}}\left(\mathbf{v_{1}},\mathbf{v_{3}}\right)+\lambda_{1}\Delta_{\mathbf{k}}\left(\mathbf{v_{2}},\mathbf{v_{3}}\right)\right]\,,$
(488)

with boundary conditions
$\lambda_{1}\mathbf{v_{1}}+\lambda_{2}\mathbf{v_{2}}+\lambda_{3}\mathbf{v_{3}}=\mathbf{0}$
and $\lambda_{1}+\lambda_{2}+\lambda_{3}\neq 0$.

After the illustration of the application of two-point splitting lemma to
partial-fractioning of products of two and three propagators, we are in the
position to extend it to the general case of the product of $n$ propagators
with $n$ being arbitrary. Therefore, we formulate the generalized reduction of
products of linear propagators:

If $\sum_{i=1}^{n}\lambda_{i}\mathbf{v_{i}}=\mathbf{0}$ with
$\lambda\equiv\sum_{i=1}^{n}\lambda_{i}\neq 0$, $n\in\mathbb{N}$,
$\lambda_{i}\in\mathbb{C}$, then it holds

$\displaystyle\prod_{i=1}^{n}\Delta_{\mathbf{k}}\left(\mathbf{v_{i}}\right)=\frac{1}{\lambda}\,\sum_{j=1}^{n}\lambda_{j}\,\prod_{i=1}^{n\neq
j}\,\Delta_{\mathbf{k}}\left(\mathbf{v_{i}}\right)\,.$ (489)

The proof of the identity (489) is based on the linearity of the propagator
and is straightforward. In particular, using ${\bf
0}=\sum_{j=1}^{n}\lambda_{j}\mathbf{v_{j}}$ one gets

$\displaystyle\lambda=\sum\limits_{j=1}^{n}\
\lambda_{j}=\sum\limits_{j=1}^{n}\ \lambda_{j}\,\Big{(}1-{\bf v_{j}}{\bf
k}\Big{)}=\sum\limits_{j=1}^{n}\
\lambda_{j}\,\Delta^{-1}_{\mathbf{k}}\left(\mathbf{v_{j}}\right)\,.$ (490)

Therefore,

$\displaystyle\Delta_{\mathbf{k}}\left(\mathbf{v_{1}},\mathbf{v_{2}},\cdots,\mathbf{v_{n}}\right)$
$\displaystyle=$
$\displaystyle\prod_{i=1}^{n}\Delta_{\mathbf{k}}\left(\mathbf{v_{i}}\right)=\frac{1}{\lambda}\,\sum_{j=1}^{n}\lambda_{j}\,\prod_{i=1}^{n}\Delta_{\mathbf{k}}\left(\mathbf{v_{i}}\right)$
(491) $\displaystyle=$
$\displaystyle\frac{1}{\lambda}\,\sum_{j=1}^{n}\lambda_{j}\,\prod_{i=1}^{n}\,\Delta^{-1}_{\mathbf{k}}\left(\mathbf{v_{j}}\right)\,\Delta_{\mathbf{k}}\left(\mathbf{v_{i}}\right)=\frac{1}{\lambda}\,\sum_{j=1}^{n}\lambda_{j}\,\prod_{i=1}^{n\neq
j}\,\Delta_{\mathbf{k}}\left(\mathbf{v_{i}}\right)$ $\displaystyle=$
$\displaystyle\frac{1}{\lambda}\,\biggl{[}\lambda_{1}\,\Delta_{\mathbf{k}}\left(\mathbf{v_{2}},\mathbf{v_{3}},\cdots,\mathbf{v_{n}}\right)\,+\,\lambda_{2}\,\Delta_{\mathbf{k}}\left(\mathbf{v_{1}},\mathbf{v_{3}},\cdots,\mathbf{v_{n}}\right)$
$\displaystyle+$
$\displaystyle\ldots\,+\,\lambda_{n}\,\Delta_{\mathbf{k}}\left(\mathbf{v_{1}},\mathbf{v_{2}},\cdots,\mathbf{v_{n-1}}\right)\biggr{]}\,.$

Geometrically Eq. (489) follows from the fact that $\mathbf{0}$ is in the
$n-1$-dimensional hyperplane spanned by the vectors $\mathbf{v_{i}}$ and, thus
it can be associated with them repeatedly using the two-point splitting lemma.
Of course, the general reduction of linear propagators contains two- and
three-point partial fractioning as mere special cases. However, we think that
the presented inductive approach is more pedagogically suited to build up an
geometric intuition.

###  D.4 Propagators to negative integer powers

In practical calculations one is also confronted with negative integer powers
$j\in\mathbb{Z}_{\leq 0}$ of massive propagators. Those can always be made
massless by simple scaling. It holds

$\displaystyle\Delta^{-1}_{\mathbf{k}}\left(\mathbf{v_{1}}\right)$
$\displaystyle=1-\mathbf{v_{1}}\cdot\mathbf{k}=1-|\mathbf{v_{1}}|+|\mathbf{v_{1}}|\,\Delta^{-1}_{\mathbf{k}}\left({\frac{\mathbf{v_{1}}}{|\mathbf{v_{1}}|}}\right)$
$\displaystyle=1-\sqrt{1-v_{11}}+\sqrt{1-v_{11}}\,\Delta^{-1}_{\mathbf{k}}\left(\mathbf{\hat{v}_{1}}\right)\,,$
(492)

with
$\mathbf{\hat{v}_{1}}=\frac{\mathbf{v_{1}}}{|\mathbf{v_{1}}|}=\frac{\mathbf{v_{1}}}{\sqrt{1-v_{11}}}$.
Contracting with another vector $\mathbf{v_{2}}$ gives the scalar product

$\displaystyle\hat{v}_{12}=1-\mathbf{\hat{v}_{1}}\cdot\mathbf{v_{2}}=1-\frac{\mathbf{v_{1}}\cdot\mathbf{v_{2}}}{\sqrt{1-v_{11}}}=1-\frac{1-v_{12}}{\sqrt{1-v_{11}}}\,.$
(493)

Eq. (492) is straightforwardly generalized to $j\in\mathbb{Z}_{\leq 0}$ by
application of the Binomial theorem

$\displaystyle\Delta^{j}_{\mathbf{k}}\left(\mathbf{v_{1}}\right)$
$\displaystyle=\left(1-\sqrt{1-v_{11}}+\sqrt{1-v_{11}}\,\Delta^{-1}_{\mathbf{k}}\left(\mathbf{\hat{v}_{1}}\right)\right)^{-j}$
$\displaystyle=\sum_{n=0}^{-j}\binom{-j}{n}\left(1-\sqrt{1-v_{11}}\,\right)^{-j-n}\,\left(1-v_{11}\right)^{n/2}\,\Delta^{-n}_{\mathbf{k}}\left(\mathbf{\hat{v}_{1}}\right)\,.$
(494)

This equation allows to generalize the two-mass splitting to all
$j,l\in\mathbb{Z}$. Another practical application is that one can reduce two-
mass angular integral to one-mass one in case of $j\in\mathbb{Z}_{\leq 0}$. In
particular, for that case it holds

$\displaystyle I_{j,l}^{(2)}(v_{12},v_{11},v_{22};\varepsilon)$
$\displaystyle=\int\mathrm{d}\Omega_{k_{1}k_{2}}\Delta^{j}_{\mathbf{k}}\left(\mathbf{v_{1}}\right)\Delta^{l}_{\mathbf{k}}\left(\mathbf{v_{2}}\right)$
$\displaystyle=\sum_{n=0}^{-j}\binom{-j}{n}\left(1-\sqrt{1-v_{11}}\,\right)^{-j-n}\left(1-v_{11}\right)^{n/2}\,\int\mathrm{d}\Omega_{k_{1}k_{2}}\Delta^{-n}_{\mathbf{k}}\left(\mathbf{\hat{v}_{1}}\right)\Delta^{l}_{\mathbf{k}}\left(\mathbf{v_{2}}\right)$
$\displaystyle=\sum_{n=0}^{-j}\binom{-j}{n}\left(1-\sqrt{1-v_{11}}\,\right)^{-j-n}\left(1-v_{11}\right)^{n/2}\,I_{l,-n}^{(1)}\left(1-\frac{1-v_{12}}{\sqrt{1-v_{11}}},v_{22};\varepsilon\right)\,.$
(495)

Equivalently for $l\in\mathbb{Z}_{\leq 0}$

$\displaystyle I_{j,l}^{(2)}(v_{12},v_{11},v_{22};\varepsilon)$
$\displaystyle=\sum_{n=0}^{-l}\binom{-l}{n}\left(1-\sqrt{1-v_{22}}\,\right)^{-l-n}\left(1-v_{22}\right)^{n/2}$
$\displaystyle\qquad\times
I_{j,-n}^{(1)}\left(1-\frac{1-v_{12}}{\sqrt{1-v_{22}}},v_{11};\varepsilon\right)\,.$
(496)

###  D.5 Connection to physical propagators

Let us consider a situation with fixed momenta $p_{1}$, $p_{2}$, $\dots$,
$p_{n}$ and a PSI over the external momenta $k_{1}$ and $k_{2}$, where
$\sum_{j=1}^{n}p_{j}=k_{1}+k_{2}$ and establish the connection of physical
propagators with the scaled propagators occurring in the derived partial
fractioning formulas. We use the CMF for the angular integration momenta
$(\mathbf{k_{1}}+\mathbf{k_{2}}=0)$ and scale all momenta by their
corresponding energy component:

$\displaystyle k_{1}=E_{k}\,\left(1,\mathbf{k}\right)\,,\qquad
k_{2}=E_{k}\,\left(1,-\mathbf{k}\right)\,,\qquad
p_{i}=E_{i}\,\left(1,\mathbf{v_{i}}\right)\,.$ (497)

Also we suppose that $k_{1,2}$ momenta are on their mass-shell and for
simplicity we assume that they correspond to the same mass:
$k_{1,2}^{2}=m_{k}^{2}$. Considering generic physical propagator depending on
external $p_{i}$ and angular loop $(k_{1},k_{2})$ momenta

$\displaystyle\Delta_{\rm
phys}(k_{1,2},p_{i},m)=\frac{1}{(k_{1,2}-p_{i})^{2}-m^{2}}\,,$ (498)

we can always express it in terms of scaled propagator
$\Delta_{\mathbf{k}}\left(\mathbf{\pm\bar{v}_{i}}\right)$:

$\displaystyle\Delta_{\rm
phys}\left(k_{1,2},p_{i},m\right)=\frac{1}{p_{i}^{2}-m^{2}+m_{k}^{2}-2E_{i}E_{k}}\,\,\Delta_{\mathbf{k}}\left(\mathbf{\pm\bar{v}_{i}}\right)\,,$
(499)

where we introduced the scaled vector

$\displaystyle{\bf\bar{v}_{i}}=\frac{{\bf
v_{i}}}{\frac{p_{i}^{2}-m^{2}+m_{k}^{2}}{2E_{i}E_{k}}-1}\,.$ (500)

Note the two vectors $v_{i}=(1,{\bf v_{i}})$ and
$\bar{v}_{i}=(1,{\bf\bar{v}_{i}})$ are differed only by spatial part [see Eq.
(500)]. So, up to an angular independent factor we can express any physical
propagator of the discussed form as a scaled propagator with a suitably chosen
scaled vector.

###  D.6 Example: Drell-Yan double real corrections kinematics

To demonstrate how we can bring our general considerations about partial
fractioning to good use in a physical situation we consider the kinematics
schematically shown in Fig. 7. This is the kinematics of the double real
corrections for the Drell-Yan process. Conservation of momentum implies that
$p_{1}+p_{2}=q+k_{1}+k_{2}$. For the PSI $P_{\text{ext}}=p_{1}+p_{2}-q$.
Hence, the angular PSI is defined in the CMF of $p_{1}$, $p_{2}$, and $-q$ (or
equivalently in the CMF of $k_{1}$ and $k_{2}$):

$\displaystyle\mathbf{p}_{1}+\mathbf{p}_{2}-\mathbf{q}=0\,.$ (501)

Figure 6: Kinematics for double real corrections of the Drell-Yan process.
$p_{1}$ and $p_{2}$ are incoming momenta, $q$ is outgoing. The PSI is
performed over the dashed momenta $k_{1}$ and $k_{2}$.

Figure 7: The picture shows the $0\equiv x_{1}\equiv\dots\equiv x_{D-2}$ slice
of Minkowski space. The lightcone is shown in light yellow. The external
momenta are depicted as red arrows, $p_{1}$ and $p_{2}$ are massless and hence
on the light-cone, $q$ has mass $Q^{2}$ and is inside the light-cone. The
scaled momenta $v$, marked by blue dots, are in the blue $x_{0}\equiv 1$
plane. The blue circle marks the intersection of the light-cone with the
$x_{0}\equiv 1$ plane. The dashed red momenta illustrate the construction of
$\bar{v}_{q}$.

These three vectors are linear dependent and thus lie in a common plane,
defining the $(x_{D-1},x_{D})$-plane. We can then depict the kinematics in a
three dimensional slice of $D$ dimensional space-time (see Fig. 7). Since all
involved particles are massless, we have propagators of the general form

$\displaystyle\frac{1}{(k_{1,2}-p)^{2}+i0}\,,$ (502)

where $p$ is one of the fixed momenta. In the rest frame of $p_{1}$, $p_{2}$,
and $-q$ with the $x_{D}$-axis chosen in the direction of $\mathbf{p_{1}}$ the
vectors can be parametrized as

$\displaystyle k_{1}$ $\displaystyle=\frac{\sqrt{s_{2}}}{2}\left(1,{\bf
v_{k}}\right)=\frac{\sqrt{s_{2}}}{2}\left(1,\dots,\sin\theta_{1}\cos\theta_{2},\cos\theta_{1}\right)\,,$
$\displaystyle k_{2}$ $\displaystyle=\frac{\sqrt{s_{2}}}{2}\left(1,{\bf
v_{k}}\right)=\frac{\sqrt{s_{2}}}{2}\left(1,\dots,-\sin\theta_{1}\cos\theta_{2},-\cos\theta_{1}\right)\,,$
$\displaystyle p_{1}$ $\displaystyle=\frac{s_{2}-u}{2\sqrt{s_{2}}}(1,{\bf
v_{1}})=\frac{s_{2}-u}{2\sqrt{s_{2}}}\left(1,\mathbf{0}_{D-3},0,1\right)\,,$
$\displaystyle p_{2}$ $\displaystyle=\frac{s_{2}-t}{2\sqrt{s_{2}}}(1,{\bf
v_{2}})=\frac{s_{2}-t}{2\sqrt{s_{2}}}\left(1,\mathbf{0}_{D-3},\sin\vartheta,\cos\vartheta\right)\,,$
$\displaystyle q$ $\displaystyle=-\frac{t+u}{2\sqrt{s_{2}}}\left(1,{\bf
v_{q}}\right)=-\frac{t+u}{2\sqrt{s_{2}}}\left(1,\mathbf{0}_{D-3},\frac{t-s_{2}}{t+u}\sin\vartheta,\frac{u-s_{2}}{t+u}+\frac{t-s_{2}}{t+u}\cos\vartheta\right)\,,$
(503)

with

$\displaystyle\cos\vartheta=1-\frac{2ss_{2}}{(s_{2}-u)(s_{2}-t)}\,,$ (504)

where

$\displaystyle s=(p_{1}+p_{2})^{2}\,,\quad t=(p_{1}-q)^{2}\,,\quad
u=(p_{2}-q)^{2}\,,\quad s_{2}=(k_{1}+k_{2})^{2}\,.$ (505)

By conservation of four momentum it holds $Q^{2}=q^{2}=s+t+u-s_{2}$. It is
also useful to write the relations between Mandelstam variables and components
of transverse momentum in light-cone frame $q=(Q^{+},Q^{-},Q_{T})$, where

$\displaystyle Q^{\pm}$ $\displaystyle=$ $\displaystyle\frac{Q^{0}\pm
Q^{3}}{\sqrt{2}}\,,\quad q^{2}\,=\,Q^{2}=2Q^{+}Q^{-}-Q^{2}_{T}\,,$
$\displaystyle Q_{T}^{2}$ $\displaystyle=$
$\displaystyle\frac{\lambda(s,s_{2},Q^{2})}{4s_{2}}\,=\,\frac{ut-
s_{2}Q^{2}}{s}\,,$ $\displaystyle\cos\vartheta$ $\displaystyle=$
$\displaystyle\frac{Q_{T}^{2}-s_{2}}{Q_{T}^{2}+s_{2}}\,,\quad\sin\vartheta\ =\
\frac{2Q_{T}\sqrt{s_{2}}}{Q_{T}^{2}+s_{2}}\,,\quad\sqrt{s_{2}}\ =\ Q_{T}\,{\rm
tan}\frac{\vartheta}{2}\,.$ (506)

