$\displaystyle-K^{2}\left(4q^{2}+k^{2}+k_{0}^{2}\right)\bigg{[}\ln\frac{k^{0}+k-2q+i\varepsilon}{k^{0}-k-2q+i\varepsilon}+\ln\frac{k^{0}-k+2q+i\varepsilon}{k^{0}+k+2q+i\varepsilon}\bigg{]}$
$\displaystyle-4qk^{0}K^{2}\bigg{[}2\ln\frac{k^{0}+k+i\varepsilon}{k^{0}-k+i\varepsilon}+\ln\frac{k^{0}-k+2q+i\varepsilon}{k^{0}+k+2q+i\varepsilon}-\ln\frac{k^{0}+k-2q+i\varepsilon}{k^{0}-k-2q+i\varepsilon}\bigg{]}\bigg{]}.$
(D.15)

The longitudinal counterpart is instead

$\displaystyle\Pi_{L}^{R}(K)_{F}=$
$\displaystyle-\frac{g^{2}T_{F}N_{F}K^{2}}{4\pi^{2}k^{3}}\int_{0}^{\infty}dq\,n_{\mathrm{F}}(q)\bigg{[}8qk$
$\displaystyle-\left(4q^{2}-k^{2}+k_{0}^{2}\right)\bigg{[}\ln\frac{k^{0}+k-2q+i\varepsilon}{k^{0}-k-2q+i\varepsilon}+\ln\frac{k^{0}-k+2q+i\varepsilon}{k^{0}+k+2q+i\varepsilon}\bigg{]}$
$\displaystyle-4qk^{0}\bigg{[}2\ln\frac{k^{0}+k+i\varepsilon}{k^{0}-k+i\varepsilon}+\ln\frac{k^{0}-k+2q+i\varepsilon}{k^{0}+k+2q+i\varepsilon}-\ln\frac{k^{0}+k-2q+i\varepsilon}{k^{0}-k-2q+i\varepsilon}\bigg{]}\bigg{]},$
(D.16)

For further use we then have

$\displaystyle\Pi_{L}^{R}(K)_{F}-\Pi_{T}^{R}(K)_{F}=$
$\displaystyle\frac{g^{2}T_{F}N_{F}}{8\pi^{2}k^{3}}\int_{0}^{\infty}dq\,n_{\mathrm{F}}(q)\bigg{[}-8qk(3k_{0}^{2}-k^{2})$
$\displaystyle+K^{2}\left(12q^{2}-k^{2}+3k_{0}^{2}\right)\bigg{[}\ln\frac{k^{0}+k-2q+i\varepsilon}{k^{0}-k-2q+i\varepsilon}+\ln\frac{k^{0}-k+2q+i\varepsilon}{k^{0}+k+2q+i\varepsilon}\bigg{]}$
$\displaystyle+12qk^{0}K^{2}\bigg{[}2\ln\frac{k^{0}+k+i\varepsilon}{k^{0}-k+i\varepsilon}+\ln\frac{k^{0}-k+2q+i\varepsilon}{k^{0}+k+2q+i\varepsilon}-\ln\frac{k^{0}+k-2q+i\varepsilon}{k^{0}-k-2q+i\varepsilon}\bigg{]}\bigg{]},$
(D.17)

We additionally need the vacuum part of the self-energy: The bare vacuum part
is [21]

$\Pi_{L}^{R}(K)_{\text{vac}}=\Pi_{T}^{R}(K)_{\text{vac}}=\frac{\alpha_{s}}{4\pi}K^{2}\bigg{[}\bigg{(}\frac{1}{\epsilon}-\gamma_{E}+\ln(4\pi)-\ln(-K^{2})\bigg{)}\bigg{(}\frac{5}{3}C_{A}-\frac{4}{3}T_{F}N_{F}\bigg{)}+\frac{31}{9}C_{A}-\frac{20}{9}T_{F}N_{F}\bigg{]}.$
(D.18)

Whereas in the $\overline{MS}$ scheme we have

$\Pi_{L}^{R}(K)_{\text{vac}\,\overline{MS}}=\Pi_{T}^{R}(K)_{\text{vac}\,\overline{MS}}=\frac{\alpha_{s}}{4\pi}K^{2}\bigg{[}-\ln\frac{k^{2}-(k^{0}+i\varepsilon)^{2}}{\bar{\mu}^{2}}\bigg{(}\frac{5}{3}C_{A}-\frac{4}{3}T_{F}N_{F}\bigg{)}+\frac{31}{9}C_{A}-\frac{20}{9}T_{F}N_{F}\bigg{]}.$
(D.19)

###  D.3 Details on Diagram $(c)$ Evaluation

We pack many of the technical details related to the computation in Sec. 6.5
into this appendix.

####  D.3.1 Fermionic Contribution: $k^{-2}$ Piece

Looking to Eq. (D.15), we start with the simplest terms, i.e. those that do
not multiply any log, i.e.

$\displaystyle\Pi^{R}_{T}(K)_{F\,\text{no log}}=$
$\displaystyle\frac{g^{2}}{16\pi^{2}k^{3}}\int_{0}^{\infty}dq\bigg{[}2T_{F}N_{F}n_{\text{F}}(q)8qk\left(k^{2}+k_{0}^{2}\right)\bigg{]}=\frac{g^{2}T_{F}N_{F}T^{2}}{12k^{2}}\left(k^{2}+k_{0}^{2}\right),$
(D.20)

Hence

$Z_{g\,k^{-2}\,F\,\text{ no
log}}^{(c)}=i\frac{g^{2}T_{F}N_{F}T^{2}}{6}\int_{K}\Big{(}1+n_{\mathrm{B}}(k^{0})\Big{)}\bigg{[}\frac{(1+k_{0}^{2}/k^{2})}{(K^{2}+i\varepsilon
k^{0})^{2}}-\text{adv.}\bigg{]}.$ (D.21)

We can do the $k^{0}$ integration using contour methods: since the terms in
the square brackets have the desired $\text{ret.}-\text{adv}$ structure, we
can close the contour in the upper (lower) half-plane to integrate over the
retarded (advanced) piece. We find

$\displaystyle Z_{g\,k^{-2}\,F\,\text{ no log}}^{(c)}=$
$\displaystyle\frac{g^{2}T_{F}N_{F}T^{2}}{6}\int_{K}\Big{(}\frac{1}{2}+n_{\mathrm{B}}(k^{0})\Big{)}\bigg{[}\frac{i(1+k_{0}^{2}/k^{2})}{(K^{2}+i\varepsilon
k^{0})^{2}}-\frac{i(1+k_{0}^{2}/k^{2})}{(K^{2}-i\varepsilon
k^{0})^{2}}\bigg{]}$
$\displaystyle=-\frac{g^{2}T_{F}N_{F}T^{2}}{6}T\sum_{n}\int_{k}\frac{1-\omega_{n}^{2}/k^{2}}{(-k^{2}-\omega_{n}^{2})^{2}}=\frac{g^{2}T_{F}N_{F}T^{2}}{24\pi^{2}}$
(D.22)

Note that in going from Eq. (D.21), we have rewritten
$\big{(}1+n_{\mathrm{B}}(k^{0})\big{)}\rightarrow\big{(}\frac{1}{2}+n_{\mathrm{B}}(k^{0})\big{)}$,
which is allowed, because the vacuum part is scaleless and vanishes in DR
anyway. The $1/2$ accompanying the statistical function is inserted, precisely
because it is the proper term for analytical continuation, and because
$1/2+n_{\mathrm{B}}(k^{0})$ is odd, thus complementing the odd expression in
square brackets.

Now we move on to the terms with logarithms, starting with the “HTL log”
contribution, which is obtained by expanding for $Q\gg K$ in Eq. (D.15)

$\displaystyle\Pi^{R}_{T}(K)_{F}\bigg{|}_{\text{HTL log}}=$
$\displaystyle-\frac{g^{2}T_{F}N_{F}}{\pi^{2}k^{3}}\int_{0}^{\infty}dq\,n_{\mathrm{F}}(q)qk^{0}K^{2}\ln\frac{k^{0}+k+i\varepsilon}{k^{0}-k+i\varepsilon}$
$\displaystyle=$
$\displaystyle-g^{2}T_{F}N_{F}T^{2}\frac{k^{0}K^{2}}{12k^{3}}\ln\frac{k^{0}+k+i\varepsilon}{k^{0}-k+i\varepsilon}.$
(D.23)

The contribution from this log to $Z_{g}$ reads

$Z_{g\,k^{-2}\,F\,\text{ HTL
log}}^{(c)}=-i\frac{g^{2}T_{F}N_{F}T^{2}}{6}\int_{K}\Big{(}\frac{1}{2}+n_{\mathrm{B}}(k^{0})\Big{)}\bigg{[}\frac{k^{0}}{k^{3}(K^{2}+i\varepsilon
k^{0})}\ln\frac{k^{0}+k+i\varepsilon}{k^{0}-k+i\varepsilon}-\text{adv.}\bigg{]}.$
(D.24)

The reason this term is dealt with separately is that the strategy of
separating out a $\delta(K^{2})$ contribution and a $\Theta$-function
contribution from the imaginary part of the log in the retarded-advanced
difference would in this case give rise to two contributions that are
separately divergent on the light-cone at $K^{2}=0$.

We can proceed by re-introducing the auxiliary angular integral and then using
analyticity to get the Matsubara modes, i.e.

$\displaystyle Z_{g\,k^{-2}\,F\,\text{ HTL log}}^{(c)}=$
$\displaystyle-i\frac{g^{2}T_{F}N_{F}T^{2}}{6}\int_{K}\int_{-1}^{1}dx\,\Big{(}\frac{1}{2}+n_{\mathrm{B}}(k^{0})\Big{)}\bigg{[}\frac{k^{0}}{k^{2}(K^{2}+i\varepsilon
k^{0})}\frac{1}{k^{0}-kx+i\varepsilon}-\text{adv.}\bigg{]}$ $\displaystyle=$
$\displaystyle\frac{g^{2}T_{F}N_{F}T^{2}}{6}T\sum_{n}\int_{k}\int_{-1}^{1}dx\frac{i\omega_{n}}{k^{2}(-\omega_{n}^{2}-k^{2})}\frac{1}{i\omega_{n}-kx}$
$\displaystyle=$
$\displaystyle-\frac{g^{2}T_{F}N_{F}T^{2}}{6}T\sum_{n}\int_{k}\int_{-1}^{1}dx\frac{i\omega_{n}}{k^{2}(\omega_{n}^{2}+k^{2})}\frac{-i\omega_{n}+kx}{\omega_{n}^{2}+k^{2}x^{2}}$
(D.25)

If we keep the even part only of the integer sum we find

$\displaystyle Z_{g\,F\,\text{HTL log}}^{(c)}=$
$\displaystyle-\frac{g^{2}T_{F}N_{F}T^{2}}{6}T\sum_{n}\int_{k}\int_{-1}^{1}dx\frac{\omega_{n}^{2}}{k^{2}(\omega_{n}^{2}+k^{2})(\omega_{n}^{2}+k^{2}x^{2})}$
$\displaystyle=$
$\displaystyle-\frac{g^{2}T_{F}N_{F}T^{2}}{24\pi^{2}}\bigg{[}\frac{\ln(2)}{\epsilon}+\left(-\ln(4)\ln\left(2\frac{4\pi
Te^{-\gamma_{E}}}{\bar{\mu}}\right)+\frac{\pi^{2}}{12}+2\ln^{2}(2)+\ln(4)\right)\bigg{]}.$
(D.26)

Let us now move on to the other $k^{-2}\Pi^{T}$-proportional terms in the
fermion contribution, i.e.

$\displaystyle Z_{g\,k^{-2}\,F\,\text{ non-HTL }}^{(c)}=$
$\displaystyle\frac{ig^{2}T_{F}N_{F}}{4\pi^{2}}\int_{K}\bigg{\\{}\frac{1+n_{\mathrm{B}}(k^{0})}{k^{3}(K^{2}+i\varepsilon
k^{0})}\int_{0}^{\infty}dq\,n_{\mathrm{F}}(q)\bigg{[}$
$\displaystyle-\left(4q^{2}+k^{2}+k_{0}^{2}\right)\bigg{(}\ln\frac{k^{0}+k-2q+i\varepsilon}{k^{0}-k-2q+i\varepsilon}+\ln\frac{k^{0}-k+2q+i\varepsilon}{k^{0}+k+2q+i\varepsilon}\bigg{)}$
$\displaystyle-4qk^{0}\bigg{(}\ln\frac{k^{0}-k+2q+i\varepsilon}{k^{0}+k+2q+i\varepsilon}-\ln\frac{k^{0}+k-2q+i\varepsilon}{k^{0}-k-2q+i\varepsilon}\bigg{)}\bigg{]}-\text{adv.}\bigg{\\}}.$
(D.27)

We can split this into a pole part ($\propto\delta(K^{2})$) and a cut part
($\propto\mathbb{P}1/K^{2}$, with $\mathbb{P}$ the principal value). The
former reads