Introducing scaled momenta

$\displaystyle v_{1}=\frac{p_{1}}{E_{1}}=(1,{\bf v_{1}})\,,\qquad
v_{2}=\frac{p_{2}}{E_{2}}=(1,{\bf v_{2}})\,,$ $\displaystyle
v_{q}=\frac{q}{E_{q}}=(1,{\bf
v_{q}})\,,\qquad\bar{v}_{q}=(1,{\bf\bar{v}_{q}})\,,$ (507)

where

$\displaystyle{\bf\bar{v}_{q}}=\frac{{\bf
v_{q}}}{1+\frac{Q^{2}}{2E_{k}E_{q}}}={\bf v_{q}}\ \frac{t+u}{t+u-2Q^{2}}\,,$
(508)

the propagators become

$\displaystyle\frac{1}{t_{1,2}}\equiv\frac{1}{(k_{1,2}-p_{1})^{2}}$
$\displaystyle=\frac{1}{-2k_{1,2}\cdot
p_{1}}=-\frac{2}{s_{2}-u}\Delta_{\mathbf{k}}\left(\mathbf{\pm
v_{1}}\right)\,,$ (509)
$\displaystyle\frac{1}{u_{1,2}}\equiv\frac{1}{(k_{1,2}-p_{2})^{2}}$
$\displaystyle=\frac{1}{-2k_{1,2}\cdot
p_{2}}=-\frac{2}{s_{2}-t}\Delta_{\mathbf{k}}\left(\mathbf{\pm
v_{2}}\right)\,,$ (510)
$\displaystyle\frac{1}{w_{1,2}}\equiv\frac{1}{(k_{1,2}+q)^{2}}$
$\displaystyle=\frac{1}{2k_{1,2}\cdot
q+Q^{2}}=\frac{2}{2Q^{2}-t-u}\Delta_{\mathbf{k}}\left(\mathbf{\pm\bar{v}_{q}}\right)\,.$
(511)

The construction of the scaled momenta is depicted in Fig. 7. The general
denominators appearing in our calculations will have the form

$\displaystyle\frac{1}{t_{1}^{n_{1}}t_{2}^{n_{2}}u_{1}^{n_{3}}u_{2}^{n_{4}}w_{1}^{n_{5}}w_{2}^{n_{6}}}=$
$\displaystyle\left(-\frac{2}{s_{2}-u}\right)^{n_{1}+n_{2}}\left(-\frac{2}{s_{2}-t}\right)^{n_{3}+n_{4}}\left(\frac{2}{2Q^{2}-t-u}\right)^{n_{5}+n_{6}}$
$\displaystyle\times$
$\displaystyle\Delta^{n_{1}}_{\mathbf{k}}\left(\mathbf{v_{1}}\right)\Delta^{n_{2}}_{\mathbf{k}}\left(\mathbf{-v_{1}}\right)\Delta^{n_{3}}_{\mathbf{k}}\left(\mathbf{v_{2}}\right)\Delta^{n_{4}}_{\mathbf{k}}\left(\mathbf{-v_{2}}\right)\Delta^{n_{5}}_{\mathbf{k}}\left(\mathbf{\bar{v}_{q}}\right)\Delta^{n_{6}}_{\mathbf{k}}\left(\mathbf{-\bar{v}_{q}}\right)\,.$
(512)

In terms of the pictures from our general considerations, the situation in the
$x_{0}\equiv 1$ plane looks as Fig. 8.

Figure 8: Drell-Yan kinematics in terms of scaled vectors. Red lines indicate
pairs of linear dependent vectors. The dashed blue lines mark contributions to
genuinely different PSIs.

We can decompose Eq. (512) into a sum of terms were each one contains a pair
of linearly independent vectors by:

  * 1.

performing two-point partial fractioning between the pairs of linearly
dependent vectors;

  * 2.

doing three-point partial fractioning between the remaining three vectors.

The $\lambda$ used for three point partial fractioning can be read of using
the condition

$\displaystyle{\bf
0}=(s_{2}-u)\mathbf{v_{1}}+(s_{2}-t)\mathbf{v_{2}}+(2Q^{2}-t-u)\mathbf{\bar{v}_{q}}=2\sqrt{s_{2}}\,\Big{(}\mathbf{p}_{1}+\mathbf{p}_{2}-\mathbf{q}\Big{)}\,,$
(513)

which is the direct consequence of Eq. (501).

Expressing the numerator originating from some Dirac trace as a polynomial of
the two corresponding Mandelstam variables we can simplify our angular
integrals. In particular, the numerator simply decreases the powers of
propagators in denominator. All that we need is to complete the calculation of
the following set of integrals

$\displaystyle I^{(0)}_{j,l}(u_{12}^{\pm};\varepsilon)$ $\displaystyle=$
$\displaystyle\int\mathrm{d}\Omega_{k_{1},k_{2}}\Delta^{j}_{\mathbf{k}}\left(\mathbf{v_{1}}\right)\Delta^{l}_{\mathbf{k}}\left(\mathbf{\mp
v_{2}}\right)=\int\mathrm{d}\Omega_{k_{1},k_{2}}\Delta^{j}_{\mathbf{k}}\left(\mathbf{-v_{1}}\right)\Delta^{l}_{\mathbf{k}}\left(\mathbf{\pm
v_{2}}\right)\,,$ $\displaystyle
I^{(1)}_{j,l}(u_{i\bar{q}}^{\pm},v_{\bar{q}\bar{q}};\varepsilon)$
$\displaystyle=$
$\displaystyle\int\mathrm{d}\Omega_{k_{1},k_{2}}\,\Delta^{j}_{\mathbf{k}}\left(\mathbf{\bar{v}_{q}}\right)\Delta^{l}_{\mathbf{k}}\left(\mathbf{\mp
v_{i}}\right)=\int\mathrm{d}\Omega_{k_{1},k_{2}}\,\Delta^{j}_{\mathbf{k}}\left(\mathbf{-\bar{v}_{q}}\right)\Delta^{l}_{\mathbf{k}}\left(\mathbf{\pm
v_{i}}\right)\,,$ (514)

where $i=1,2$ and

$\displaystyle u_{12}^{\pm}=1\pm(1-v_{12})\,,\qquad
u_{i\bar{q}}^{\pm}=1\pm(1-v_{i\bar{q}})\,,$ $\displaystyle
v_{12}=1-\cos\vartheta=\frac{2ss_{2}}{(s_{2}-u)(s_{2}-t)}\,,\qquad
v_{\bar{q}\bar{q}}=\frac{4sQ^{2}}{(s+Q^{2}-s_{2})^{2}}\,,$ $\displaystyle
v_{1\bar{q}}=2\biggl{[}1+\frac{su}{(s_{2}-u)(s+Q^{2}-s_{2})}\biggr{]}\,,\qquad
v_{2\bar{q}}=2\biggl{[}1+\frac{st}{(s_{2}-t)(s+Q^{2}-s_{2})}\biggr{]}\,,$
(515)

for integer valued $j$ and $l$. These are exactly the massless and single-
massive two-denominator integrals presented in the main text. Note that the
integrals depending on the arguments $v_{1\bar{q}}$ and $v_{2\bar{q}}$ are
related by a crossing symmetry, i.e. by interchange $t\leftrightarrow u$. All
in all, we were able to write down all potentially appearing integrals for the
process without any concrete calulation.

##  Appendix E Consistency of the hypergeometric representation of angular
integrals

In this appendix we present further details regarding hypergeometric
representation of the general two denominator angular integrals in $D$
dimensions. First, we check consistency of the master formula for the two
denominator integral in terms of the Lauricella function derived in Eq. (153)
with known special results, e.g. derived in Ref. Somogyi:2011ir .

###  E.1 Double massive integral in four dimensions

To perform the check of Eq. (153) we start with the double massive integral in
four dimensions. In particular, for the special case $j=l=1$ and
$\varepsilon=0$ we recover the well-known four-dimensional expression for
$I_{1,1}$ using the chain of the reduction identities

$\displaystyle
F_{B}^{(3)}(a_{1},a_{2},a_{3},b_{1},b_{2},0,c,x_{1},x_{2},x_{3})=F_{3}(a_{1},a_{2},b_{1},b_{2},c,x_{1},x_{2})\,,$
(516) $\displaystyle
F_{3}(a_{1},a_{2},b_{1},b_{2},a_{1}+a_{2},x_{1},x_{2})=(1-x_{2})^{-b_{2}}\,F_{1}\left(a_{1},b_{1},b_{2},a_{1}+a_{2},x_{1},\frac{x_{2}}{x_{2}-1}\right)\,,$
(517) $\displaystyle
F_{1}(a,b_{1},b_{2},b_{1}+b_{2},x_{1},x_{2})=(1-x_{2})^{-a}\,\phantom{}_{2}F_{1}\left(a,b_{1},b_{1}+b_{2},\frac{x_{1}-x_{2}}{1-x_{2}}\right)\,,$
(518)

the quadratic hypergeometric transformation

$\phantom{}{}_{2}F_{1}\left(\frac{a}{2},\frac{a+1}{2},b+\frac{1}{2},x\right)=(1+\sqrt{x})^{-a}\,\phantom{}_{2}F_{1}\left(a,b,2b,2-\frac{2}{1+\sqrt{x}}\right)\,,$
(519)

and

$\phantom{}{}_{2}F_{1}(1,1,2,x)=\sum_{n=0}^{\infty}\frac{\left(1\right)_{n}\left(1\right)_{n}}{\left(2\right)_{n}}\,\frac{x^{n}}{n!}=\sum_{n=0}^{\infty}\frac{x^{n}}{n+1}=-\frac{\log(1-x)}{x}\,.$
(520)

Starting with formula (153), where we interchange the symmetric parameters
$a_{1}$ and $b_{1}$, we straightforwardly obtain

$\displaystyle I_{1,1}(v_{12},v_{11},v_{22};0)$
$\displaystyle\overset{(\ref{eq:Hypergeometric representation of
Ijl})}{=}\frac{2\pi}{v_{12}}\,F_{B}^{(3)}\left(1,\frac{1}{2},\frac{1}{2},\frac{1}{2},1,0,\frac{3}{2};1-\frac{v_{11}}{v_{12}},1-\frac{v_{22}}{v_{12}},1-v_{12}\right)$
$\displaystyle\overset{(\ref{eq:lauricellatoappellthree})}{=}\frac{2\pi}{v_{12}}\,F_{3}\left(1,\frac{1}{2},\frac{1}{2},1,\frac{3}{2},1-\frac{v_{11}}{v_{12}},1-\frac{v_{22}}{v_{12}}\right)$
$\displaystyle\overset{(\ref{eq:appellthreetoappell})}{=}\frac{2\pi}{v_{12}}\,\frac{v_{12}}{v_{22}}\,F_{1}\left(1,\frac{1}{2},1,\frac{3}{2},1-\frac{v_{11}}{v_{12}},1-\frac{v_{12}}{v_{22}}\right)$
$\displaystyle\overset{(\ref{eq:appelltoghy})}{=}\frac{2\pi}{v_{12}}\,\phantom{}_{2}F_{1}\left(1,\frac{1}{2},\frac{3}{2};1-\frac{v_{11}v_{22}}{v_{12}^{2}}\right)$
$\displaystyle\overset{(\ref{eq:ghy quadratic
transformation})}{=}\frac{2\pi}{v_{12}+\sqrt{v_{12}^{2}-v_{11}v_{22}}}\,\phantom{}_{2}F_{1}\left(1,1,2,1-\frac{v_{12}-\sqrt{v_{12}^{2}-v_{11}v_{22}}}{v_{12}+\sqrt{v_{12}^{2}-v_{11}v_{22}}}\right)$
$\displaystyle\overset{(\ref{eq:ghytolog})}{=}\frac{\pi}{\sqrt{v_{12}^{2}-v_{11}v_{22}}}\,\log\left(\frac{v_{12}+\sqrt{v_{12}^{2}-v_{11}v_{22}}}{v_{12}-\sqrt{v_{12}^{2}-v_{11}v_{22}}}\right)\,.$
(521)

###  E.2 Single massive integral

In consideration of single massive integral we use two important indentities

$\displaystyle
F_{1}\left(a,b,b,c,\frac{y+\sqrt{y+(1-y)x}}{y-1},\frac{y-\sqrt{y+(1-y)x}}{y-1}\right)$
$\displaystyle=\frac{(1-y)^{b}}{B(a,c-a)}\
\int_{0}^{1}\mathrm{d}t\,t^{a-1}(1-t)^{c-a-1}(1-xt^{2}-y(1-t)^{2})^{-b}$
$\displaystyle=(1-y)^{b}\sum_{m,n=0}^{\infty}\frac{\left(a\right)_{2m}\left(c-a\right)_{2n}\left(b\right)_{m+n}}{\left(c\right)_{2m+2n}}\frac{x^{m}}{m!}\frac{y^{n}}{n!}\,,$
(522)

and

$\displaystyle\sum_{n=0}^{\infty}\frac{\left(l\right)_{2n}4^{-n}}{\left(\frac{3}{2}-\varepsilon\right)_{m+n+p}n!}=\phantom{}_{2}F_{1}\left(\frac{l}{2},\frac{l+1}{2},\frac{3}{2}-\varepsilon,1\right)\frac{\left(1-l-\varepsilon\right)_{m+p}}{\left(2-l-2\varepsilon\right)_{2m+2p}}\,4^{m+p}\,.$
(523)

The identity (522) is derived using integral representations for the Appell
and gamma functions and the bivariate generating function

$\displaystyle\Big{(}1-xt^{2}-y(1-t)^{2}\Big{)}^{-b}=\sum_{m,n=0}^{\infty}\,\left(b\right)_{m+n}\,\frac{x^{m}}{m!}\frac{y^{n}}{n!}\,t^{2m}\,(1-t)^{2n}\,.$
(524)

and then using duplication formula for the Pochhammer symbols

$\displaystyle(x)_{n}\,\biggl{(}x+\frac{1}{2}\biggr{)}_{n}=2^{-2n}\,(2x)_{2n}\,.$
(525)

The second identity (523) is proved using the Pochhammer shift identities

$\displaystyle(x)_{m+n}$ $\displaystyle=$ $\displaystyle(x+m)_{n}\,(x)_{m}\,,$
(526) $\displaystyle(x)_{m+n}\,(x-n)_{n}$ $\displaystyle=$
$\displaystyle(x+m)_{n}\,(x-n)_{m+n}\,,$ (527)

duplication identity (525), and the Pochhammer-Gauss summation formula

$\displaystyle\sum\limits_{n=0}^{\infty}\,\frac{(a)_{n}\,(b)_{n}}{(c)_{n}n!}=\phantom{}_{2}F_{1}(a,b,c,1)=\frac{(c-a)_{a}}{(c-a-b)_{a}}\,.$
(528)

Here, Eq. (527) is a generalization of Eq. (526) derived by multiplying of
l.h.s. and r.h.s. of Eq. (526) by $(x-n)_{n}$:

$\displaystyle(x)_{m+n}\,(x-n)_{n}=(x+m)_{n}\,\underbrace{(x)_{m}\,(x-n)_{n}}_{=(x-n)_{m+n}}=(x+m)_{n}\,(x-n)_{m+n}\,.$
(529)

Eq. (528) is the combination of two formulas — power expansion of Gauss
function (103) and Gauss summation theorem (109).

In particular, Eq. (522) is derived as:

$\displaystyle\sum_{n=0}^{\infty}\frac{\left(l\right)_{2n}4^{-n}}{\left(\frac{3}{2}-\varepsilon\right)_{m+n+p}n!}$
$\displaystyle\overset{\rm(\ref{eq:pochhammer
doubling}),\rm(\ref{eq:pochhammer
shift})}{=}\frac{1}{\left(\frac{3}{2}-\varepsilon\right)_{m+p}}\
\sum_{n=0}^{\infty}\frac{\left(\frac{l}{2}\right)_{n}\left(\frac{l+1}{2}\right)_{n}}{\left(\frac{3}{2}-\varepsilon+m+p\right)_{n}n!}$
$\displaystyle\overset{\rm(\ref{Pochhamer_Gauss_Theorem})}{=}\frac{1}{\left(\frac{3}{2}-\varepsilon\right)_{m+p}}\
\frac{\left(\frac{3}{2}-\frac{l}{2}+m+p-\varepsilon\right)_{\frac{l}{2}}}{\left(1-l+m+p-\varepsilon\right)_{\frac{l}{2}}}$
$\displaystyle\overset{\rm(\ref{eq:pochhammer
shift2})}{=}\frac{\left(\frac{3}{2}-\frac{l}{2}-\varepsilon\right)_{\frac{l}{2}}}{\left(1-l-\varepsilon\right)_{\frac{l}{2}}}\
\frac{\left(1-l-\varepsilon\right)_{m+p}}{\left(\frac{3}{2}-\frac{l}{2}-\varepsilon\right)_{m+p}\left(1-\frac{l}{2}-\varepsilon\right)_{m+p}}$
$\displaystyle\overset{\rm(\ref{eq:pochhammer
doubling})}{=}\phantom{}_{2}F_{1}\left(\frac{l}{2},\frac{l+1}{2},\frac{3}{2}-\varepsilon,1\right)\
\frac{\left(1-l-\varepsilon\right)_{m+p}}{\left(2-l-2\varepsilon\right)_{2m+2p}}\,4^{m+p}\,.$
(530)

Note, in the step from the second to the third line of Eq. (E.2) we apply the
Pochhammer shift identity (527) twice.