$\displaystyle Z_{g\,k^{-2}\,F\,\text{pole}}^{(c)}=$
$\displaystyle\frac{g^{2}T_{F}N_{F}}{4\pi^{2}}\int_{K}\frac{\frac{1}{2}+n_{\mathrm{B}}(|k^{0}|)}{k^{3}}2\pi\delta(K^{2})\int_{0}^{\infty}dq\,n_{\mathrm{F}}(q)\bigg{[}$
$\displaystyle-\left(4q^{2}+k^{2}+k_{0}^{2}\right)\bigg{(}\ln\left|\frac{k^{0}+k-2q}{k^{0}-k-2q}\right|+\ln\left|\frac{k^{0}-k+2q}{k^{0}+k+2q}\right|\bigg{)}$
$\displaystyle-4qk^{0}\bigg{(}\ln\left|\frac{k^{0}-k+2q}{k^{0}+k+2q}\right|-\ln\left|\frac{k^{0}+k-2q}{k^{0}-k-2q}\right|\bigg{)}\bigg{]}.$
(D.28)

This becomes

$\displaystyle Z_{gk^{-2}\,F\,\text{pole}}^{(c)}=$
$\displaystyle\frac{g^{2}T_{F}N_{F}}{2\pi^{2}}\int_{k}\frac{\frac{1}{2}+n_{\mathrm{B}}(k)}{k^{4}}\int_{0}^{\infty}dq\,n_{\mathrm{F}}(q)\bigg{[}\left(2q^{2}+k^{2}\right)\ln\frac{k+q}{|k-q|}+2qk\ln\frac{|k^{2}-q^{2}|}{q^{2}}\bigg{]}.$
(D.29)

Let us split this into regions

$\displaystyle Z_{gk^{-2}\,F\,\text{pole}}^{(c)}=$
$\displaystyle\frac{g^{2}T_{F}N_{F}}{2\pi^{2}}\int_{0}^{\infty}dq\,n_{\mathrm{F}}(q)\int_{k}\frac{\frac{1}{2}+n_{\mathrm{B}}(k)}{k^{4}}$
$\displaystyle\times\bigg{\\{}\Theta(q-k)\bigg{[}\left(2q^{2}+k^{2}\right)\ln\frac{q+k}{q-k}+2qk\ln\left(1-\frac{k^{2}}{q^{2}}\right)\mp\frac{4qkT}{k\left(\frac{1}{2}+n_{\mathrm{B}}(k)\right)}\bigg{]}$
$\displaystyle+\Theta(k-q)\bigg{[}\left(2q^{2}+k^{2}\right)\ln\frac{k+q}{k-q}+2qk\ln\frac{k^{2}-q^{2}}{q^{2}}\bigg{]}\bigg{\\}}.$
(D.30)

The strategy is as follows: on the second line we have introduced a
subtraction term, $-4qk$, which renders the integration IR finite, so that we
can do it numerically. We then add back
$+4qkT/k\left(\frac{1}{2}+n_{\mathrm{B}}(k)\right)$ and do it in DR. For the
third line, we instead do the thermal part numerically and the vacuum part
analytically in DR. We then find

$\displaystyle
Z_{gk^{-2}\,F\,\text{pole}}^{(c)}=\frac{g^{2}T_{F}N_{F}T^{2}}{96\pi^{2}}\bigg{[}\underbrace{4.4713}_{q>k\text{
num}}\quad\underbrace{-\frac{96\ln 2}{\pi^{2}}}_{q>k\text{
subtr}}\quad\underbrace{+3.7933}_{k>q\text{ therm.}}$
$\displaystyle+\frac{1}{\epsilon^{2}}+\frac{2\ln\frac{\bar{\mu}}{8\pi T}+24\ln
A+1}{\epsilon}+(48\ln(A)+2)\ln\frac{\bar{\mu}}{8\pi
T}+\frac{12\zeta^{\prime\prime}(2)}{\pi^{2}}+2\ln^{2}\frac{\bar{\mu}}{8\pi
T}-\frac{\pi^{2}}{12}$
$\displaystyle-2\Big{[}2\gamma_{E}(-12\ln(A)+\ln(2\pi))-12(1+2\ln(2\pi))\ln(A)-1+\gamma_{E}^{2}+\ln(2)(\ln(8)-1)$
$\displaystyle+\ln(\pi)\ln(4\pi)\Big{]}\bigg{]}.$ (D.31)

Here $\ln(A)$ is the logarithm of Glaisher’s constant and the terms on the
second, third and and fourth lines come from the DR-evaluated vacuum
contribution for $k>q$.

Going back to Eq. (D.27) the cut piece on the other hand reads

$\displaystyle Z_{gk^{-2}\,F\,\text{cut}}^{(c)}=$
$\displaystyle\frac{g^{2}T_{F}N_{F}}{4\pi^{2}}\int_{K}(2\pi)\mathbb{P}\frac{1+n_{\mathrm{B}}(k^{0})}{k^{3}K^{2}}\int_{0}^{\infty}dq\,n_{\mathrm{F}}(q)\bigg{[}$
$\displaystyle-\left(4q^{2}+k^{2}+k_{0}^{2}\right)\bigg{(}\Theta(k^{2}-(2q-k^{0})^{2})-\Theta(k^{2}-(2q+k^{0})^{2})\bigg{)}$
$\displaystyle+4qk^{0}\bigg{(}\Theta(k^{2}-(2q+k^{0})^{2})+\Theta(k^{2}-(2q-k^{0})^{2})\bigg{)}\bigg{]}$
$\displaystyle=$
$\displaystyle\frac{g^{2}T_{F}N_{F}}{4\pi^{2}}\int_{K}(2\pi)\mathbb{P}\frac{\frac{1}{2}+n_{\mathrm{B}}(k^{0})}{k^{3}K^{2}}\int_{0}^{\infty}dq\,n_{\mathrm{F}}(q)\bigg{[}$
$\displaystyle-\Theta(k^{2}-(2q-k^{0})^{2})\left((2q-k^{0})^{2}+k^{2}\right)+\Theta(k^{2}-(2q+k^{0})^{2})\left((2q+k^{0})^{2}+k^{2}\right)\bigg{]}.$
(D.32)

For the cut piece, the strategy is as follows: we exploit the frequency
evenness to restrict the frequency integral to twice the positive range.