Now we are in the position to derive the formula for the angular single
massive integral

$\displaystyle
I_{j,l}(v_{12},v_{11},0,\varepsilon)=\frac{2\pi}{1-2\varepsilon}\,v_{12}^{1-j-l-\varepsilon}$
$\displaystyle\times
F_{B}^{(3)}\left(\frac{j}{2},\frac{l}{2},\frac{3-j-l-2\varepsilon}{2},\frac{j+1}{2},\frac{l+1}{2},\frac{2-j-l-2\varepsilon}{2},\frac{3}{2}-\varepsilon,1-\frac{v_{11}}{v_{12}},1,1-v_{12}\right)$
$\displaystyle=\frac{2\pi}{1-2\varepsilon}\,v_{12}^{1-j-l-\varepsilon}\sum_{m,n,p=0}^{\infty}\frac{\left(j\right)_{2m}\left(l\right)_{2n}\left(2-j-l-2\varepsilon\right)_{2p}}{\left(\frac{3}{2}-\varepsilon\right)_{m+n+p}m!n!p!}\,4^{-m-n-p}\left(1-\frac{v_{11}}{v_{12}}\right)^{m}(1-v_{12})^{p}$
$\displaystyle=\frac{2\pi}{1-2\varepsilon}\,v_{12}^{1-j-l-\varepsilon}\,\phantom{}{}_{2}F_{1}\left(\frac{l}{2},\frac{l+1}{2},\frac{3}{2}-\varepsilon,1\right)$
$\displaystyle\times\sum_{m,p=0}^{\infty}\frac{\left(j\right)_{2m}\left(2-j-l-2\varepsilon\right)_{2p}\left(1-l-\varepsilon\right)_{m+p}}{\left(2-l-2\varepsilon\right)_{2m+2p}m!p!}\,\left(1-\frac{v_{11}}{v_{12}}\right)^{m}(1-v_{12})^{p}$
$\displaystyle\overset{(\ref{eq:Appell function bb
sum})}{=}I_{l}^{(0)}(\varepsilon)\,v_{12}^{-j}F_{1}\left(j,1-l-\varepsilon,1-l-\varepsilon,2-l-2\varepsilon,1-\frac{1+\sqrt{1-v_{11}}}{v_{12}},1-\frac{1-\sqrt{1-v_{11}}}{v_{12}}\right)$
$\displaystyle=I_{j,l}^{(1)}(v_{12},v_{11},\varepsilon)\,,$ (531)

which is in the exact agreement with the known analytic result for the single
massive integral in $D$ dimensions Somogyi:2011ir .

###  E.3 Massless integral

We perform the check of the massless integral starting from the single massive
result (E.2), where we put $v_{11}=0$. Using the identities

$\displaystyle F_{1}(a,b_{1},b_{2},c,x,1)$
$\displaystyle=\phantom{}_{2}F_{1}(a,b_{2},c,1)\,\phantom{}_{2}F_{1}(a,b_{1},c-b_{2},x)\,,$
(532) $\phantom{}{}_{2}F_{1}(a,b,c,x)$
$\displaystyle=(1-x)^{-a}\,\phantom{}_{2}F_{1}\left(a,c-b,c,\frac{x}{x-1}\right)\,,$
(533)

the Gauss theorem (109)

$\phantom{}{}_{2}F_{1}(j,1-l-\varepsilon,2-l-2\varepsilon,1)$ $\displaystyle=$
$\displaystyle\frac{\Gamma(2-l-2\varepsilon)\Gamma(1-j-\varepsilon)}{\Gamma(2-l-j-2\varepsilon)\Gamma(1-\varepsilon)}=\frac{\left(2-j-l-2\varepsilon\right)_{j}}{\left(1-j-\varepsilon\right)_{j}}$
(534) $\displaystyle=$
$\displaystyle\frac{\left(2-j-l-2\varepsilon\right)_{j+l}}{\left(1-j-\varepsilon\right)_{j}\,\left(2-l-2\varepsilon\right)_{l}}\,,$

and Eq. (3.2.4) we obtain the formula

$\displaystyle I_{j,l}(v_{12},0,0;\varepsilon)$
$\displaystyle=I_{l}^{(0)}(\varepsilon)\,v_{12}^{-j}\,F_{1}\left(j,1-l-\varepsilon,1-l-\varepsilon,2-l-2\varepsilon,1-\frac{2}{v_{12}},1\right)$
$\displaystyle=I_{l}^{(0)}(\varepsilon)\,v_{12}^{-j}\phantom{}{}_{2}F_{1}(j,1-l-\varepsilon,2-l-2\varepsilon,1)\,\phantom{}_{2}F_{1}\left(j,1-l-\varepsilon,1-\varepsilon,1-\frac{2}{v_{12}}\right)$
$\displaystyle=\frac{I_{l}^{(0)}(\varepsilon)}{2^{j}}\phantom{}_{2}F_{1}(j,1-l-\varepsilon,2-l-2\varepsilon,1)\,\phantom{}_{2}F_{1}\left(j,l,1-\varepsilon,1-\frac{v_{12}}{2}\right)$
$\displaystyle=\frac{I^{(0)}(\varepsilon)}{2^{j+l}}\,\frac{\left(2-j-l-2\varepsilon\right)_{j+l}}{\left(1-j-\varepsilon\right)_{j}\left(1-l-\varepsilon\right)_{l}}\,\phantom{}_{2}F_{1}\left(j,l,1-\varepsilon,1-\frac{v_{12}}{2}\right)$
$\displaystyle=I^{(0)}_{j,l}(v_{12};\varepsilon)\,,$ (535)

which is again in complete agreement with the result of Ref. Somogyi:2011ir .

##  Appendix F Double Nielsen polylogarithms

###  F.1 Basic properties of double Nielsen polylogarithms

Basic properties of the double Nielsen polylogarithm

$\displaystyle
S_{n,p_{1},p_{2}}(x,y)\equiv\frac{(-1)^{n+p_{1}+p_{2}-1}}{(n-1)!p_{1}!p_{2}!}\int_{0}^{1}\frac{\mathrm{d}t}{t}\,\log^{n-1}t\log^{p_{1}}(1-xt)\log^{p_{2}}(1-yt)\,.$
(536)

are:

(1) symmetry $S_{n,p_{1},p_{2}}(x,y)=S_{n,p_{2},p_{1}}(y,x)$,

(2) reduction to single Nielsen polylogarithm
$S_{n,p_{1},0}(x,y)=S_{n,p_{1}}(x)$ and $S_{n,0,p_{2}}(x,y)=S_{n,p_{2}}(y)$.

With expansion

$\displaystyle\log^{m}(1+x)=m!\sum_{k=m}^{\infty}S_{k}^{(m)}\frac{x^{k}}{k!}\quad\text{for
}|x|<1\,,$ (537)

where

$\displaystyle
S_{k}^{(m)}=\sum_{i=0}^{k-m}\frac{1}{i!}\binom{k-1+i}{k-m+i}\binom{2k-m}{k-m-i}\sum_{j=0}^{i}(-1)^{i}\binom{i}{j}^{k-m+i}\,,$
(538)

are Stirling numbers of the 1st kind and integral

$\displaystyle\int\limits_{0}^{1}\mathrm{d}tt^{m-1}(\log
t)^{n-1}=-(-m)^{-n}(n-1)!\,,$ (539)

one obtains the series

$\displaystyle
S_{n,p_{1},p_{2}}(x,y)=\sum_{k,l=0}^{\infty}\frac{(-1)^{k+l}S_{p_{1}+k}^{(p_{1})}S_{p_{2}+l}^{(p_{2})}}{(p_{1}+k)!(p_{2}+l)!(p_{1}+p_{2}+k+l)^{n}}\,x^{p_{1}+k}y^{p_{2}+l}\quad\text{for
}|x|\leq 1\,,\,|y|\leq 1\,.$ (540)

###  F.2 Relation of Goncharov and Nielsen polylogarithms

Goncharov polylogarithms are a powerful tool for working with a wide range of
generalized logarithms Goncharov:2001iea . They are recursively defined via
the iterated integral

$\displaystyle
G(a_{1},\dots,a_{n};z)\equiv\int_{0}^{1}\frac{\mathrm{d}t}{t-a_{1}}\,G(a_{2},\dots,a_{n};t)$
(541)

with $G(z)=1$. It holds with
$\vec{a}_{n}=(\underbrace{a,\dots,a}_{n\text{-times}})$:

$\displaystyle G(\vec{0}_{n};z)=\frac{1}{n!}\log^{n}z\,,\quad
G(\vec{a}_{n};z)=\frac{1}{n!}\log^{n}\left(1-\frac{z}{a}\right)\,,$
$\displaystyle
G(\vec{0}_{n-1},a;z)=-\mathrm{Li}_{n}\left(\frac{z}{a}\right),\quad
G(\vec{0}_{n-k},\vec{a}_{k};z)=(-1)^{k}S_{n-k,k}\left(\frac{z}{a}\right)\,.$
(542)

Goncharov polylogarithms are related to other polylogarithms by the relations:

(1) with Nielsen polylogarithms by
$S_{n,p}(z)=(-1)^{p}G(\overbrace{0,\dots,0}^{n},\overbrace{1,\dots,1}^{p};z)$;

(2) with Harmonic polylogarithms by $H(a_{1},\dots
a_{n};z)=(-1)^{p}G(a_{1},\dots,a_{n};z)$,

where $a_{i}\in\\{0,1\\}$

(3) with Multiple polylogarithms by

$G(\vec{0}_{m_{1}-1},a_{1},\dots,\vec{0}_{m_{k}-1},a_{k};z)=(-1)^{k}\mathrm{Li}_{m_{k},\dots,m_{1}}\left(\frac{a_{k-1}}{a_{k}},\dots,\frac{a_{1}}{a_{2}},\frac{z}{a_{1}}\right)\,,$

where

$\displaystyle\mathrm{Li}_{m_{1},\dots,m_{k}}=\sum_{0<n_{1}<\dots<n_{k}}\frac{z_{1}^{n_{1}}z_{2}^{n_{2}}\dots
z_{k}^{n_{k}}}{n_{1}^{m_{1}}n_{2}^{m_{2}}\dots n_{k}^{m_{k}}}\,.$ (543)

Goncharov polylogarithms satisfy the shuffle algebra

$\displaystyle
G(\vec{a};z)G(\vec{b},z)=\sum_{\vec{c}\in\vec{a}\sqcup\mathchoice{\mkern-8.0mu}{\mkern-8.0mu}{\mkern-3.5mu}{\mkern-3.5mu}\sqcup\vec{b}}G(\vec{c};z)\,,$
(544)

where
$\vec{a}\sqcup\mathchoice{\mkern-8.0mu}{\mkern-8.0mu}{\mkern-3.5mu}{\mkern-3.5mu}\sqcup\vec{b}$
denotes the set of all shuffles of the vectors $\vec{a}$ and $\vec{b}$, i.e.
all the ways of interlacing them while keeping the order within $\vec{a}$ and
$\vec{b}$. Goncharov polylogarithm has the integration by parts property

$\displaystyle G(z_{1},z_{2},\dots,z_{n};1)$
$\displaystyle=G(z_{1};1)G(z_{2},\dots,z_{n};1)-G(z_{2},z_{1};1)G(z_{3},\dots,z_{n};1)$
$\displaystyle+\dots-(-1)^{n}G(z_{n},\dots,z_{2},z_{1};1)\,.$ (545)

Another useful identity is the logarithmic integral

$\displaystyle
G(\vec{0}_{m},z_{1},z_{2},\dots,z_{n};1)=\frac{(-1)^{m}}{m!}\int_{0}^{1}\frac{\mathrm{d}t\,\log^{m}t}{t-z_{1}}\,G(z_{2},\dots,z_{n};t)\,.$
(546)

###  F.3 Double Nielsen polylogarithms in terms of multiple polylogarithms

Using the logarithmic integral Eq. (546) and the shuffle algebra we can
represent the double Nielsen polylogarithm in terms of multiple
polylogarithms. Starting from Eq. (536) we find

$\displaystyle S_{n,p_{1},p_{2}}(x,y)$
$\displaystyle=\frac{(-1)^{n+p_{1}+p_{2}-1}}{(n-1)!p_{1}!p_{2}!}\int_{0}^{1}\frac{\mathrm{d}t}{t}\,\log^{n-1}t\log^{p_{1}}(1-xt)\log^{p_{2}}(1-yt)$
$\displaystyle=\frac{(-1)^{n+p_{1}+p_{2}-1}}{(n-1)!}\int_{0}^{1}\frac{\mathrm{d}t}{t}\,\log^{n-1}t\,G\left(\vec{\left(\frac{1}{x}\right)}_{p_{1}};t\right)\,G\left(\vec{\left(\frac{1}{y}\right)}_{p_{2}};t\right)$
$\displaystyle=\frac{(-1)^{n+p_{1}+p_{2}-1}}{(n-1)!}\int_{0}^{1}\frac{\mathrm{d}t}{t}\,\log^{n-1}t\,\sum_{\vec{z}\in\vec{\frac{1}{x}}\sqcup\mathchoice{\mkern-8.0mu}{\mkern-8.0mu}{\mkern-3.5mu}{\mkern-3.5mu}\sqcup\vec{\frac{1}{y}}}G(z_{1},\dots,z_{p_{1}+p_{2}};t)$
$\displaystyle=(-1)^{p_{1}+p_{2}}\sum_{\vec{z}\in\vec{\frac{1}{x}}\sqcup\mathchoice{\mkern-8.0mu}{\mkern-8.0mu}{\mkern-3.5mu}{\mkern-3.5mu}\sqcup\vec{\frac{1}{y}}}G(\vec{0}_{n-1},0,z_{1},\dots,z_{p_{1}+p_{2}};1)$
$\displaystyle=\sum_{\vec{z}\in\vec{x}_{p_{1}}\sqcup\mathchoice{\mkern-8.0mu}{\mkern-8.0mu}{\mkern-3.5mu}{\mkern-3.5mu}\sqcup\vec{y}_{p_{2}}}\mathrm{Li}_{1,1,\dots,1,n+1}\left(\frac{z_{p_{1}+p_{2}}}{z_{p_{1}+p_{2}-1}},\dots,\frac{z_{2}}{z_{1}},z_{1}\right)\,.$
(547)

Therefore, double Nielsen polylogarithms can be calculated from multiple
polylogarithms using the identity

$S_{n,p_{1},p_{2}}(x,y)=\sum_{\vec{z}\in\vec{x}_{p_{1}}\sqcup\mathchoice{\mkern-8.0mu}{\mkern-8.0mu}{\mkern-3.5mu}{\mkern-3.5mu}\sqcup\vec{y}_{p_{2}}}\mathrm{Li}_{1,1,\dots,1,n+1}\left(\frac{z_{p_{1}+p_{2}}}{z_{p_{1}+p_{2}-1}},\dots,\frac{z_{2}}{z_{1}},z_{1}\right)\,.$
(548)

For the reduction and calculation of multiple polylogarithms there exist
publicly available tools Frellesvig:2016ske ; Duhr:2019tlz . Some explicit
examples for the shuffle sums are

$\displaystyle S_{n,1,1}(x,y)$ $\displaystyle=\sum_{\vec{z}\in
x\sqcup\mathchoice{\mkern-8.0mu}{\mkern-8.0mu}{\mkern-3.5mu}{\mkern-3.5mu}\sqcup
y}\mathrm{Li}_{1,n+1}\left(\frac{z_{2}}{z_{1}},z_{1}\right)=\mathrm{Li}_{1,n+1}\left(\frac{y}{x},x\right)+\mathrm{Li}_{1,n+1}\left(\frac{x}{y},y\right)\,,$
(549) $\displaystyle S_{n,2,1}(x,y)$
$\displaystyle=\sum_{\vec{z}\in\vec{x}_{2}\sqcup\mathchoice{\mkern-8.0mu}{\mkern-8.0mu}{\mkern-3.5mu}{\mkern-3.5mu}\sqcup
y}\mathrm{Li}_{1,1,n+1}\left(\frac{z_{3}}{z_{2}},\frac{z_{2}}{z_{1}},z_{1}\right)$
$\displaystyle=\mathrm{Li}_{1,1,n+1}\left(\frac{y}{x},1,x\right)+\mathrm{Li}_{1,1,n+1}\left(\frac{x}{y},\frac{y}{x},x\right)+\mathrm{Li}_{1,1,n+1}\left(1,\frac{x}{y},y\right)\,,$
(550) $\displaystyle S_{n,2,2}(x,y)$
$\displaystyle=\sum_{\vec{z}\in\vec{x}_{2}\sqcup\mathchoice{\mkern-8.0mu}{\mkern-8.0mu}{\mkern-3.5mu}{\mkern-3.5mu}\sqcup\vec{y}_{2}}\mathrm{Li}_{1,1,1,n+1}\left(\frac{z_{4}}{z_{3}},\frac{z_{3}}{z_{2}},\frac{z_{2}}{z_{1}},z_{1}\right)$
$\displaystyle=\mathrm{Li}_{1,1,1,n+1}\left(1,\frac{y}{x},1,x\right)+\mathrm{Li}_{1,1,1,n+1}\left(\frac{y}{x},\frac{x}{y},\frac{y}{x},x\right)$
$\displaystyle+\mathrm{Li}_{1,1,1,n+1}\left(\frac{x}{y},1,\frac{y}{x},1,x\right)+\mathrm{Li}_{1,1,1,n+1}\left(\frac{y}{x},1,\frac{x}{y},y\right)$
$\displaystyle+\mathrm{Li}_{1,1,1,n+1}\left(\frac{x}{y},\frac{y}{x},\frac{x}{y},y\right)+\mathrm{Li}_{1,1,1,n+1}\left(1,\frac{x}{y},1,y\right)\,.$
(552)

##  Appendix G Tables of angular integrals

###  G.1 Massless integral with one denominator

$\displaystyle I^{(0)}_{-4}(\varepsilon)$
$\displaystyle=4I^{(0)}(\varepsilon)\,\frac{(3-\varepsilon)(4-\varepsilon)}{(3-2\varepsilon)(5-2\varepsilon)}$
$\displaystyle=\frac{32\pi}{5}+\frac{1192\pi\varepsilon}{75}+\frac{38152\pi\varepsilon^{2}}{1125}+\frac{1168912\pi\varepsilon^{3}}{16875}+\frac{35313472\pi\varepsilon^{4}}{253125}+\mathcal{O}\left(\varepsilon^{5}\right)\,,$
(553) $\displaystyle I^{(0)}_{-3}(\varepsilon)$
$\displaystyle=2I^{(0)}(\varepsilon)\,\frac{3-\varepsilon}{3-2\varepsilon}$
$\displaystyle=4\pi+\frac{28\pi\varepsilon}{3}+\frac{176\pi\varepsilon^{2}}{9}+\frac{1072\pi\varepsilon^{3}}{27}+\frac{6464\pi\varepsilon^{4}}{81}+\mathcal{O}\left(\varepsilon^{5}\right)\,,$
(554) $\displaystyle I^{(0)}_{-2}(\varepsilon)$
$\displaystyle=2I^{(0)}(\varepsilon)\,\frac{2-\varepsilon}{3-2\varepsilon}$
$\displaystyle=\frac{8\pi}{3}+\frac{52\pi\varepsilon}{9}+\frac{320\pi\varepsilon^{2}}{27}+\frac{1936\pi\varepsilon^{3}}{81}+\frac{11648\pi\varepsilon^{4}}{243}+\mathcal{O}\left(\varepsilon^{5}\right)\,,$
(555) $\displaystyle I^{(0)}_{0}(\varepsilon)$
$\displaystyle=I^{(0)}_{-1}(\varepsilon)=I^{(0)}(\varepsilon)=\frac{2\pi}{1-2\varepsilon}=2\pi+4\pi\varepsilon+8\pi\varepsilon^{2}+16\pi\varepsilon^{3}+32\pi\varepsilon^{4}+\mathcal{O}\left(\varepsilon^{5}\right)\,,$
(556) $\displaystyle I^{(0)}_{1}(\varepsilon)$
$\displaystyle=-I^{(0)}(\varepsilon)\,\frac{1-2\varepsilon}{2\varepsilon}=-\frac{\pi}{\varepsilon}\,,$
(557) $\displaystyle I^{(0)}_{2}(\varepsilon)$
$\displaystyle=-I^{(0)}(\varepsilon)\,\frac{1-2\varepsilon}{2(1+\varepsilon)}=-\frac{\pi}{1+\varepsilon}=-\pi+\pi\varepsilon-\pi\varepsilon^{2}+\pi\varepsilon^{3}-\pi\varepsilon^{4}+\mathcal{O}\left(\varepsilon^{5}\right)\,.$
(558)

###  G.2 Massive integral with one denominator

$\displaystyle I^{(1)}_{-4}(v_{11};\varepsilon)$
$\displaystyle=I^{(0)}(\varepsilon)\,\frac{2(6-3v_{11}-2\varepsilon)(4-\varepsilon)-3v_{11}(4-v_{11}-2\varepsilon)}{(3-2\varepsilon)(5-2\varepsilon)}$
$\displaystyle=\frac{2}{5}\pi\left(16-12v_{11}+v_{11}^{2}\right)+\frac{4}{75}\pi\left(298-246v_{11}+23v_{11}^{2}\right)\varepsilon+\mathcal{O}\left(\varepsilon^{2}\right)\,,$
(559) $\displaystyle I^{(1)}_{-3}(v_{11};\varepsilon)$
$\displaystyle=I^{(0)}(\varepsilon)\,\frac{6-3v_{11}-2\varepsilon}{3-2\varepsilon}$
$\displaystyle=2\pi(2-v_{11})+\frac{4}{3}\pi(7-4v_{11})\varepsilon+\mathcal{O}\left(\varepsilon^{2}\right)\,,$
(560) $\displaystyle I^{(1)}_{-2}(v_{11};\varepsilon)$
$\displaystyle=I^{(0)}(\varepsilon)\,\frac{4-v_{11}-2\varepsilon}{3-2\varepsilon}$
$\displaystyle=\frac{2}{3}\pi(4-v_{11})+\frac{4}{9}\pi(13-4v_{11})\varepsilon+\mathcal{O}\left(\varepsilon^{2}\right)\,,$
(561) $\displaystyle I^{(1)}_{0}(v_{11};\varepsilon)$
$\displaystyle=I^{(1)}_{-1}(v_{11};\varepsilon)=I^{(0)}(\varepsilon)=\frac{2\pi}{1-2\varepsilon}=2\pi+4\pi\varepsilon+\mathcal{O}\left(\varepsilon^{2}\right)\,,$
(562) $\displaystyle I^{(1)}_{1}(v_{11};\varepsilon)$
$\displaystyle=\frac{\pi\log\left(\frac{1+\sqrt{1-v_{11}}}{1-\sqrt{1-v_{11}}}\right)}{\sqrt{1-v_{11}}}-\frac{\pi\varepsilon\left(\log^{2}\left(\frac{1+\sqrt{1-v_{11}}}{1-\sqrt{1-v_{11}}}\right)+4\mathrm{Li}_{2}\left(\frac{2\sqrt{1-v_{11}}}{\sqrt{1-v_{11}}-1}\right)\right)}{2\sqrt{1-v_{11}}}$
$\displaystyle+\mathcal{O}\left(\varepsilon^{2}\right)\,,$ (563)
$\displaystyle I^{(1)}_{2}(v_{11};\varepsilon)$
$\displaystyle=\frac{2\varepsilon}{v_{11}}\,I^{(1)}_{1}(v_{11};\varepsilon)+\frac{1-2\varepsilon}{v_{11}}\,I^{(0)}=\frac{2\pi}{v_{11}}\,\biggl{[}1+\frac{\varepsilon\log\left(\frac{1+\sqrt{1-v_{11}}}{1-\sqrt{1-v_{11}}}\right)}{\sqrt{1-v_{11}}}\biggr{]}$
$\displaystyle+\mathcal{O}\left(\varepsilon^{2}\right)\,.$ (564)

###  G.3 Massless integral with two denominators

$\displaystyle I^{(0)}_{-4,0}(v_{12};\varepsilon)$
$\displaystyle=I^{(0)}_{-4}(\varepsilon)\,,$ (565) $\displaystyle
I^{(0)}_{-4,-1}(v_{12};\varepsilon)$
$\displaystyle=\frac{1}{4(3-\varepsilon)}\,\biggl{[}(7-2\varepsilon)I^{(0)}_{-4}(\varepsilon)+(13-2\varepsilon)I^{(0)}_{-3,-1}(v_{12};\varepsilon)-5v_{12}I^{(0)}_{-3}(\varepsilon)\biggr{]}$
$\displaystyle=\frac{32\pi}{15}\left(5-2v_{12}\right)+\frac{8\pi}{225}\left(785-338v_{12}\right)\varepsilon+\mathcal{O}\left(\varepsilon^{2}\right)\,,$
(566) $\displaystyle I^{(0)}_{-4,-2}(v_{12};\varepsilon)$
$\displaystyle=\frac{1}{7-2\varepsilon}\,\biggl{[}(5-\varepsilon)I^{(0)}_{-4,-1}(v_{12};\varepsilon)+(7-\varepsilon)I^{(0)}_{-3,-2}(v_{12};\varepsilon)-3v_{12}I^{(0)}_{-3,-1}(v_{12};\varepsilon)\biggr{]}$
$\displaystyle=\frac{64\pi}{105}\left(30-20v_{12}+3v_{12}^{2}\right)+\frac{16\pi}{11025}\left(34470-24380v_{12}+3909v_{12}^{2}\right)\varepsilon$
$\displaystyle+\mathcal{O}\left(\varepsilon^{2}\right)\,,$ (567)
$\displaystyle I^{(0)}_{-4,-3}(v_{12};\varepsilon)$
$\displaystyle=\frac{1}{4(4-\varepsilon)}\,\biggl{[}(13-2\varepsilon)I^{(0)}_{-4,-2}(v_{12};\varepsilon)+(15-2\varepsilon)I^{(0)}_{-3,-3}(v_{12};\varepsilon)$
$\displaystyle-7v_{12}I^{(0)}_{-3,-2}(v_{12};\varepsilon)\biggr{]}=\frac{16\pi}{35}\left(70-60v_{12}+15v_{12}^{2}-v_{12}^{3}\right)$
$\displaystyle+\frac{16\pi}{3675}\left(20895-18810v_{12}+4965v_{12}^{2}-352v_{12}^{3}\right)\varepsilon+\mathcal{O}\left(\varepsilon^{2}\right)\,,$
(568) $\displaystyle I^{(0)}_{-4,-4}(v_{12};\varepsilon)$
$\displaystyle=\frac{1}{9-\varepsilon}\,\biggl{[}(8-\varepsilon)\Big{(}I^{(0)}_{-4,-3}(v_{12};\varepsilon)+I^{(0)}_{-3,-4}(v_{12};\varepsilon)\Big{)}-4v_{12}I^{(0)}_{-3,-3}(v_{12};\varepsilon)\biggr{]}$
$\displaystyle=\frac{16\pi}{315}\left(1120-1120v_{12}+360v_{12}^{2}-40v_{12}^{3}+v_{12}^{4}\right)$
$\displaystyle+\frac{32\pi}{99225}\left(518630-540680v_{12}+181890v_{12}^{2}-21260v_{12}^{3}+563v_{12}^{4}\right)\varepsilon$
$\displaystyle+\mathcal{O}\left(\varepsilon^{2}\right)\,,$ (569)
$\displaystyle I^{(0)}_{-4,1}(v_{12};\varepsilon)$
$\displaystyle=(1-v_{12})I^{(0)}_{-3}(\varepsilon)$
$\displaystyle+\frac{v_{12}}{2(2-\varepsilon)}\,\biggl{[}3I^{(0)}_{-2}(\varepsilon)+(7-2\varepsilon)I^{(0)}_{-3,1}(v_{12};\varepsilon)-3v_{12}I^{(0)}_{-2,1}(v_{12};\varepsilon)\biggr{]}$
$\displaystyle=-\frac{\pi}{\varepsilon}v_{12}^{4}+\frac{\pi}{6}\left(24+16v_{12}+12v_{12}^{2}+12v_{12}^{3}-25v_{12}^{4}\right)$
$\displaystyle+\frac{\pi}{36}\left(336+256v_{12}+228v_{12}^{2}+300v_{12}^{3}-415v_{12}^{4}\right)\varepsilon+\mathcal{O}\left(\varepsilon^{2}\right)\,,$
(570) $\displaystyle I^{(0)}_{-4,2}(v_{12};\varepsilon)$
$\displaystyle=(1-v_{12})I^{(0)}_{-3,1}(v_{12};\varepsilon)$
$\displaystyle+\frac{v_{12}}{3-2\varepsilon}\,\biggl{[}3I^{(0)}_{-2,1}(v_{12};\varepsilon)+(7-2\varepsilon)I^{(0)}_{-3,2}(v_{12};\varepsilon)-3v_{12}I^{(0)}_{-2,2}(v_{12};\varepsilon)\biggr{]}$
$\displaystyle=-\frac{2\pi}{\varepsilon}\left(8-5v_{12}\right)v_{12}^{3}+\frac{4\pi}{3}\left(2+4v_{12}+9v_{12}^{2}-44v_{12}^{3}+20v_{12}^{4}\right)$
$\displaystyle+\frac{2\pi}{9}\,\left(26+64v_{12}+198v_{12}^{2}-680v_{12}^{3}+305v_{12}^{4}\right)\,\varepsilon+\mathcal{O}\left(\varepsilon^{2}\right)\,,$
(571) $\displaystyle I^{(0)}_{-3,0}(v_{12};\varepsilon)$
$\displaystyle=I^{(0)}_{-3}(\varepsilon)\,,$ (572) $\displaystyle
I^{(0)}_{-3,-1}(v_{12};\varepsilon)$
$\displaystyle=\frac{1}{(5-2\varepsilon)}\,\biggl{[}(3-\varepsilon)I^{(0)}_{-3}(\varepsilon)+(5-\varepsilon)I^{(0)}_{-2,-1}(v_{12};\varepsilon)-2v_{12}I^{(0)}_{-2}(\varepsilon)\biggr{]}$
$\displaystyle=\frac{4\pi}{5}\left(8-3v_{12}\right)+\frac{4\pi}{75}\left(298-123v_{12}\right)\varepsilon+\mathcal{O}\left(\varepsilon^{2}\right)\,,$
(573)