For the $n_{\mathrm{B}}(k^{0})$-proportional part we first perform the $k$
integration analytically, using the PV prescription for the $k=k^{0}$
singularities. The $k^{0}$ and $q$ integrals can then be done numerically and
are free of IR divergences. The $1/2$-proportional part is carried out by
first performing the $k^{0}$ integration, to be then followed by the $k$ and
$q$ ones. The $k$ integration is divergent and needs to be regulated
appropriately. This then yields

$\displaystyle Z_{gk^{-2}\,F\,\text{cut}}^{(c)}=$
$\displaystyle\frac{g^{2}T_{F}N_{F}T^{2}}{96\pi^{2}}\bigg{\\{}-\frac{1}{\epsilon^{2}}-\frac{2\ln\frac{A^{12}\bar{\mu}}{32\pi
T}+1}{\epsilon}-2\ln^{2}\frac{A^{12}\bar{\mu}}{32\pi
T}-2\ln\frac{A^{12}\bar{\mu}}{32\pi T}$
$\displaystyle-\frac{12\zeta^{\prime\prime}(2)}{\pi^{2}}+\frac{72\zeta^{\prime}(2)^{2}}{\pi^{4}}+\frac{7\pi^{2}}{12}-6+8\ln^{2}2-4\ln
2+1.70118\bigg{\\}}.$ (D.33)

Here all terms come from the $1/2$-proportional vacuum part with the exception
of the final term, coming from the numerical integrations. Summing the cut and
pole contribution yields

$\displaystyle Z_{g\,k^{-2}\,F\,\text{ non-HTL}}^{(c)}=$
$\displaystyle\frac{g^{2}T_{F}N_{F}T^{2}}{96\pi^{2}}\bigg{\\{}\underbrace{4.4713}_{q>k\text{
num}}\quad\underbrace{-\frac{96\ln 2}{\pi^{2}}}_{q>k\text{
subtr}}\quad\underbrace{+3.7933}_{k>q\text{ therm.}}+1.70118$
$\displaystyle+\frac{4\ln 2}{\epsilon}+8\ln 2\ln\frac{A^{12}\bar{\mu}}{8\pi
T}+\frac{\pi^{2}}{2}-4+2\ln(2)(1-2\ln 2)\bigg{\\}}$ (D.34)

The total contribution from the $k^{-2}$-proportional part in the fermion
contribution is, from Eqs. (D.22), (D.26), and (D.34) therefore

$\displaystyle Z_{g\,k^{-2}\,F}^{(c)}=$
$\displaystyle\frac{g^{2}T_{F}N_{F}T^{2}}{96\pi^{2}}\bigg{\\{}-3\left(4\ln(2)\ln\left(\frac{2}{A^{8}}\right)+2\ln(2)\right)+\frac{\pi^{2}}{6}-8\gamma_{E}\ln
2$ $\displaystyle+4.4713-\frac{96\ln 2}{\pi^{2}}+3.7933+1.70118\bigg{\\}}.$
(D.35)

We can thus see that the entire $k^{-2}$-proportional contribution is finite.
The result has been confirmed, by numerically performing the sum over the
Matusbara modes associated with $K$.

####  D.3.2 Fermionic Contribution: $\Pi_{L}-\Pi_{T}$ Piece

Let us start from the HTL subset of the full contribution

$Z_{g\,F\,\text{ L-T
HTL}}^{(c)}=i\int_{K}\frac{1+n_{\mathrm{B}}(k^{0})}{(k^{-}-i\varepsilon)^{2}}\bigg{[}\frac{k_{\perp}^{2}(\Pi_{L}^{R}(K)-\Pi_{T}^{R}(K))_{\mathrm{HTL}}}{k^{2}(K^{2}+i\varepsilon
k^{0})}-\text{adv}\bigg{]}.$ (D.36)

From Eq. (D.17) we then have

$\displaystyle\Pi_{L}^{R}(K)_{F\,\mathrm{HTL}}-\Pi_{T}^{R}(K)_{F\,\mathrm{HTL}}=$
$\displaystyle\frac{g^{2}T_{F}N_{F}T^{2}}{12}\bigg{[}-\frac{(6k_{0}^{2}-4k^{2})}{k^{2}}+3\frac{k^{0}K^{2}}{k^{3}}\ln\frac{k^{0}+k+i\varepsilon}{k^{0}-k+i\varepsilon}\bigg{]}.$
(D.37)

The pole contribution is then

$Z_{g\,F\,\text{ L-T HTL
pole}}^{(c)}=-\frac{g^{2}T_{F}N_{F}T^{2}}{12}\int_{K}\frac{\frac{1}{2}+n_{\mathrm{B}}(|k^{0}|)}{(k^{-}-i\varepsilon)^{2}}\frac{(6k_{0}^{2}-4k^{2})}{k^{2}}\bigg{[}\frac{k_{\perp}^{2}}{k^{2}}2\pi\delta(K^{2})\bigg{]}.$
(D.38)

The vacuum part is scale-free, so in DR, we have

$Z_{g\,F\,\text{ L-T HTL
pole}}^{(c)}=-\frac{g^{2}T_{F}N_{F}T^{2}}{6}\int_{k}\frac{n_{\mathrm{B}}(k)}{2k}\bigg{[}\frac{1}{(k-k^{z}-i\varepsilon)^{2}}+\frac{1}{(-k-k^{z}-i\varepsilon)^{2}}\bigg{]}\frac{k_{\perp}^{2}}{k^{2}}.$
(D.39)

We drop the $i\varepsilon$ prescription, because DR can make the $x$
integration convergent in the appropriate number of dimensions, i.e.

$\displaystyle Z_{g\,F\,\text{ L-T HTL pole}}^{(c)}=$
$\displaystyle-\frac{g^{2}T_{F}N_{F}T^{2}}{6}\Big{(}\frac{\bar{\mu}^{2}e^{\gamma_{E}}}{4\pi}\Big{)}^{\epsilon}\frac{2\pi^{d/2-1/2}}{\Gamma(d/2-1/2)(2\pi)^{d}}\int_{0}^{\infty}dk\,k^{d-4}n_{\mathrm{B}}(k)$
$\displaystyle\times\int_{-1}^{1}dx(1-x^{2})^{d/2-3/2}\frac{1+x^{2}}{1-x^{2}}.$
$\displaystyle=$
$\displaystyle-\frac{g^{2}T_{F}N_{F}T^{2}}{6}\Big{(}\frac{\bar{\mu}^{2}e^{\gamma_{E}}}{4\pi}\Big{)}^{\epsilon}\frac{2\pi^{d/2-1/2}}{\Gamma(d/2-1/2)(2\pi)^{d}}T^{d-3}\zeta(d-3)\Gamma(d-3)$
$\displaystyle\frac{\sqrt{\pi}(d-1)\Gamma\left(\frac{d-3}{2}\right)}{2\Gamma\left(\frac{d}{2}\right)}$
$\displaystyle=\frac{g^{2}T_{F}N_{F}T^{2}}{48\pi^{2}}\bigg{[}\frac{1}{\epsilon^{2}}+\frac{2\ln\frac{\bar{\mu}\exp(\gamma_{E}+1/2)}{4\pi
T}}{\epsilon}+2\ln^{2}\frac{\bar{\mu}\exp(\gamma_{E}+1/2)}{4\pi
T}-4\gamma_{1}+\frac{\pi^{2}}{4}-2\gamma_{E}^{2}+\frac{3}{2}\bigg{]}.$ (D.40)