$\displaystyle I^{(0)}_{-3,-2}(v_{12};\varepsilon)$
$\displaystyle=\frac{1}{4(3-\varepsilon)}\,\biggl{[}(9-2\varepsilon)I^{(0)}_{-3,-1}(v_{12};\varepsilon)+(11-2\varepsilon)I^{(0)}_{-2,-2}(v_{12};\varepsilon)$
$\displaystyle-5v_{12}I^{(0)}_{-2,-1}(v_{12};\varepsilon)\biggr{]}$
$\displaystyle=\frac{4\pi}{15}\left(40-24v_{12}+3v_{12}^{2}\right)+\frac{8\pi}{225}\left(785-507v_{12}+69v_{12}^{2}\right)\varepsilon+\mathcal{O}\left(\varepsilon^{2}\right)\,,$
(574) $\displaystyle I^{(0)}_{-3,-3}(v_{12};\varepsilon)$
$\displaystyle=\frac{1}{7-2\varepsilon}\,\biggl{[}(6-\varepsilon)\Big{(}I^{(0)}_{-3,-2}(v_{12};\varepsilon)+I^{(0)}_{-2,-3}(v_{12};\varepsilon)\Big{)}-3v_{12}I^{(0)}_{-2,-2}(v_{12};\varepsilon)\biggr{]}$
$\displaystyle=\frac{4\pi}{35}\left(160-120v_{12}+24v_{12}^{2}-v_{12}^{3}\right)$
$\displaystyle+\frac{8\pi}{3675}\left(22980-18285v_{12}+3909v_{12}^{2}-176v_{12}^{3}\right)\varepsilon+\mathcal{O}\left(\varepsilon^{2}\right)\,,$
(575) $\displaystyle I^{(0)}_{-3,1}(v_{12};\varepsilon)$
$\displaystyle=(1-v_{12})I^{(0)}_{-2}(\varepsilon)$
$\displaystyle+\frac{v_{12}}{(3-2\varepsilon)}\,\biggl{[}2I^{(0)}_{-1}(\varepsilon)+(5-2\varepsilon)I^{(0)}_{-2,1}(v_{12};\varepsilon)-2v_{12}I^{(0)}_{-1,1}(v_{12};\varepsilon)\biggr{]}$
$\displaystyle=-\frac{\pi}{\varepsilon}v_{12}^{3}+\frac{\pi}{3}\left(8+6v_{12}+6v_{12}^{2}-11v_{12}^{3}\right)$
$\displaystyle+\frac{\pi}{9}\left(52+48v_{12}+66v_{12}^{2}-85v_{12}^{3}\right)\varepsilon+\mathcal{O}\left(\varepsilon^{2}\right)\,,$
(576) $\displaystyle I^{(0)}_{-3,2}(v_{12};\varepsilon)$
$\displaystyle=(1-v_{12})I^{(0)}_{-2,1}(v_{12};\varepsilon)$
$\displaystyle+\frac{v_{12}}{(1-\varepsilon)}\,\biggl{[}I^{(0)}_{-1,1}(v_{12};\varepsilon)+(2-\varepsilon)I^{(0)}_{-2,2}(v_{12};\varepsilon)-v_{12}I^{(0)}_{-1,2}(v_{12};\varepsilon)\biggr{]}$
$\displaystyle=-\frac{3\pi}{\varepsilon}\left(3-2v_{12}\right)v_{12}^{2}+\pi\left(2+6v_{12}-27v_{12}^{2}+12v_{12}^{3}\right)$
$\displaystyle+\pi\left(4+18v_{12}-63v_{12}^{2}+30v_{12}^{3}\right)\varepsilon+\mathcal{O}\left(\varepsilon^{2}\right)\,,$
(577) $\displaystyle I^{(0)}_{-2,0}(v_{12};\varepsilon)$
$\displaystyle=I^{(0)}_{-2}(\varepsilon)\,,$ (578) $\displaystyle
I^{(0)}_{-2,-1}(v_{12};\varepsilon)$
$\displaystyle=\frac{1}{4(2-\varepsilon)}\,\biggl{[}(7-2\varepsilon)I^{(0)}_{-2}(\varepsilon)+(5-2\varepsilon)I^{(0)}_{-1,-1}(v_{12};\varepsilon)-3v_{12}I^{(0)}(\varepsilon)\biggr{]}$
$\displaystyle=\frac{4\pi}{3}\left(3-v_{12}\right)+\frac{4\pi}{9}\left(21-8v_{12}\right)\varepsilon+\mathcal{O}\left(\varepsilon^{2}\right)\,,$
(579) $\displaystyle I^{(0)}_{-2,-2}(v_{12};\varepsilon)$
$\displaystyle=\frac{1}{5-2\varepsilon}\,\biggl{[}(4-\varepsilon)\Big{(}I^{(0)}_{-2,-1}(v_{12};\varepsilon)+I^{(0)}_{-1,-2}(v_{12};\varepsilon)\Big{)}-2v_{12}I^{(0)}_{-1,-1}(v_{12};\varepsilon)\biggr{]}$
$\displaystyle=\frac{4\pi}{15}\left(24-12v_{12}+v_{12}^{2}\right)+\frac{8\pi}{225}\left(447-246v_{12}+23v_{12}^{2}\right)\varepsilon+\mathcal{O}\left(\varepsilon^{2}\right)\,,$
(580) $\displaystyle I^{(0)}_{-2,1}(v_{12};\varepsilon)$
$\displaystyle=(1-v_{12})I^{(0)}(\varepsilon)$
$\displaystyle+\frac{v_{12}}{2(1-\varepsilon)}\,\biggl{[}I^{(0)}(\varepsilon)+(3-2\varepsilon)I^{(0)}_{-1,1}(v_{12};\varepsilon)-v_{12}I^{(0)}_{1}(\varepsilon)\biggr{]}$
$\displaystyle=-\frac{\pi}{\varepsilon}v_{12}^{2}+\pi\left(2+2v_{12}-3v_{12}^{2}\right)+\pi\left(4+6v_{12}-7v_{12}^{2}\right)\varepsilon+\mathcal{O}\left(\varepsilon^{2}\right)\,,$
(581) $\displaystyle I^{(0)}_{-2,2}(v_{12};\varepsilon)$
$\displaystyle=(1-v_{12})I^{(0)}_{-1,1}(v_{12};\varepsilon)$
$\displaystyle+\frac{v_{12}}{(1-2\varepsilon)}\,\biggl{[}I^{(0)}_{1}(\varepsilon)+2(1-\varepsilon)I^{(0)}_{-1,2}(v_{12};\varepsilon)-v_{12}I^{(0)}_{2}(\varepsilon)\biggr{]}$
$\displaystyle=-\frac{\pi}{\varepsilon}v_{12}\left(4-3v_{12}\right)+\pi\left(2-8v_{12}+3v_{12}^{2}\right)+\pi\left(4-16v_{12}+9v_{12}^{2}\right)\varepsilon$
$\displaystyle+\mathcal{O}\left(\varepsilon^{2}\right)\,,$ (582)

$\displaystyle I^{(0)}_{-1,0}(v_{12};\varepsilon)$
$\displaystyle=I^{(0)}(\varepsilon)\,,$ (583) $\displaystyle
I^{(0)}_{-1,-1}(v_{12};\varepsilon)$
$\displaystyle=\frac{4-v_{12}-2\varepsilon}{3-2\varepsilon}\,I^{(0)}$
$\displaystyle=\frac{2\pi}{3}\left(4-v_{12}\right)+\frac{4\pi}{9}\left(13-4v_{12}\right)\varepsilon+\mathcal{O}\left(\varepsilon^{2}\right)\,,$
(584) $\displaystyle I^{(0)}_{-1,1}(v_{12};\varepsilon)$
$\displaystyle=(1-v_{12})I^{(0)}(\varepsilon)+v_{12}I^{(0)}_{1}(\varepsilon)$
$\displaystyle=-\frac{\pi}{\varepsilon}v_{12}+2\pi\left(1-v_{12}\right)+4\pi\left(1-v_{12}\right)\varepsilon+\mathcal{O}\left(\varepsilon^{2}\right)\,,$
(585) $\displaystyle I^{(0)}_{-1,2}(v_{12};\varepsilon)$
$\displaystyle=(1-v_{12})I^{(0)}_{1}(\varepsilon)+v_{12}I^{(0)}_{2}(\varepsilon)$
$\displaystyle=-\frac{\pi}{\varepsilon}\left(1-v_{12}\right)-\pi v_{12}+\pi
v_{12}\varepsilon+\mathcal{O}\left(\varepsilon^{2}\right)\,,$ (586)
$\displaystyle I^{(0)}_{1,1}(v_{12};\varepsilon)$
$\displaystyle=\frac{2\pi}{v_{12}}\,\biggl{[}-\frac{1}{\varepsilon}+\log\left(\frac{v_{12}}{2}\right)-\varepsilon\biggl{(}\frac{1}{2}\log^{2}\left(\frac{v_{12}}{2}\right)+\mathrm{Li}_{2}\left(1-\frac{v_{12}}{2}\right)\biggr{)}\biggr{]}$
$\displaystyle+\mathcal{O}\left(\varepsilon^{2}\right)\,,$ (587)
$\displaystyle I^{(0)}_{1,2}(v_{12};\varepsilon)$
$\displaystyle=\frac{1}{v_{12}}\,\biggl{[}(1+2\varepsilon)I^{(0)}_{1,1}(v_{12};\varepsilon)-2\varepsilon
I^{(0)}_{1}(\varepsilon)+I^{(0)}_{2}(\varepsilon)\biggr{]}$
$\displaystyle=\frac{\pi}{v_{12}^{2}}\,\biggl{[}-\frac{2}{\varepsilon}-4+v_{12}+2\log\left(\frac{v_{12}}{2}\right)$
$\displaystyle+\varepsilon\biggl{(}v_{12}+4\log\left(\frac{v_{12}}{2}\right)-\log^{2}\left(\frac{v_{12}}{2}\right)-2\mathrm{Li}_{2}\left(1-\frac{v_{12}}{2}\right)\biggr{)}\biggr{]}+\mathcal{O}\left(\varepsilon^{2}\right)\,,$
(588) $\displaystyle I^{(0)}_{2,2}(v_{12};\varepsilon)$
$\displaystyle=\frac{1}{v_{12}^{2}}\,\biggl{[}(1+2\varepsilon)(1-v_{12})I^{(0)}_{1,1}(v_{12};\varepsilon)-(1+2\varepsilon)I^{(0)}_{2}(\varepsilon)+(3+2\varepsilon)v_{12}I^{(0)}_{1,2}(v_{12};\varepsilon)\biggr{]}$
$\displaystyle=-\frac{2\pi}{v_{12}^{3}}\,\biggl{[}\frac{4-v_{12}}{\varepsilon}+10-4v_{12}-(4-v_{12})\log\left(\frac{v_{12}}{2}\right)$
$\displaystyle+\varepsilon\biggl{(}4-3v_{12}-2\left(5-v_{12}\right)\log\left(\frac{v_{12}}{2}\right)+\left(2-\frac{v_{12}}{2}\right)\log^{2}\left(\frac{v_{12}}{2}\right)$
$\displaystyle+\left(4-v_{12}\right)\mathrm{Li}_{2}\left(1-\frac{v_{12}}{2}\right)\biggr{)}\biggr{]}+\mathcal{O}\left(\varepsilon^{2}\right)\,.$
(589)

###  G.4 Single massive integral with two denominators

$\displaystyle I^{(1)}_{-4,-1}(v_{12},v_{11};\varepsilon)$
$\displaystyle=\frac{1}{4(3-\varepsilon)}\,\biggl{[}(7-2\varepsilon)I^{(1)}_{-4}(v_{11};\varepsilon)+(13-2\varepsilon)I^{(1)}_{-3,-1}(v_{12},v_{11};\varepsilon)$
$\displaystyle-3v_{11}I^{(1)}_{-2,-1}(v_{12},v_{11};\varepsilon)-5v_{12}I^{(1)}_{-3}(v_{11};\varepsilon)\biggr{]}$
$\displaystyle=\frac{2\pi}{15}\Big{(}80-32v_{12}-48v_{11}+3v_{11}^{2}+12v_{12}v_{11}\Big{)}$
$\displaystyle+\frac{4\pi}{225}\Big{(}1570-676v_{12}-1014v_{11}+69v_{11}^{2}+276v_{12}v_{11}\Big{)}\varepsilon$
$\displaystyle+\mathcal{O}\left(\varepsilon^{2}\right)\,,$ (590)

$\displaystyle I^{(1)}_{-4,-2}(v_{12},v_{11};\varepsilon)$
$\displaystyle=\frac{1}{2(7-2\varepsilon)}\,\biggl{[}2(5-\varepsilon)I^{(1)}_{-4,-1}(v_{12},v_{11};\varepsilon)+2(7-\varepsilon)I^{(1)}_{-3,-2}(v_{12},v_{11};\varepsilon)$
$\displaystyle-3v_{11}I^{(1)}_{-2,-2}(v_{12},v_{11};\varepsilon)-6v_{12}I^{(1)}_{-3,-1}(v_{12},v_{11};\varepsilon)\biggr{]}$
$\displaystyle=\frac{8\pi}{105}\Big{(}240-160v_{12}-120v_{11}+6v_{11}^{2}+48v_{12}v_{11}+24v_{12}^{2}-3v_{12}^{2}v_{11}\Big{)}$
$\displaystyle+\frac{4\pi}{11025}\Big{(}137880-97520v_{12}-73140v_{11}+3909v_{11}^{2}+31272v_{12}v_{11}$
$\displaystyle+15636v_{12}^{2}-2112v_{12}^{2}v_{11}\Big{)}\varepsilon+\mathcal{O}\left(\varepsilon^{2}\right)\,,$
(591) $\displaystyle I^{(1)}_{-4,-3}(v_{12},v_{11};\varepsilon)$
$\displaystyle=\frac{1}{4(4-\varepsilon)}\,\biggl{[}(13-2\varepsilon)I^{(1)}_{-4,-2}(v_{12},v_{11};\varepsilon)+(15-2\varepsilon)I^{(1)}_{-3,-3}(v_{12},v_{11};\varepsilon)$
$\displaystyle-3v_{11}I^{(1)}_{-2,-3}(v_{12},v_{11};\varepsilon)-7v_{12}I^{(1)}_{-3,-2}(v_{12},v_{11};\varepsilon)\biggr{]}$
$\displaystyle=\frac{4\pi}{35}\Big{(}280-240v_{12}-120v_{11}+5v_{11}^{2}+60v_{11}v_{12}+60v_{12}^{2}$
$\displaystyle-4v_{12}^{3}-6v_{11}v_{12}^{2}\Big{)}+\frac{4\pi}{3675}\Big{(}83580-75240v_{12}-37620v_{11}+1655v_{11}^{2}$
$\displaystyle+19860v_{11}v_{12}+19860v_{12}^{2}-1408v_{12}^{3}-2112v_{11}v_{12}^{2}\Big{)}\varepsilon+\mathcal{O}\left(\varepsilon^{2}\right)\,,$
(592) $\displaystyle I^{(1)}_{-4,-4}(v_{12},v_{11};\varepsilon)$
$\displaystyle=\frac{1}{2(9-2\varepsilon)}\,\biggl{[}2(8-\varepsilon)I^{(1)}_{-4,-3}(v_{12},v_{11};\varepsilon)+2(8-\varepsilon)I^{(1)}_{-3,-4}(v_{12},v_{11};\varepsilon)$
$\displaystyle-3v_{11}I^{(1)}_{-2,-4}(v_{12},v_{11};\varepsilon)-8v_{12}I^{(1)}_{-3,-3}(v_{12},v_{11};\varepsilon)\biggr{]}$
$\displaystyle=\frac{16\pi}{315}\Big{(}1120-1120v_{12}-420v_{11}+15v_{11}^{2}+240v_{11}v_{12}+360v_{12}^{2}$
$\displaystyle-30v_{11}v_{12}^{2}-40v_{12}^{3}+v_{12}^{4}\Big{)}+\frac{8\pi}{99225}\Big{(}2074520-2162720v_{12}$
$\displaystyle-811020v_{11}+30315v_{11}^{2}+727560v_{12}^{2}+485040v_{11}v_{12}-63780v_{11}v_{12}^{2}$
$\displaystyle-85040v_{12}^{3}+2252v_{12}^{4}\Big{)}\varepsilon+\mathcal{O}\left(\varepsilon^{2}\right)\,,$
(593) $\displaystyle I^{(1)}_{-4,1}(v_{12},v_{11};\varepsilon)$
$\displaystyle=(1-v_{12})I^{(1)}_{-3}(v_{11};\varepsilon)+\frac{1}{2(2-\varepsilon)}\,\biggl{[}3(v_{12}-v_{11})I^{(1)}_{-2}(v_{11};\varepsilon)$
$\displaystyle-3v_{12}^{2}I^{(1)}_{-2,1}(v_{11},v_{12};\varepsilon)+(7-2\varepsilon)I^{(1)}_{-3,1}(v_{12},v_{11};\varepsilon)\biggr{]}$
$\displaystyle=-\frac{\pi}{\varepsilon}v_{12}^{4}+\frac{\pi}{6}\Big{(}24+16v_{12}-24v_{11}+3v_{11}^{2}-12v_{11}v_{12}+12v_{12}^{2}$
$\displaystyle-6v_{11}v_{12}^{2}+12v_{12}^{3}-25v_{12}^{4}\Big{)}+\frac{\pi}{36}\Big{(}336+256v_{12}-384v_{11}+57v_{11}^{2}$
$\displaystyle-228v_{11}v_{12}+228v_{12}^{2}-150v_{11}v_{12}^{2}+300v_{12}^{3}-415v_{12}^{4}\Big{)}\varepsilon$
$\displaystyle+\mathcal{O}\left(\varepsilon^{2}\right)\,,$ (594)
$\displaystyle I^{(1)}_{-4,2}(v_{12},v_{11};\varepsilon)$
$\displaystyle=(1-v_{12})I^{(1)}_{-3,1}(v_{12},v_{11};\varepsilon)+\frac{1}{(3-2\varepsilon)}\,\biggl{[}3(v_{12}-v_{11})I^{(1)}_{-2,1}(v_{11},v_{12};\varepsilon)$
$\displaystyle-3v_{12}^{2}I^{(1)}_{-2,2}(v_{11},v_{12};\varepsilon)+2(3-\varepsilon)I^{(1)}_{-3,2}(v_{12},v_{11};\varepsilon)\biggr{]}$
$\displaystyle=\frac{2\pi}{\varepsilon}v_{12}^{2}\Big{(}3v_{11}-8v_{12}+5v_{12}^{2}\Big{)}+\frac{\pi}{3}\Big{(}8+16v_{12}-12v_{11}+3v_{11}^{2}-24v_{11}v_{12}$
$\displaystyle+36v_{12}^{2}+66v_{11}v_{12}^{2}-176v_{12}^{3}+80v_{12}^{4}\Big{)}+\frac{\pi}{9}\Big{(}52+128v_{12}-96v_{11}$
$\displaystyle+33v_{11}^{2}-264v_{11}v_{12}+396v_{12}^{2}+510v_{11}v_{12}^{2}-1360v_{12}^{3}+610v_{12}^{4}\Big{)}\varepsilon$
$\displaystyle+\mathcal{O}\left(\varepsilon^{2}\right)\,,$ (595)
$\displaystyle I^{(1)}_{-3,-1}(v_{12},v_{11};\varepsilon)$
$\displaystyle=\frac{1}{5-2\varepsilon}\,\biggl{[}(3-\varepsilon)I^{(1)}_{-3}(v_{11};\varepsilon)+(5-\varepsilon)I^{(1)}_{-2,-1}(v_{11},v_{12};\varepsilon)$
$\displaystyle-
v_{11}I^{(1)}_{-1,-1}(v_{11},v_{12};\varepsilon)-2v_{12}I^{(1)}_{-2}(v_{11};\varepsilon)\biggr{]}$
$\displaystyle=\frac{2\pi}{5}\Big{(}16-6v_{12}-6v_{11}+v_{12}v_{11}\Big{)}$
$\displaystyle+\frac{4\pi}{75}\Big{(}298-123v_{12}-123v_{11}+23v_{11}v_{12}\Big{)}\varepsilon$
$\displaystyle+\mathcal{O}\left(\varepsilon^{2}\right)\,,$ (596)
$\displaystyle I^{(1)}_{-3,-2}(v_{12},v_{11};\varepsilon)$
$\displaystyle=\frac{1}{4(3-\varepsilon)}\,\biggl{[}(9-2\varepsilon)I^{(1)}_{-3,-1}(v_{11},v_{12};\varepsilon)+(11-2\varepsilon)I^{(1)}_{-2,-2}(v_{11},v_{12};\varepsilon)$
$\displaystyle-2v_{11}I^{(1)}_{-1,-2}(v_{11},v_{12};\varepsilon)-5v_{12}I^{(1)}_{-2,-1}(v_{11},v_{12};\varepsilon)\biggr{]}$
$\displaystyle=\frac{4\pi}{15}\Big{(}40-24v_{12}-12v_{11}+3v_{12}^{2}+3v_{11}v_{12}\Big{)}$
$\displaystyle+\frac{4\pi}{225}\Big{(}1570-1014v_{12}-507v_{11}+138v_{12}v_{11}+138v_{12}^{2}\Big{)}\varepsilon$
$\displaystyle+\mathcal{O}\left(\varepsilon^{2}\right)\,,$ (597)
$\displaystyle I^{(1)}_{-3,-3}(v_{12},v_{11};\varepsilon)$
$\displaystyle=\frac{1}{7-2\varepsilon}\,\biggl{[}(6-\varepsilon)\Big{(}I^{(1)}_{-3,-2}(v_{12},v_{11};\varepsilon)+I^{(1)}_{-2,-3}(v_{12},v_{11};\varepsilon)\Big{)}$
$\displaystyle-
v_{11}I^{(1)}_{-1,-3}(v_{12},v_{11};\varepsilon)-3v_{12}I^{(1)}_{-2,-2}(v_{12},v_{11};\varepsilon)\biggr{]}$
$\displaystyle=\frac{4\pi}{35}\Big{(}160-120v_{12}-40v_{11}+12v_{11}v_{12}+24v_{12}^{2}-v_{12}^{3}\Big{)}$
$\displaystyle+\frac{4\pi}{3675}\Big{(}45960-36570v_{12}-12190v_{11}+3909v_{11}v_{12}$
$\displaystyle+7818v_{12}^{2}-352v_{12}^{3}\Big{)}\varepsilon+\mathcal{O}\left(\varepsilon^{2}\right)\,,$
(598)

$\displaystyle I^{(1)}_{-3,-4}(v_{12},v_{11};\varepsilon)$
$\displaystyle=\frac{1}{4(4-\varepsilon)}\,\biggl{[}(15-2\varepsilon)I^{(1)}_{-3,-3}(v_{12},v_{11};\varepsilon)+(13-2\varepsilon)I^{(1)}_{-2,-4}(v_{12},v_{11};\varepsilon)$
$\displaystyle-2v_{11}I^{(1)}_{-1,-4}(v_{12},v_{11};\varepsilon)-7v_{12}I^{(1)}_{-2,-3}(v_{12},v_{11};\varepsilon)\biggr{]}$
$\displaystyle=\frac{16\pi}{35}\Big{(}70-60v_{12}-15v_{11}+5v_{11}v_{12}+15v_{12}^{2}-v_{12}^{3}\Big{)}$
$\displaystyle+\frac{8\pi}{3675}\Big{(}41790-37620v_{12}-9405v_{11}+3310v_{11}v_{12}+9930v_{12}^{2}$
$\displaystyle-704v_{12}^{3}\Big{)}\varepsilon+\mathcal{O}\left(\varepsilon^{2}\right)\,,$
(599) $\displaystyle I^{(1)}_{-3,1}(v_{12},v_{11};\varepsilon)$
$\displaystyle=(1-v_{12})I^{(1)}_{-2}(v_{11};\varepsilon)+\frac{1}{3-2\varepsilon}\,\biggl{[}2(v_{12}-v_{11})I^{(0)}(\varepsilon)$
$\displaystyle-2v_{12}^{2}I^{(1)}_{-1,-1}(v_{11},v_{12};\varepsilon)+(5-2\varepsilon)v_{12}I^{(1)}_{-2,1}(v_{11},v_{12};\varepsilon)\biggr{]}$
$\displaystyle=-\frac{\pi}{\varepsilon}v_{12}^{3}+\frac{\pi}{3}\Big{(}8+6v_{12}-6v_{11}-3v_{11}v_{12}+6v_{12}^{2}-11v_{12}^{3}\Big{)}$
$\displaystyle+\frac{\pi}{9}\Big{(}52+48v_{12}-48v_{11}-33v_{11}v_{12}+66v_{12}^{2}-85v_{12}^{3}\Big{)}\varepsilon$
$\displaystyle+\mathcal{O}\left(\varepsilon^{2}\right)\,,$ (600)
$\displaystyle I^{(1)}_{-3,2}(v_{12},v_{11};\varepsilon)$
$\displaystyle=(1-v_{12})I^{(1)}_{-2,1}(v_{11},v_{12};\varepsilon)+\frac{1}{1-\varepsilon}\,\biggl{[}(v_{12}-v_{11})I^{(1)}_{-1,1}(v_{11},v_{12};\varepsilon)$
$\displaystyle-
v_{12}^{2}I^{(1)}_{-1,2}(v_{11},v_{12};\varepsilon)+(2-\varepsilon)v_{12}I^{(1)}_{-2,2}(v_{11},v_{12};\varepsilon)\biggr{]}$
$\displaystyle=\frac{3\pi}{\varepsilon}v_{12}\Big{(}v_{11}-3v_{12}+2v_{12}^{2}\Big{)}$
$\displaystyle+\pi\Big{(}2+6v_{12}-3v_{11}+9v_{11}v_{12}-27v_{12}^{2}+12v_{12}^{3}\Big{)}$
$\displaystyle+\pi\Big{(}4+18v_{12}-9v_{11}+21v_{11}v_{12}-63v_{12}^{2}+30v_{12}^{3}\Big{)}\varepsilon+\mathcal{O}\left(\varepsilon^{2}\right)\,,$
(601) $\displaystyle I^{(1)}_{-2,-1}(v_{12},v_{11};\varepsilon)$
$\displaystyle=\frac{1}{4(2-\varepsilon)}\,\biggl{[}(5-2\varepsilon)I^{(1)}_{-2}(v_{11};\varepsilon)+(7-2\varepsilon)I^{(1)}_{-1,-1}(v_{12},v_{11};\varepsilon)$
$\displaystyle-(v_{11}+3v_{12})I^{(0)}(\varepsilon)\biggr{]}$
$\displaystyle=\frac{2\pi}{3}\Big{(}6-2v_{12}-v_{11}\Big{)}+\frac{4\pi}{9}\pi\Big{(}21-8v_{12}-4v_{11}\Big{)}\varepsilon+\mathcal{O}\left(\varepsilon^{2}\right)\,,$
(602) $\displaystyle I^{(1)}_{-2,-2}(v_{12},v_{11};\varepsilon)$
$\displaystyle=\frac{1}{2(5-2\varepsilon)}\,\biggl{[}2(4-\varepsilon)(I^{(1)}_{-2,-1}(v_{12},v_{11};\varepsilon)+I^{(1)}_{-1,-2}(v_{12},v_{11};\varepsilon))$
$\displaystyle-
v_{11}I^{(0)}_{-2}(\varepsilon)-4v_{12}I^{(1)}_{-1,-1}(v_{12},v_{11};\varepsilon)\biggr{]}$
$\displaystyle=\frac{4\pi}{15}\Big{(}24-12v_{12}-3v_{11}+v_{12}^{2}\Big{)}$
$\displaystyle+\frac{4\pi}{225}\Big{(}894-492v_{12}-123v_{11}+46v_{12}^{2}\Big{)}\varepsilon+\mathcal{O}\left(\varepsilon^{2}\right)\,,$
(603)

$\displaystyle I^{(1)}_{-2,-3}(v_{12},v_{11};\varepsilon)$
$\displaystyle=\frac{1}{4(3-\varepsilon)}\,\biggl{[}(11-2\varepsilon)I^{(1)}_{-2,-2}(v_{12},v_{11};\varepsilon)+(9-2\varepsilon)I^{(1)}_{-1,-3}(v_{12},v_{11};\varepsilon))$
$\displaystyle-
v_{11}I^{(0)}_{-3}(\varepsilon)-5v_{12}I^{(1)}_{-1,-2}(v_{12},v_{11};\varepsilon)\biggr{]}$
$\displaystyle=\frac{4\pi}{15}\Big{(}40-24v_{12}-4v_{11}+3v_{12}^{2}\Big{)}$
$\displaystyle+\frac{4\pi}{225}\Big{(}1570-1014v_{12}-169v_{11}+138v_{12}^{2}\Big{)}\varepsilon+\mathcal{O}\left(\varepsilon^{2}\right)\,,$
(604) $\displaystyle I^{(1)}_{-2,-4}(v_{12},v_{11};\varepsilon)$
$\displaystyle=\frac{1}{2(7-2\varepsilon)}\,\biggl{[}2(7-\varepsilon)I^{(1)}_{-2,-3}(v_{12},v_{11};\varepsilon)+2(5-\varepsilon)I^{(1)}_{-1,-4}(v_{12},v_{11};\varepsilon))$
$\displaystyle-
v_{11}I^{(0)}_{-4}(\varepsilon)-6v_{12}I^{(1)}_{-1,-3}(v_{12},v_{11};\varepsilon)\biggr{]}$
$\displaystyle=\frac{32\pi}{105}\Big{(}60-40v_{12}-5v_{11}+6v_{12}^{2}\Big{)}$
$\displaystyle+\frac{8\pi}{11025}\Big{(}68940-48760v_{12}-6095v_{11}+7818v_{12}^{2}\Big{)}\varepsilon+\mathcal{O}\left(\varepsilon^{2}\right)\,,$
(605) $\displaystyle I^{(1)}_{-2,1}(v_{12},v_{11};\varepsilon)$
$\displaystyle=(1-v_{12})I^{(0)}(\varepsilon)+\frac{1}{2(1-\varepsilon)}\,\biggl{[}(v_{12}-v_{11})I^{(0)}(\varepsilon)$
$\displaystyle-
v_{12}^{2}I^{(0)}_{1}(\varepsilon)+(3-2\varepsilon)v_{12}I^{(1)}_{-1,1}(v_{11},v_{12};\varepsilon)\biggr{]}$
$\displaystyle=-\frac{\pi}{\varepsilon}v_{12}^{2}+\pi\Big{(}2-v_{11}+2v_{12}-3v_{12}^{2}\Big{)}$
$\displaystyle+\pi\Big{(}4-3v_{11}+6v_{12}-7v_{12}^{2}\Big{)}\varepsilon+\mathcal{O}\left(\varepsilon^{2}\right)\,,$
(606) $\displaystyle I^{(1)}_{-2,2}(v_{12},v_{11};\varepsilon)$
$\displaystyle=(1-v_{12})I^{(1)}_{-1,1}(v_{11},v_{12};\varepsilon)+\frac{1}{2(1-\varepsilon)}\,\biggl{[}(v_{12}-v_{11})I^{(0)}_{1}(\varepsilon)$
$\displaystyle-
v_{12}^{2}I^{(0)}_{2}(\varepsilon)+(3-2\varepsilon)v_{12}I^{(1)}_{-1,2}(v_{11},v_{12};\varepsilon)\biggr{]}$
$\displaystyle=\frac{\pi\left(v_{11}-4v_{12}+3v_{12}^{2}\right)}{\varepsilon}+\pi\left(2+2v_{11}-8v_{12}+3v_{12}^{2}\right)$
$\displaystyle+\pi\left(4+4v_{11}-16v_{12}+9v_{12}^{2}\right)\varepsilon+\mathcal{O}\left(\varepsilon^{2}\right)\,.$
(607) $\displaystyle I^{(1)}_{-1,-1}(v_{12},v_{11};\varepsilon)$
$\displaystyle=\frac{4-v_{12}-2\varepsilon}{3-2\varepsilon}\,I^{(0)}(\varepsilon)$
$\displaystyle=\frac{2}{3}\pi\left(4-v_{12}\right)+\frac{4}{9}\pi\left(13-4v_{12}\right)\varepsilon+\mathcal{O}\left(\varepsilon^{2}\right)\,,$
(608) $\displaystyle I^{(1)}_{-1,-2}(v_{12},v_{11};\varepsilon)$
$\displaystyle=\frac{1}{4(2-\varepsilon)}\,\biggl{[}(7-2\varepsilon)I^{(1)}_{-1,-1}(v_{12},v_{11};\varepsilon)+(5-2\varepsilon)I^{(0)}_{-2}(\varepsilon)-3v_{12}I^{(0)}(\varepsilon)\biggr{]}$
$\displaystyle=\frac{4\pi}{3}\left(3-v_{12}\right)+\frac{4\pi}{9}\left(21-8v_{12}\right)\varepsilon+\mathcal{O}\left(\varepsilon^{2}\right)\,,$
(609)

$\displaystyle I^{(1)}_{-1,-3}(v_{12},v_{11};\varepsilon)$
$\displaystyle=\frac{1}{5-2\varepsilon}\,\biggl{[}(5-\varepsilon)I^{(1)}_{-1,-2}(v_{12},v_{11};\varepsilon)+(3-\varepsilon)I^{(0)}_{-3}(\varepsilon)-2v_{12}I^{(0)}_{-2}(\varepsilon)\biggr{]}$
$\displaystyle=\frac{4\pi}{5}\left(8-3v_{12}\right)+\frac{4\pi}{75}\left(298-123v_{12}\right)\varepsilon+\mathcal{O}\left(\varepsilon^{2}\right)\,,$
(610) $\displaystyle I^{(1)}_{-1,-4}(v_{12},v_{11};\varepsilon)$
$\displaystyle=\frac{1}{4(3-\varepsilon)}\,\biggl{[}(13-2\varepsilon)I^{(1)}_{-1,-3}(v_{12},v_{11};\varepsilon)+(7-2\varepsilon)I^{(0)}_{-4}(\varepsilon)-5v_{12}I^{(0)}_{-3}(\varepsilon)\biggr{]}$
$\displaystyle=\frac{32\pi}{15}\left(5-2v_{12}\right)+\frac{8\pi}{225}\left(785-338v_{12}\right)\varepsilon+\mathcal{O}\left(\varepsilon^{2}\right)\,,$
(611) $\displaystyle I^{(1)}_{-1,1}(v_{12},v_{11};\varepsilon)$
$\displaystyle=(1-v_{12})I^{(0)}(\varepsilon)+v_{12}I^{(0)}_{1}(\varepsilon)$
$\displaystyle=-\frac{\pi}{\varepsilon}v_{12}+2\pi\left(1-v_{12}\right)(1+2\varepsilon)+\mathcal{O}\left(\varepsilon^{2}\right)\,,$
(612) $\displaystyle I^{(1)}_{-1,2}(v_{12},v_{11};\varepsilon)$
$\displaystyle=(1-v_{12})I^{(0)}_{1}(\varepsilon)+v_{12}I^{(0)}_{2}(\varepsilon)\,$
$\displaystyle=-\frac{\pi\left(1-v_{12}\right)}{\varepsilon}-\pi
v_{12}\left(1-\varepsilon\right)+\mathcal{O}\left(\varepsilon^{2}\right)\,,$
(613) $\displaystyle I^{(1)}_{1,1}(v_{12},v_{11};\varepsilon)$
$\displaystyle=\frac{\pi}{v_{12}}\biggl{[}-\frac{1}{\varepsilon}-\log\frac{v_{11}}{v_{12}^{2}}-\frac{\varepsilon}{2}\,L(v_{11},v_{12})+\mathcal{O}(\varepsilon^{2})\biggr{]}\,,$
(614) $\displaystyle I^{(1)}_{1,2}(v_{12},v_{11};\varepsilon)$
$\displaystyle=\frac{\pi}{v_{12}^{3}}\biggl{[}\frac{v_{11}-v_{12}}{\varepsilon}+2v_{11}-4v_{12}+v_{12}^{2}+(v_{11}-v_{12})\log\frac{v_{11}}{v_{12}^{2}}$
$\displaystyle+\frac{\varepsilon}{2}\biggl{(}2v_{12}^{2}+4(v_{11}-v_{12})\log\frac{v_{11}}{v_{12}^{2}}-4v_{12}\sqrt{1-v_{11}}\log\biggl{(}\frac{1+\sqrt{1-v_{11}}}{1-\sqrt{1-v_{11}}}\biggr{)}$
$\displaystyle+(v_{11}-v_{12})L(v_{11},v_{12})\biggr{)}\biggr{]}+\mathcal{O}(\varepsilon^{2})\,,$
(615) $\displaystyle I^{(1)}_{2,1}(v_{12},v_{11};\varepsilon)$
$\displaystyle=\frac{\pi}{v_{12}^{2}v_{11}}\biggl{[}-\frac{v_{11}}{\varepsilon}+2v_{12}-2v_{11}-v_{11}\log\left(\frac{v_{11}}{v_{12}^{2}}\right)$
$\displaystyle-\varepsilon\biggl{(}2v_{11}\log\frac{v_{11}}{v_{12}^{2}}-2v_{12}\sqrt{1-v_{11}}\log\left(\frac{1+\sqrt{1-v_{11}}}{1-\sqrt{1-v_{11}}}\right)$
$\displaystyle+\frac{v_{11}}{2}\,L(v_{11},v_{12})\biggr{)}\biggr{]}+\mathcal{O}(\varepsilon^{2})\,,$
(616) $\displaystyle I^{(1)}_{2,2}(v_{12},v_{11};\varepsilon)$
$\displaystyle=\frac{\pi}{v_{12}^{4}v_{11}}\biggl{[}\frac{1}{\varepsilon}\left(3v_{11}^{2}-4v_{11}v_{12}+v_{11}v_{12}^{2}\right)+8v_{11}^{2}+5v_{11}v_{12}^{2}-16v_{11}v_{12}+2v_{12}^{2}$
$\displaystyle+(3v_{11}^{2}-4v_{11}v_{12}+v_{11}v_{12}^{2})\log\frac{v_{11}}{v_{12}^{2}}+\varepsilon\biggl{(}4v_{11}^{2}-8v_{11}v_{12}+5v_{11}v_{12}^{2}$
$\displaystyle+(8v_{11}^{2}-10v_{11}v_{12}+2v_{11}v_{12}^{2})\log\frac{v_{11}}{v_{12}^{2}}$
$\displaystyle-(6v_{11}v_{12}-2v_{12}^{2})\sqrt{1-v_{11}}\log\left(\frac{1+\sqrt{1-v_{11}}}{1-\sqrt{1-v_{11}}}\right)$
$\displaystyle+\frac{1}{2}(3v_{11}^{2}-4v_{11}v_{12}+v_{11}v_{12}^{2})\,L(v_{11},v_{12})\biggr{)}\biggr{]}+\mathcal{O}(\varepsilon^{2})\,,$
(617)