The cut contribution of the HTL piece is instead

$Z_{g\,F\,\text{ L-T HTL
cut}}^{(c)}=\frac{g^{2}T_{F}N_{F}T^{2}}{4}\int_{K}\frac{1+n_{\mathrm{B}}(k^{0})}{(k^{-}-i\varepsilon)^{2}}\frac{k_{\perp}^{2}k^{0}}{k^{5}}2\pi\Theta(k^{2}-k_{0}^{2}).$
(D.41)

To simplify things, let us symmetrise the frequency integration

$Z_{g\,F\,\text{ L-T HTL
cut}}^{(c)}=\frac{g^{2}T_{F}N_{F}T^{2}}{4}\int_{K}\frac{k_{\perp}^{2}k^{0}}{k^{5}}2\pi\Theta(k^{2}-k_{0}^{2})\frac{1}{2}\bigg{[}\frac{1+n_{\mathrm{B}}(k^{0})}{(k^{0}-k^{z}-i\varepsilon)^{2}}-\frac{-n_{\mathrm{B}}(k^{0})}{(-k^{0}-k^{z}-i\varepsilon)^{2}}\bigg{]}.$
(D.42)

As this integral is now even, we can restrict over twice the positive range,
i.e.

$Z_{g\,F\,\text{ L-T HTL
cut}}^{(c)}=\frac{g^{2}T_{F}N_{F}T^{2}}{4}\int_{0}^{\infty}\frac{dk^{0}}{2\pi}\int_{k}\frac{k_{\perp}^{2}k^{0}}{k^{5}}2\pi\Theta(k^{2}-k_{0}^{2})\bigg{[}\frac{1+n_{\mathrm{B}}(k^{0})}{(k^{0}-k^{z}-i\varepsilon)^{2}}+\frac{n_{\mathrm{B}}(k^{0})}{(k^{0}+k^{z}+i\varepsilon)^{2}}\bigg{]}.$
(D.43)

The $n_{\mathrm{B}}$-independent part is now scale-free, while the rest can be
done in a somewhat similar way to the pole piece

$\displaystyle Z_{g\,F\,\text{ L-T HTL
cut}}^{(c)}=\frac{g^{2}T_{F}N_{F}T^{2}}{4}\Big{(}\frac{\bar{\mu}^{2}e^{\gamma_{E}}}{4\pi}\Big{)}^{\epsilon}\frac{2\pi^{d/2-1/2}}{\Gamma(d/2-1/2)(2\pi)^{d-1}}\int_{0}^{\infty}\frac{dk^{0}}{2\pi}k^{0}n_{\mathrm{B}}(k^{0})$
$\displaystyle\times\int_{k_{0}}^{\infty}dkk^{d-4}\int_{-1}^{1}dx(1-x^{2})^{d/2-1/2}\bigg{[}\frac{1}{(k^{0}-kx-i\varepsilon)^{2}}+\frac{1}{(k^{0}+kx+i\varepsilon)^{2}}\bigg{]}.$
(D.44)

Our strategy here is to do the following subtraction:
$n_{\mathrm{B}}(k^{0})\to
n_{\mathrm{B}}(k^{0})\mp(T/k^{0}-\Theta(T-k^{0})/2)$. In this way, the
subtraction is finite, and then we re-evaluate the addition in DR. For the
finite part, we can first perform the angular and momentum integration, i.e.

$\displaystyle Z_{g\,F\,\text{ L-T HTL cut
finite}}^{(c)}=\frac{g^{2}T_{F}N_{F}T^{2}}{4}\frac{1}{2\pi}\int_{0}^{\infty}\frac{dk^{0}}{2\pi}k^{0}\bigg{[}n_{\mathrm{B}}(k^{0})-\frac{T}{k^{0}}+\frac{\Theta(T-k^{0})}{2}\bigg{]}$
$\displaystyle\times\int_{k^{0}}^{\infty}\frac{dk}{k}\int_{-1}^{1}dx(1-x^{2})\bigg{[}\frac{1}{(k^{0}-kx-i\varepsilon)^{2}}+\frac{1}{(k^{0}+kx+i\varepsilon)^{2}}\bigg{]}$
$\displaystyle=$
$\displaystyle\frac{g^{2}T_{F}N_{F}T^{2}}{4}\frac{1}{2\pi}\int_{0}^{\infty}\frac{dk^{0}}{2\pi}k^{0}\bigg{[}n_{\mathrm{B}}(k^{0})-\frac{T}{k^{0}}+\frac{\Theta(T-k^{0})}{2}\bigg{]}$
$\displaystyle\times\int_{k^{0}}^{\infty}\frac{dk}{k}\bigg{[}\frac{4k^{0}\tanh^{-1}\left(\frac{2kk^{0}}{k^{2}+k_{0}^{2}}\right)}{k^{3}}-\frac{8}{k^{2}}\bigg{]}$
$\displaystyle=$ $\displaystyle\frac{2g^{2}T_{F}N_{F}T^{2}}{3}(\ln
2-1)\frac{1}{(2\pi)^{2}}\int_{0}^{\infty}\frac{dk^{0}}{k^{0}}\bigg{[}n_{\mathrm{B}}(k^{0})-\frac{T}{k^{0}}+\frac{\Theta(T-k^{0})}{2}\bigg{]}$
$\displaystyle=$
$\displaystyle\frac{g^{2}T_{F}N_{F}T^{2}}{12\pi^{2}}\ln\frac{T}{\omega_{\text{T}}}(\ln
2-1),$ (D.45)

where $\omega_{\text{T}}=2\pi T\exp(-\gamma_{E})$. For the DR part subtracted
part we can instead do frequency first and forget about the zero-mode
subtraction, because it will be scale-free. Then

$\displaystyle Z_{g\,F\,\text{ L-T HTL cut
div}}^{(c)}=-\frac{g^{2}T_{F}N_{F}T^{2}}{4}\Big{(}\frac{\bar{\mu}^{2}e^{\gamma_{E}}}{4\pi}\Big{)}^{\epsilon}\frac{2\pi^{d/2-1/2}}{\Gamma(d/2-1/2)(2\pi)^{d-1}}\int_{0}^{\infty}dkk^{d-4}$
$\displaystyle\times\int_{0}^{k}\frac{dk^{0}}{2\pi}k^{0}\frac{\Theta(T-k^{0})}{2}\int_{-1}^{1}dx(1-x^{2})^{d/2-1/2}\bigg{[}\frac{1}{(k^{0}-kx-i\varepsilon)^{2}}+\frac{1}{(k^{0}+kx+i\varepsilon)^{2}}\bigg{]}.$
(D.46)

We can continue splitting this in the $k>T$ and $k<T$ ranges, i.e.