$\displaystyle I^{(1)}_{1,3}(v_{12},v_{11};\varepsilon)$
$\displaystyle=\frac{\pi}{2v_{12}^{5}}\biggl{[}\frac{1}{\varepsilon}\left(-3v_{11}^{2}+6v_{11}v_{12}-v_{11}v_{12}^{2}-2v_{12}^{2}\right)-8v_{11}^{2}+22v_{11}v_{12}-5v_{11}v_{12}^{2}$
$\displaystyle-12v_{12}^{2}+2v_{12}^{3}+\frac{v_{12}^{4}}{2}-(3v_{11}^{2}-6v_{11}v_{12}+v_{11}v_{12}^{2}+2v_{12}^{2})\log\frac{v_{11}}{v_{12}^{2}}$
$\displaystyle+\varepsilon\biggl{(}-4v_{11}^{2}+12v_{11}v_{12}\vphantom{\frac{v_{11}}{v_{12}^{2}}}-5v_{11}v_{12}^{2}-8v_{12}^{2}+4v_{12}^{3}+\frac{3v_{12}^{4}}{4}$
$\displaystyle-(8v_{11}^{2}+6v_{12}^{2}-16v_{11}v_{12}+2v_{11}v_{12}^{2})\log\frac{v_{11}}{v_{12}^{2}}$
$\displaystyle+6v_{12}(v_{11}-v_{12})\sqrt{1-v_{11}}\log\left(\frac{1+\sqrt{1-v_{11}}}{1-\sqrt{1-v_{11}}}\right)$
$\displaystyle-\frac{1}{2}\,(3v_{11}^{2}-6v_{11}v_{12}+v_{11}v_{12}^{2}+2v_{12}^{2})\,L(v_{11},v_{12})\biggr{)}\biggr{]}+\mathcal{O}(\varepsilon^{2})\,,$
(618) $\displaystyle I^{(1)}_{3,1}(v_{12},v_{11};\varepsilon)$
$\displaystyle=\frac{\pi}{v_{12}^{3}v_{11}^{2}}\bigg{[}-\frac{v_{11}^{2}}{\varepsilon}-3v_{11}^{2}+2v_{11}v_{12}+2v_{12}^{2}-v_{11}v_{12}^{2}-v_{11}^{2}\log\frac{v_{11}}{v_{12}^{2}}$
$\displaystyle+\frac{\varepsilon}{2}\biggl{(}4v_{11}v_{12}+4v_{12}^{2}-4v_{11}^{2}-4v_{11}v_{12}^{2}-6v_{11}^{2}\log\frac{v_{11}}{v_{12}^{2}}+4v_{12}(v_{11}+v_{12})$
$\displaystyle\times\sqrt{1-v_{11}}\log\left(\frac{1+\sqrt{1-v_{11}}}{1-\sqrt{1-v_{11}}}\right)-v_{11}^{2}\,L(v_{11},v_{12})\biggr{)}\biggr{]}+\mathcal{O}(\varepsilon^{2})\,,$
(619) $\displaystyle I^{(1)}_{2,3}(v_{12},v_{11};\varepsilon)$
$\displaystyle=\frac{\pi}{8v_{12}^{6}v_{11}}\biggl{[}\frac{12}{\varepsilon}\,v_{11}\left(-5v_{11}^{2}+12v_{11}v_{12}-6v_{12}^{2}-3v_{11}v_{12}^{2}+2v_{12}^{3}\right)$
$\displaystyle-184v_{11}^{3}+552v_{11}^{2}v_{12}-384v_{11}v_{12}^{2}-156v_{11}^{2}v_{12}^{2}+16v_{12}^{3}+160v_{11}v_{12}^{3}$
$\displaystyle-6v_{11}v_{12}^{4}-12v_{11}\left(5v_{11}^{2}-12v_{11}v_{12}+6v_{12}^{2}+3v_{11}v_{12}^{2}-2v_{12}^{3}\right)\log\frac{v_{11}}{v_{12}^{2}}$
$\displaystyle+\varepsilon\biggl{(}-144v_{11}^{3}+448v_{11}^{2}v_{12}-320v_{11}v_{12}^{2}-172v_{11}^{2}v_{12}^{2}+200v_{11}v_{12}^{3}-13v_{11}v_{12}^{4}$
$\displaystyle+8v_{11}\left(-23v_{11}^{2}+54v_{11}v_{12}-27v_{12}^{2}-12v_{11}v_{12}^{2}+8v_{12}^{3}\right)\log\frac{v_{11}}{v_{12}^{2}}$
$\displaystyle+8v_{12}\left(15v_{11}^{2}-21v_{11}v_{12}+2v_{12}^{2}+4v_{11}v_{12}^{2}\right)\sqrt{1-v_{11}}\log\left(\frac{1+\sqrt{1-v_{11}}}{1-\sqrt{1-v_{11}}}\right)$
$\displaystyle-6v_{11}\left(5v_{11}^{2}-12v_{11}v_{12}+6v_{12}^{2}+3v_{11}v_{12}^{2}-2v_{12}^{3}\right)\,L(v_{11},v_{12})\biggr{)}\biggr{]}$
$\displaystyle+\mathcal{O}(\varepsilon^{2})\,,$ (620) $\displaystyle
I^{(1)}_{3,2}(v_{12},v_{11};\varepsilon)$
$\displaystyle=\frac{\pi}{v_{12}^{5}v_{11}^{2}}\biggl{[}\frac{3}{\varepsilon}\,v_{11}^{2}\left(2v_{11}-3v_{12}+v_{12}^{2}\right)+19v_{11}^{3}+6v_{11}v_{12}^{2}-39v_{11}^{2}v_{12}$
$\displaystyle+2v_{12}^{3}-3v_{11}v_{12}^{3}+14v_{11}^{2}v_{12}^{2}+3v_{11}^{2}\left(2v_{11}-3v_{12}+v_{12}^{2}\right)\log\frac{v_{11}}{v_{12}^{2}}$
$\displaystyle+\frac{\varepsilon}{2}\biggl{(}32v_{11}^{3}+4v_{12}^{3}+8v_{11}v_{12}^{2}-68v_{11}^{2}v_{12}-8v_{11}v_{12}^{3}+34v_{11}^{2}v_{12}^{2}$
$\displaystyle+2v_{11}^{2}\left(19v_{11}-27v_{12}+8v_{12}^{2}\right)\log\frac{v_{11}}{v_{12}^{2}}$
$\displaystyle-4v_{12}\left(6v_{11}^{2}-3v_{11}v_{12}-v_{12}^{2}+v_{11}v_{12}^{2}\right)\sqrt{1-v_{11}}\log\left(\frac{1+\sqrt{1-v_{11}}}{1-\sqrt{1-v_{11}}}\right)$
$\displaystyle+3v_{11}^{2}\left(2v_{11}-3v_{12}+v_{12}^{2}\right)\,L(v_{11},v_{12})\biggr{)}\biggr{]}+\mathcal{O}(\varepsilon^{2})\,,$
(621)

$\displaystyle I^{(1)}_{3,3}(v_{12},v_{11};\varepsilon)$
$\displaystyle=\frac{\pi}{4v_{12}^{7}v_{11}^{2}}\biggl{[}-\frac{6}{\varepsilon}\,v_{11}^{2}\left(15v_{11}^{2}-40v_{11}v_{12}+24v_{12}^{2}+12v_{11}v_{12}^{2}-12v_{12}^{3}+v_{12}^{4}\right)$
$\displaystyle-306v_{11}^{4}+976v_{11}^{3}v_{12}-768v_{11}^{2}v_{12}^{2}-312v_{11}^{3}v_{12}^{2}+48v_{11}v_{12}^{3}+420v_{11}^{2}v_{12}^{3}$
$\displaystyle+8v_{12}^{4}-24v_{11}v_{12}^{4}-43v_{11}^{2}v_{12}^{4}-6v_{11}^{2}\Big{(}15v_{11}^{2}-40v_{11}v_{12}+12v_{11}v_{12}^{2}$
$\displaystyle+24v_{12}^{2}-12v_{12}^{3}+v_{12}^{4}\Big{)}\log\frac{v_{11}}{v_{12}^{2}}+\frac{\varepsilon}{2}\biggl{(}-616v_{11}^{4}+2024v_{11}^{3}v_{12}-1616v_{11}^{2}v_{12}^{2}$
$\displaystyle-784v_{11}^{3}v_{12}^{2}+48v_{11}v_{12}^{3}+1120v_{11}^{2}v_{12}^{3}+16v_{12}^{4}-48v_{11}v_{12}^{4}-145v_{11}^{2}v_{12}^{4}$
$\displaystyle+2v_{11}^{2}\left(-306v_{11}^{2}+796v_{11}v_{12}-468v_{12}^{2}-222v_{11}v_{12}^{2}+216v_{12}^{3}-16v_{12}^{4}\right)\log\frac{v_{11}}{v_{12}^{2}}$
$\displaystyle+8v_{12}\left(21v_{11}^{2}v_{12}^{2}+12v_{11}v_{12}^{2}+2v_{12}^{3}+45v_{11}^{3}-75v_{11}^{2}v_{12}-5v_{11}v_{12}^{3}\right)$
$\displaystyle\times\sqrt{1-v_{11}}\log\left(\frac{1+\sqrt{1-v_{11}}}{1-\sqrt{1-v_{11}}}\right)-6v_{11}^{2}\Big{(}15v_{11}^{2}-40v_{11}v_{12}+24v_{12}^{2}$
$\displaystyle+12v_{11}v_{12}^{2}-12v_{12}^{3}+v_{12}^{4}\Big{)}\,L(v_{11},v_{12})\biggr{)}\biggr{]}+\mathcal{O}(\varepsilon^{2})\,,$
(622)

where

$\displaystyle
L(v_{11},v_{12})=\log^{2}\frac{v_{11}}{v_{12}^{2}}+4\mathrm{Li}_{2}\left(1-\frac{v_{12}}{1-\sqrt{1-v_{11}}}\right)+4\mathrm{Li}_{2}\left(1-\frac{v_{12}}{1+\sqrt{1-v_{11}}}\right)\,.$
(623)

###  G.5 Double massive integral with two denominators

In this section we briefly review how we proceed with double massive integrals
with two denominators. As we showed in Sec. 3.4.4, the $\varepsilon$ expansion
of the double massive master integral with two denominators
$I_{1,1}^{(2)}(v_{12},v_{11},v_{22};\varepsilon)$ is generated by analytically
using Eq. (3.4.4). Eq. (3.4.4) explicitly illustrates the $\varepsilon$
expansion of this integral up to second order. All other double massive
integrals are obtained using recursive relations (222), (224), (226), and
(229) and $\varepsilon$-expansion of the master integrals $I^{(0)}$,
$I_{1,0}^{(1)}(v_{11};\varepsilon)$, and
$I_{1,1}^{(2)}(v_{12},v_{11},v_{22};\varepsilon)$.

###  G.6 Mirkes type III integrals

Mirkes type III integrals are defined as

$\displaystyle
I^{j,l}_{D}(A,B,C)=\int\mathrm{d}\Omega_{k_{1}k_{2}}\frac{(-\cos\theta_{1})^{l}}{\left(A+B\cos\theta_{1}+C\sin\theta_{1}\cos\theta_{2}\right)^{j}}=A^{-j}\,I^{j,l}_{D}(b,\rho^{2})\,,$
(624)

where $b=B/A$ , $c=C/A$ , and $\rho^{2}=b^{2}+c^{2}$ .

By application of the recursion relations (295)-(298) we find the integrals
with specific values of indices $j=1,2$ and $l=0,\ldots,4$:

$\displaystyle I^{1,0}_{D}(b,\rho^{2})$ $\displaystyle=$ $\displaystyle
I^{(1)}_{1}\,,$ (625) $\displaystyle I^{1,1}_{D}(b,\rho^{2})$ $\displaystyle=$
$\displaystyle\frac{b}{\rho^{2}}\,\Big{[}-I^{(0)}+I^{(1)}_{1}\Big{]}\,,$ (626)
$\displaystyle I^{1,2}_{D}(b,\rho^{2})$ $\displaystyle=$
$\displaystyle\frac{1}{2\rho^{4}(1-\varepsilon)}\,\biggl{[}I^{(0)}\,\Big{(}\rho^{2}-b^{2}(3-2\varepsilon)\Big{)}$
(627) $\displaystyle-$ $\displaystyle
I^{(1)}_{1}\,\Big{(}\rho^{2}(1-\rho^{2})-b^{2}(3-\rho^{2}-2\varepsilon)\Big{)}\biggr{]}\,,$
$\displaystyle I^{1,3}_{D}(b,\rho^{2})$ $\displaystyle=$
$\displaystyle\frac{b}{2\rho^{6}\,(1-\varepsilon)}\,\biggl{[}\frac{I^{(0)}}{3-2\varepsilon}\,\Big{(}\rho^{2}(1-\varepsilon)(9-6\rho^{2}+4b^{2}-2\varepsilon)$
(628) $\displaystyle-$ $\displaystyle
b^{2}(3-2\varepsilon)(5-2\varepsilon)\Big{)}-I^{(1)}_{1}\,\Big{(}3\rho^{2}(1-\rho^{2})-b^{2}(5-3\rho^{2}-2\varepsilon)\Big{)}\biggr{]}\,,$
$\displaystyle I^{1,4}_{D}(b,\rho^{2})$ $\displaystyle=$
$\displaystyle\frac{1}{4\rho^{8}(1-\varepsilon)(2-\varepsilon)}\,\biggl{[}\frac{I^{(0)}}{3-2\varepsilon}\,\Big{(}-b^{4}(3-2\varepsilon)(5-2\varepsilon)(7-2\varepsilon)$
(629) $\displaystyle-$ $\displaystyle
3\rho^{4}(3-2\varepsilon)+3\rho^{6}(5-4\varepsilon)+b^{4}\rho^{2}(5-2\varepsilon)(11-10\varepsilon)$
$\displaystyle-$ $\displaystyle
6b^{2}\rho^{4}(13-16\varepsilon+4\varepsilon^{2})+6b^{2}\rho^{2}(3-2\varepsilon)(5-2\varepsilon)\Big{)}$
$\displaystyle+$ $\displaystyle
I^{(1)}_{1}\,\Big{(}3\rho^{4}(1-\rho^{2}+b^{2})^{2}-6b^{2}\rho^{2}(1-\rho^{2}+b^{2})(5-2\varepsilon)$
$\displaystyle+$ $\displaystyle
b^{4}(5-2\varepsilon)(7-2\varepsilon)\Big{)}\biggr{]}\,,$ $\displaystyle
I^{2,0}_{D}(b,\rho^{2})$ $\displaystyle=$
$\displaystyle\frac{1}{1-\rho^{2}}\biggl{[}(1-2\varepsilon)\,I^{(0)}+2\varepsilon\,I^{(1)}_{1}\biggr{]}\,,$
(630) $\displaystyle I^{2,1}_{D}(b,\rho^{2})$ $\displaystyle=$
$\displaystyle\frac{b}{\rho^{2}\,(1-\rho^{2})}\,\Big{[}(1-2\varepsilon)\,I^{(0)}\,-\,(1-\rho^{2}-2\varepsilon)\,I^{(1)}_{1}\Big{]}\,,$
(631) $\displaystyle I^{2,2}_{D}(b,\rho^{2})$ $\displaystyle=$ $\displaystyle
I^{(0)}\,\biggl{[}-\frac{1}{\rho^{2}}+\frac{b^{2}(3-2\rho^{2}-2\varepsilon)}{\rho^{4}(1-\rho^{2})}\biggr{]}$
(632) $\displaystyle+$ $\displaystyle
I^{(1)}_{1}\,\biggl{[}\frac{\rho^{2}-3b^{2}}{\rho^{4}}+\frac{2b^{2}\varepsilon}{\rho^{4}(1-\rho^{2})}\biggr{]}\,,$
$\displaystyle I^{2,3}_{D}(b,\rho^{2})$ $\displaystyle=$
$\displaystyle\frac{b}{2\rho^{6}\,(1-\rho^{2})\,(1-\varepsilon)}\,\biggl{[}I^{(0)}\,\Big{(}-3\rho^{2}(1-\rho^{2})(3-2\varepsilon)$
(633) $\displaystyle+$ $\displaystyle
b^{2}(3-2\varepsilon)(5-2\varepsilon)-b^{2}\rho^{2}(13-10\varepsilon)\Big{)}$
$\displaystyle+$ $\displaystyle
I^{(1)}_{1}\,\Big{(}3\rho^{2}(1-\rho^{2})(3-\rho^{2}-2\varepsilon)$
$\displaystyle-$ $\displaystyle
b^{2}((3-2\varepsilon)(5-2\varepsilon)-3\rho^{2}(6-\rho^{2}-4\varepsilon))\Big{)}\biggr{]}\,,$
$\displaystyle I^{2,4}_{D}(b,\rho^{2})$ $\displaystyle=$
$\displaystyle\frac{I^{(0)}}{2\rho^{2}}\,\biggl{[}\frac{3}{\rho^{2}(1-\varepsilon)}-\frac{6}{3-2\varepsilon}+\frac{6b^{2}}{\rho^{2}}\,\biggl{(}\frac{4}{3-2\varepsilon}-\frac{5-2\varepsilon}{\rho^{2}(1-\varepsilon)}\biggr{)}$
(634) $\displaystyle+$
$\displaystyle\frac{b^{4}}{\rho^{2}(1-\rho^{2})}\,\biggl{(}\frac{16}{3-2\varepsilon}+\frac{(5-2\varepsilon)(7-2\varepsilon)}{\rho^{4}(1-\varepsilon)}-\frac{(5-2\varepsilon)(23-18\varepsilon)}{\rho^{2}(1-\varepsilon)(3-2\varepsilon)}\biggr{)}\biggr{]}$
$\displaystyle+$
$\displaystyle\frac{I^{(1)}_{1}}{2\rho^{8}}\,\biggl{[}4b^{2}(3\beta^{2}-5b^{2})-\frac{3(1-\rho^{2})(b^{2}-\rho^{2})(5b^{2}-\rho^{2})}{1-\varepsilon}+\frac{4b^{4}\varepsilon}{1-\rho^{2}}\biggr{]}\,.$

## References

  * (1) G. ’t Hooft and M. J. G. Veltman, Nucl. Phys. B 44, 189 (1972).
  * (2) G. ’t Hooft and M. J. G. Veltman, Nucl. Phys. B 153, 365 (1979).
  * (3) G. Passarino and M. J. G. Veltman, Nucl. Phys. B 160, 151 (1979).
  * (4) R. K. Ellis, D. A. Ross, and A. E. Terrano, Nucl. Phys. B 178, 421 (1981).
  * (5) W. L. van Neerven and J. A. M. Vermaseren, Phys. Lett. 137B, 241 (1984).
  * (6) R. G. Stuart, Comput. Phys. Commun. 48, 367 (1988).
  * (7) G. J. van Oldenborgh and J. A. M. Vermaseren, Z. Phys. C 46, 425 (1990).
  * (8) J. G. Korner and P. Sieben, Nucl. Phys. B 363, 65 (1991).
  * (9) J. G. Korner, J. H. Kuhn, M. Krammer and H. Schneider, Nucl. Phys. B 229, 115 (1983).
  * (10) E. E. Boos and A. I. Davydychev, Theor. Math. Phys. 89, 1052 (1991) [Teor. Mat. Fiz. 89, 56 (1991)].
  * (11) A. I. Davydychev, Phys. Lett. B 263, 107 (1991).
  * (12) Z. Bern, L. J. Dixon, and D. A. Kosower, Nucl. Phys. B 412, 751 (1994) [hep-ph/9306240].
  * (13) Z. Bern, L. J. Dixon, D. C. Dunbar, and D. A. Kosower, Nucl. Phys. B 425, 217 (1994) [hep-ph/9403226].
  * (14) Z. Bern, L. J. Dixon, D. C. Dunbar, and D. A. Kosower, Nucl. Phys. B 435, 59 (1995) [hep-ph/9409265].
  * (15) J. M. Campbell, E. W. N. Glover, and D. J. Miller, Nucl. Phys. B 498, 397 (1997) [hep-ph/9612413].
  * (16) R. Pittau, Comput. Phys. Commun. 104, 23 (1997) [hep-ph/9607309].
  * (17) G. Devaraj and R. G. Stuart, Nucl. Phys. B 519, 483 (1998) [hep-ph/9704308].
  * (18) O. V. Tarasov, Phys. Rev. D 54, 6479 (1996) [hep-th/9606018].
  * (19) J. Fleischer, F. Jegerlehner, and O. V. Tarasov, Nucl. Phys. B 566, 423 (2000) [hep-ph/9907327].
  * (20) G. Duplancic and B. Nizic, Eur. Phys. J. C 20, 357 (2001) [hep-ph/0006249].
  * (21) S. Laporta, Int. J. Mod. Phys. A 15, 5087 (2000) hep-ph/0102033].
  * (22) F. Cachazo, P. Svrcek, and E. Witten, JHEP 0410, 074 (2004) [hep-th/0406177].
  * (23) R. Britto, F. Cachazo, and B. Feng, Nucl. Phys. B 725, 275 (2005) [hep-th/0412103].
  * (24) V. A. Smirnov, “Feynman integral calculus,” Berlin, Germany: Springer (2006) 283p.
  * (25) V. A. Smirnov, “Analytic tools for Feynman integrals,” Springer Tracts Mod. Phys. 250, 1 (2012).
  * (26) A. Denner and S. Dittmaier, Nucl. Phys. B 734, 62 (2006) [hep-ph/0509141].
  * (27) G. Ossola, C. G. Papadopoulos, and R. Pittau, Nucl. Phys. B 763, 147 (2007) [hep-ph/0609007].
  * (28) R. K. Ellis, W. T. Giele, and Z. Kunszt, JHEP 0803, 003 (2008) [arXiv:0708.2398 [hep-ph]].
  * (29) A. Faessler, T. Gutsche, M. A. Ivanov, J. G. Korner, V. E. Lyubovitskij, D. Nicmorus, and K. Pumsa-ard, Phys. Rev. D 73, 094013 (2006) [hep-ph/0602193].
  * (30) T. Gutsche, M. A. Ivanov, J. G. Körner, V. E. Lyubovitskij, and Z. Tyulemissov, Phys. Rev. D 99, 056013 (2019) [arXiv:1812.09212 [hep-ph]].
  * (31) D. Forde, Phys. Rev. D 75, 125019 (2007) [arXiv:0704.1835 [hep-ph]].
  * (32) W. T. Giele, Z. Kunszt, and K. Melnikov, JHEP 0804, 049 (2008) [arXiv:0801.2237 [hep-ph]].
  * (33) A. S. Zhevlakov, M. Gorchtein, A. N. Hiller Blin, T. Gutsche, and V. E. Lyubovitskij, Phys. Rev. D 99, 031703 (2019) [arXiv:1812.00171 [hep-ph]].
  * (34) M. Y. Kalmykov, JHEP 0604, 056 (2006) [hep-th/0602028].
  * (35) R. K. Ellis and G. Zanderighi, JHEP 0802, 002 (2008) [arXiv:0712.1851 [hep-ph]].
  * (36) A. Denner and S. Dittmaier, Nucl. Phys. B 844, 199 (2011) [arXiv:1005.2076 [hep-ph]].
  * (37) R. K. Ellis, Z. Kunszt, K. Melnikov, and G. Zanderighi, Phys. Rept. 518, 141 (2012) [arXiv:1105.4319 [hep-ph]].
  * (38) P. Mastrolia, T. Peraro, and A. Primo, JHEP 1608, 164 (2016) [arXiv:1605.03157 [hep-ph]].
  * (39) T. Peraro and L. Tancredi, Phys. Rev. D 103, 054042 (2021) [arXiv:2012.00820 [hep-ph]].
  * (40) J. Davies, G. Mishima, M. Steinhauser, and D. Wellmann, JHEP 2004, 024 (2020) [arXiv:2002.05558 [hep-ph]].
  * (41) M. Kalmykov, V. Bytev, B. Kniehl, S. O. Moch, B. Ward, and S. Yost, arXiv:2012.14492 [hep-th].
  * (42) D. Boer and W. Vogelsang, Phys. Rev. D 74, 014004 (2006) [hep-ph/0604177].
  * (43) E. L. Berger, J. W. Qiu, and R. A. Rodriguez-Pedraza, Phys. Rev. D 76, 074006 (2007) [arXiv:0708.0578 [hep-ph]].
  * (44) W. L. van Neerven, Nucl. Phys. B 268, 453 (1986).
  * (45) A. Devoto, D. W. Duke, J. D. Kimel, and G. A. Sowell, Phys. Rev. D 30, 541 (1984).
  * (46) J. Smith, D. Thomas, and W. L. van Neerven, Z. Phys. C 44, 267 (1989).
  * (47) J. Smith, Preprint YITP-SB-09-13, 2009.
  * (48) G. Somogyi, J. Math. Phys. 52, 083501 (2011) [arXiv:1101.3557 [hep-ph]].
  * (49) E. Mirkes, Nucl. Phys. B 387, 3 (1992).
  * (50) C. Anastasiou and K. Melnikov, Nucl. Phys. B 646, 220 (2002) [hep-ph/0207004].
  * (51) A. Mitov, Phys. Lett. B 643, 366 (2006) [arXiv:hep-ph/0511340 [hep-ph]].
  * (52) D. Bonocore, E. Laenen, and R. Rietkerk, JHEP 05, 079 (2016) [arXiv:1603.05252 [hep-ph]].
  * (53) M. Argeri and P. Mastrolia, Int. J. Mod. Phys. A 22, 4375 (2007) [arXiv:0707.4037 [hep-ph]].
  * (54) S. Moch and C. Schneider, PoS ACAT (2007), 083 [arXiv:0709.1769 [math-ph]].
  * (55) A. N. Schellekens, “Perturbative QCD and lepton Pair Production”, PhD Thesis, RX-953, Nijmegen (1981).
  * (56) T. Matsuura, S. C. van der Marck, and W. L. van Neerven, Nucl. Phys. B 319, 570 (1989).
  * (57) T. Matsuura, R. Hamberg, and W. L. van Neerven, Nucl. Phys. B 345, 331 (1990).
  * (58) R. Hamberg, W. L. van Neerven, and T. Matsuura, Nucl. Phys. B 359, 343 (1991), B 644, 403(E) (2002).
  * (59) N. Bahjat-Abbas, J. Sinninghe Damsté, L. Vernazza, and C. D. White, JHEP 1810, 144 (2018) [arXiv:1807.09246 [hep-ph]].
  * (60) D. W. Duke and J. F. Owens, Phys. Rev. D 26, 1600 (1982); D 28, 1227(E) (1983).
  * (61) D. Anderle, D. de Florian, and Y. Rotstein Habarnau, Phys. Rev. D 95, 034027 (2017) [arXiv:1612.01293 [hep-ph]].
  * (62) B. Wang, J. O. Gonzalez-Hernandez, T. C. Rogers, and N. Sato, Phys. Rev. D 99, 094029 (2019) [arXiv:1903.01529 [hep-ph]].
  * (63) L. E. Gordon and W. Vogelsang, Phys. Rev. D 48, 3136 (1993).
  * (64) R. K. Ellis, M. A. Furman, H. E. Haber, and I. Hinchliffe, Nucl. Phys. B 173, 397 (1980).
  * (65) W. Beenakker, H. Kuijf, W. L. van Neerven, and J. Smith, Phys. Rev. D 40, 54 (1989).
  * (66) I. Bojak, hep-ph/0005120.
  * (67) G. Isidori, S. Nabeebaccus, and R. Zwicky, JHEP 2012, 104 (2020) [arXiv:2009.00929 [hep-ph]].
  * (68) P. Bolzoni, G. Somogyi, and Z. Trocsanyi, JHEP 1101, 059 (2011) [arXiv:1011.1909 [hep-ph]].
  * (69) S. Lionetti, “Subtraction of Infrared Singularities at Higher Orders in QCD”, PhD Thesis, ETH Zürich, 2018.
  * (70) C. Specchia, “Perturbative Corrections to Inclusive and Differential Cross Sections for Higgs Production at the LHC”, PhD Thesis, ETH Zürich, 2018.
  * (71) M. Höschele, “Phasenraum-Masterintegrale zur Berechnung der Higgsproduktion in Gluonfusion”, PhD Thesis, KIT Karlsruhe, 2018.
  * (72) C. Anastasiou, C. Duhr, F. Dulat, and B. Mistlberger, JHEP 1307, 003 (2013) [arXiv:1302.4379 [hep-ph]].
  * (73) D. Baranowski, Eur. Phys. J. C 80, 523 (2020) [arXiv:2004.03285 [hep-ph]].
  * (74) F. M. Ringer, “Threshold Resummation and Higher Order Effects in QCD”, PhD Thesis, Tübingen University, 2015.
  * (75) F. Ringer and W. Vogelsang, Phys. Rev. D 91, 094033 (2015) [arXiv:1503.07052 [hep-ph]].
  * (76) M. Schlegel, Phys. Rev. D 87, 034006 (2013) [arXiv:1211.3579 [hep-ph]].
  * (77) W. Kotlarski, JHEP 1702, 027 (2017) [arXiv:1608.00915 [hep-ph]].
  * (78) B. Lillard, T. M. P. Tait, and P. Tanedo, Phys. Rev. D 94, 054012 (2016) [arXiv:1602.08622 [hep-ph]].
  * (79) P. Hinderer, “Higher Order Studies in Perturbative QCD”, PhD Thesis, Tübingen University, 2017.
  * (80) F. Hekhorn, “Next-to-Leading Order QCD Corrections to Heavy-Flavour Production in Neutral Current DIS”, PhD Thesis, Tübingen University, 2019.
  * (81) J. Blümlein, A. De Freitas, C. Raab, and K. Schönwald, Nucl. Phys. B 956, 115055 (2020) [arXiv:2003.14289 [hep-ph]].
  * (82) T. Huber and D. Maitre, Comput. Phys. Commun. 175, 122 (2006) [hep-ph/0507094].
  * (83) K. S. Kolbig, J. A. Mignoco, and E. Remiddi, Preprint CERN-DD-CO-69-5, 1969.
  * (84) K. S. Kolbig, SIAM J. Math. Anal. 17, 1232 (1986).
  * (85) A. Erdelyi, “Higher Transcendental Functions”, McGraw-Hill Book Company, Inc., 1953, 317p.
  * (86) G. F. Sterman, “An Introduction to quantum field theory”, Cambridge University Press, 1993, 592 pp.
  * (87) A. B. Goncharov, math/0103059 [math.AG].
  * (88) H. Frellesvig, D. Tommasini, and C. Wever, JHEP 1603, 189 (2016) [arXiv:1601.02649 [hep-ph]].
  * (89) C. Duhr and F. Dulat, JHEP 1908, 135 (2019) [arXiv:1904.07279 [hep-th]].
  * (90) C. Duhr, F. Dulat, and B. Mistlberger, Phys. Rev. Lett. 125, 172001 (2020) [arXiv:2001.07717 [hep-ph]].
  * (91) C. Duhr, F. Dulat, and B. Mistlberger, [arXiv:2007.13313 [hep-ph]].