$\displaystyle Z_{g\,F\,\text{ L-T HTL cut
div}}^{(c)}=-\frac{g^{2}T_{F}N_{F}T^{2}}{8}\Big{(}\frac{\bar{\mu}^{2}e^{\gamma_{E}}}{4\pi}\Big{)}^{\epsilon}\frac{2\pi^{d/2-1/2}}{\Gamma(d/2-1/2)(2\pi)^{d}}$
$\displaystyle\times\int_{0}^{\infty}dkk^{d-4}\int_{-1}^{1}dx(1-x^{2})^{d/2-1/2}\bigg{\\{}\Theta(T-k)\bigg{[}-\mathbb{P}\frac{2}{1-x^{2}}+\ln\frac{1-x^{2}}{x^{2}}\bigg{]}$
$\displaystyle+\Theta(k-T)\bigg{[}\ln\frac{\left|T^{2}-k^{2}x^{2}\right|}{k^{2}x^{2}}-\mathbb{P}\frac{2T^{2}}{T^{2}-k^{2}x^{2}}\bigg{]}\bigg{\\}},.$
(D.47)

where we have used the fact that the imaginary parts of the frequency integral
are odd in $x$ and thus vanish. The principal value arises from the
$\varepsilon\to 0$ limit. The $k>T$ slice is finite, since the $k$ integration
is IR divergent, while the $T>k$ part is IR divergent and needs
regularisation. We then have

$\displaystyle Z_{g\,F\,\text{ L-T HTL cut
div}}^{(c)}=-\frac{g^{2}T_{F}N_{F}T^{2}}{8}\Big{(}\frac{\bar{\mu}^{2}e^{\gamma_{E}}}{4\pi}\Big{)}^{\epsilon}\frac{2\pi^{d/2-1/2}}{\Gamma(d/2-1/2)(2\pi)^{d}}$
$\displaystyle\times\bigg{\\{}\int_{-1}^{1}dx(1-x^{2})^{d/2-1/2}\frac{T^{d-3}}{d-3}\bigg{[}-\mathbb{P}\frac{2}{1-x^{2}}+\ln\frac{1-x^{2}}{x^{2}}\bigg{]}$
$\displaystyle+\int_{-1}^{1}dx(1-x^{2})\bigg{[}\frac{\text{Li}_{2}\left(x^{2}\right)}{2}+\ln\left(\frac{x^{2}}{1-x^{2}}\right)+\ln^{2}|x|-\frac{\pi^{2}}{6}\bigg{]}\bigg{\\}},$
(D.48)

which yields

$\displaystyle Z_{g\,F\,\text{ L-T HTL cut
div}}^{(c)}=-\frac{g^{2}T_{F}N_{F}T^{2}}{8}\Big{(}\frac{\bar{\mu}^{2}e^{\gamma_{E}}}{4\pi}\Big{)}^{\epsilon}\frac{2\pi^{d/2-1/2}}{\Gamma(d/2-1/2)(2\pi)^{d}}$
$\displaystyle\times\bigg{\\{}\frac{T^{d-3}}{d-3}\frac{\sqrt{\pi}\Gamma\left(\frac{d}{2}+\frac{1}{2}\right)\left(\psi^{(0)}\left(\frac{d}{2}-\frac{1}{2}\right)+\gamma-2+2\ln(2)\right)}{\Gamma\left(\frac{d}{2}+1\right)}$
$\displaystyle+\frac{1}{9}\left(-8-\pi^{2}+8\ln 2)\right)\bigg{\\}}$
$\displaystyle=$
$\displaystyle-\frac{g^{2}T_{F}N_{F}T^{2}}{8}\bigg{[}\frac{1-\ln
2}{3\pi^{2}\epsilon}+\frac{(1-\ln
2)(2\ln\frac{\bar{\mu}}{2T}+1)}{3\pi^{2}}\bigg{]},$ (D.49)

so that finally

$Z_{g\,F\,\text{ L-T HTL
cut}}^{(c)}=\frac{g^{2}T_{F}N_{F}T^{2}}{24\pi^{2}}(\ln(2)-1)\bigg{[}\frac{1}{\epsilon}+2\ln\frac{\bar{\mu}\exp(\gamma_{E}+1/2)}{4\pi
T}\bigg{]}.$ (D.50)

Let us now consider the non-HTL piece, i.e

$\displaystyle\delta\Pi_{L}^{R}(K)_{F}-\delta\Pi_{T}^{R}(K)_{F}=\frac{g^{2}T_{F}N_{F}K^{2}}{8\pi^{2}k^{3}}\int_{0}^{\infty}dq\,n_{\mathrm{F}}(q)\bigg{[}24qk$
$\displaystyle+\left(12q^{2}-k^{2}+3k_{0}^{2}\right)\bigg{[}\ln\frac{k^{0}+k-2q+i\varepsilon}{k^{0}-k-2q+i\varepsilon}+\ln\frac{k^{0}-k+2q+i\varepsilon}{k^{0}+k+2q+i\varepsilon}\bigg{]}$
$\displaystyle+12qk^{0}\bigg{[}\ln\frac{k^{0}-k+2q+i\varepsilon}{k^{0}+k+2q+i\varepsilon}-\ln\frac{k^{0}+k-2q+i\varepsilon}{k^{0}-k-2q+i\varepsilon}\bigg{]}\bigg{]},$
(D.51)

where $\delta\Pi=\Pi-\Pi_{\text{HTL}}$. This then gives

$\displaystyle Z_{g\,F\,\text{ L-T non-
HTL}}^{(c)}=\frac{g^{2}T_{F}N_{F}}{4\pi}\int_{0}^{\infty}dq\,n_{\mathrm{F}}(q)\int_{K}\frac{\frac{1}{2}+n_{\mathrm{B}}(k^{0})}{(k^{-}-i\varepsilon)^{2}}\frac{k_{\perp}^{2}}{k^{5}}$
$\displaystyle\times\bigg{[}\Theta(k^{2}-(k^{0}-2q)^{2})\left(3(2q-k^{0})^{2}-k^{2}\right)-\Theta(k^{2}-(k^{0}+2q)^{2})\left(3(2q+k^{0})^{2}-k^{2}\right)\bigg{]}.$
(D.52)

Here we used the fact that the 1/2 piece vanishes to restrict to the purely
odd $1/2+n_{\mathrm{B}}(k^{0})$ component. Let us again symmetrise and
restrict to twice the positive frequency range:

$\displaystyle Z_{g\,F\,\text{ L-T non-
HTL}}^{(c)}=\frac{g^{2}T_{F}N_{F}}{4\pi}\int_{0}^{\infty}dq\,n_{\mathrm{F}}(q)\int_{0}^{\infty}\frac{dk^{0}}{2\pi}\int_{k}\frac{k_{\perp}^{2}}{k^{5}}\bigg{[}\frac{\frac{1}{2}+n_{\mathrm{B}}(k^{0})}{(k^{0}-k^{z}-i\varepsilon)^{2}}+\frac{\frac{1}{2}+n_{\mathrm{B}}(k^{0})}{(k^{0}+k^{z}+i\varepsilon)^{2}}\bigg{]}$
$\displaystyle\times\bigg{[}\Theta(k^{2}-(k^{0}-2q)^{2})\left(3(2q-k^{0})^{2}-k^{2}\right)-\Theta(k^{2}-(k^{0}+2q)^{2})\left(3(2q+k^{0})^{2}-k^{2}\right)\bigg{]}.$
(D.53)

This corresponds to

$\displaystyle Z_{g\,F\,\text{ L-T non-
HTL}}^{(c)}=\frac{g^{2}T_{F}N_{F}}{4\pi}\int_{0}^{\infty}dq\,n_{\mathrm{F}}(q)\int_{0}^{\infty}\frac{dk^{0}}{2\pi}\int_{k}\frac{k_{\perp}^{2}}{k^{5}}\bigg{[}\frac{1/2+n_{\mathrm{B}}(k^{0})}{(k^{0}-k^{z}-i\varepsilon)^{2}}+\frac{1/2+n_{\mathrm{B}}(k^{0})}{(k^{0}+k^{z}+i\varepsilon)^{2}}\bigg{]}$
$\displaystyle\times\bigg{\\{}\Theta(k-2q)\bigg{[}\left(3(2q-k^{0})^{2}-k^{2}\right)\Theta(2q+k-k^{0})-\left(3(2q+k^{0})^{2}-k^{2}\right)\Theta(k-2q-k^{0})\bigg{]}$
$\displaystyle+\Theta(2q-k)\left(3(2q-k^{0})^{2}-k^{2}\right)\Theta(2q+k-k^{0}))\Theta(k^{0}+k-2q)\bigg{\\}}.$
(D.54)

We start by dealing with the vacuum part

$\displaystyle Z_{g\,F\,\text{ L-T non-HTL
vac}}^{(c)}=\frac{g^{2}T_{F}N_{F}}{8\pi}\int_{0}^{\infty}dq\,n_{\mathrm{F}}(q)\int_{0}^{\infty}\frac{dk^{0}}{2\pi}\int_{k}\frac{k_{\perp}^{2}}{k^{5}}\bigg{[}\frac{1}{(k^{0}-k^{z}-i\varepsilon)^{2}}+\frac{1}{(k^{0}+k^{z}+i\varepsilon)^{2}}\bigg{]}$
$\displaystyle\times\bigg{\\{}\Theta(k-2q)\bigg{[}\left(3(2q-k^{0})^{2}-k^{2}\right)\Theta(k+2q-k^{0})-\left(3(2q+k^{0})^{2}-k^{2}\right)\Theta(k-2q-k^{0})\bigg{]}$
$\displaystyle+\Theta(2q-k)\left(3(2q-k^{0})^{2}-k^{2}\right)\Theta(2q+k-k^{0})\Theta(k^{0}-2q+k)\bigg{\\}},$
(D.55)

which we can simplify as

$\displaystyle Z_{g\,F\,\text{ L-T non-HTL
vac}}^{(c)}=\frac{g^{2}T_{F}N_{F}}{8\pi}\int_{0}^{\infty}dq\,n_{\mathrm{F}}(q)\int_{0}^{\infty}\frac{dk^{0}}{2\pi}\int_{k}\frac{k_{\perp}^{2}}{k^{5}}\bigg{[}\frac{1}{(k^{0}-k^{z}-i\varepsilon)^{2}}+\frac{1}{(k^{0}+k^{z}+i\varepsilon)^{2}}\bigg{]}$
$\displaystyle\times\bigg{\\{}\Theta(k-2q)\bigg{[}\left(3(2q-k^{0})^{2}-k^{2}\right)\Theta(k+2q-k^{0})\Theta(k^{0}-k+2q)-24qk^{0}\Theta(k-2q-k^{0})\bigg{]}$
$\displaystyle+\Theta(2q-k)\left(3(2q-k^{0})^{2}-k^{2}\right)\Theta(2q+k-k^{0})\Theta(k^{0}-2q+k)\bigg{\\}},$
(D.56)

The strategy here is the following: we further symmetrise the expression in
$k^{z}$, to perform the $k^{0}$ integrations. After that one can take
$\varepsilon\to 0$. Logarithmic pieces will see their argument change to its
absolute value, while possible imaginary parts will cancel out due to $k^{z}$
symmetrisation (in essence, the imaginary part is odd in $k^{z}$ and thus
vanishing). Non-logarithmic pieces present poles on the $k^{z}$ integration
range. If we restrict to twice the positive $k^{z}$ range, these occur at
$-|2q-k|+k^{z}=0$ and are turned into a PV prescription by the $\varepsilon\to
0$ limit. We can thus proceed to the $k^{z}$ integral in the two ranges $2q>k$
and $k>2q$. In the latter case, we find two subranges, one for $k_{\perp}<2q$
and one for $k_{\perp}>2q$. This last one has to be carried out in DR, since
it contains a UV divergence. We then find

$\displaystyle Z_{g\,F\,\text{ L-T non-HTL
vac}}=\frac{g^{2}T_{F}N_{F}T^{2}}{48\pi^{2}}\bigg{\\{}\frac{1}{\epsilon^{2}}+\frac{2}{\epsilon}\ln\frac{\bar{\mu}A^{12}e^{-1/2}}{16\pi
T}+2\ln^{2}\frac{\bar{\mu}A^{12}e^{-1/2}}{16\pi T}$
$\displaystyle+48\ln(A)(\gamma_{E}+\ln(2\pi)-6\ln(A))+\frac{12\zeta^{\prime\prime}(2)}{\pi^{2}}+\frac{1}{2}\text{Li}_{2}\left(\frac{2}{3}\right)+\frac{1}{4}\text{Li}_{2}\left(\frac{3}{4}\right)$
$\displaystyle-\frac{1}{2}-\frac{11}{2}\ln^{2}(2)-2\ln(\pi)\ln(4\pi)-2\gamma_{E}\left(\gamma_{E}+\ln\left(4\pi^{2}\right)\right)-\frac{1}{2}\ln(3)\coth^{-1}(7)\bigg{\\}}.$
(D.57)

For the thermal part we have instead

$\displaystyle Z_{g\,F\,\text{ L-T non-HTL
th}}^{(c)}=\frac{g^{2}T_{F}N_{F}}{2\pi}\int_{0}^{\infty}dq\,n_{\mathrm{F}}(q)\int_{0}^{\infty}\frac{dk^{0}}{2\pi}\int_{0}^{\infty}\frac{dk}{4\pi^{2}}\frac{n_{\mathrm{B}}(k^{0})}{k}$
$\displaystyle\times\bigg{[}\frac{2k^{0}\tanh^{-1}\left(\frac{2kk^{0}}{k^{2}+k_{0}^{2}}\right)}{k^{3}}-\frac{4}{k^{2}}\bigg{]}$
$\displaystyle\times\bigg{\\{}\Theta(k-2q)\bigg{[}\left(3(2q-k^{0})^{2}-k^{2}\right)\Theta(2q+k-k^{0})-\left(3(2q+k^{0})^{2}-k^{2}\right)\Theta(k-2q-k^{0})\bigg{]}$
$\displaystyle+\Theta(2q-k)\left(3(2q-k^{0})^{2}-k^{2}\right)\Theta(2q+k-k^{0}))\Theta(k^{0}+k-2q)\bigg{\\}}.$
(D.58)

Here we instead perform $k$ first analytically, followed by $k^{0}$ and $q$
numerically. We find

$Z_{g\,F\,\text{ L-T non-HTL
th}}^{(c)}=\frac{g^{2}T_{F}N_{F}T^{2}}{48\pi^{2}}\bigg{[}\underbrace{-0.438129}_{k<2q}+\underbrace{1.97382}_{k>2q}\bigg{]}.$
(D.59)

Putting everything together we find

$\displaystyle Z_{g\,F\,\text{ L-T
}}^{(c)}=\frac{g^{2}T_{F}N_{F}T^{2}}{48\pi^{2}}\bigg{\\{}\frac{2}{\epsilon^{2}}+\frac{2\ln\frac{\bar{\mu}\exp(\gamma_{E}+1/2)}{4\pi
T}+2(\ln(2)-1)+2\ln\frac{\bar{\mu}A^{12}e^{-1/2}}{16\pi T}}{\epsilon}$
$\displaystyle+2\ln^{2}\frac{\bar{\mu}\exp(\gamma_{E}+1/2)}{4\pi
T}-4\gamma_{1}+\frac{\pi^{2}}{4}-2\gamma_{E}^{2}+\frac{3}{2}$
$\displaystyle+2(\ln(2)-1)\bigg{[}2\ln\frac{\bar{\mu}\exp(\gamma_{E}+1/2)}{4\pi
T}\bigg{]}$ $\displaystyle+2\ln^{2}\frac{\bar{\mu}A^{12}e^{-1/2}}{16\pi T}$
$\displaystyle+48\ln(A)(\gamma_{E}+\ln(2\pi)-6\ln(A))+\frac{12\zeta^{\prime\prime}(2)}{\pi^{2}}+\frac{1}{2}\text{Li}_{2}\left(\frac{2}{3}\right)+\frac{1}{4}\text{Li}_{2}\left(\frac{3}{4}\right)$
$\displaystyle-\frac{1}{2}-\frac{11}{2}\ln^{2}(2)-2\ln(\pi)\ln(4\pi)-2\gamma_{E}\left(\gamma_{E}+\ln\left(4\pi^{2}\right)\right)-\frac{1}{2}\ln(3)\coth^{-1}(7)$
$\displaystyle\underbrace{-0.438129}_{k<2q}+\underbrace{1.97382}_{k>2q}\bigg{\\}}.$
(D.60)

In contrast to the $k^{-2}$ piece, this contribution is indeed divergent.
Moreover, we hammer home the point that these divergences _are not_ cancelled
by opposing ones coming from the soft sector, as EQCD is a theory of gluons
only.

####  D.3.3 Gluonic Contribution: $k^{-2}$ Piece

We persevere with the full contribution to diagram $(c)$, starting with the
gluonic $k^{-2}$ piece, as we did for the fermionic counterpart in Sec. D.3.1.

Our starting point is Eq. (6.47): we need to be aware that there may be
divergences, other than the zero-mode contribution, Eq. (6.49). Upon computing
the contribution of the non-zero Matsubara modes numerically, we see a
(logarithmic) divergence appearing. In order to isolate the finite
contribution, we perform a subtraction on Eq. (6.47) so that we are
essentially left with the equivalent fermionic expression (up to the factor of
$n_{B}(k)$), which was finite. This leaves us with

$\displaystyle Z^{(c)}_{g\,k^{-2}\,G\,n\neq
0\,\text{finite}}=-\frac{g^{2}C_{A}T}{8\pi^{2}}\sum_{n\neq
0}\int_{k}\int_{0}^{\infty}dq\frac{1}{k^{3}}n_{\text{B}}(q)\frac{1}{(-\omega_{n}^{2}-k^{2})^{2}}\Bigg{(}8qk(k^{2}-\omega_{n}^{2})$
$\displaystyle-(-\omega_{n}^{2}-k^{2})(4q^{2}+k^{2}-\omega_{n}^{2})\left(\ln\frac{i\omega_{n}+k-2q}{i\omega_{n}-k-2q}+\ln\frac{i\omega_{n}-k+2q}{i\omega_{n}+k+2q}\right)$
$\displaystyle-4iq\omega_{n}(-\omega_{n}^{2}-k^{2})\left(2\ln\frac{i\omega_{n}+k}{i\omega_{n}-k}+\ln\frac{i\omega_{n}-k+2q}{i\omega_{n}+k+2q}-\ln\frac{i\omega_{n}+k-2q}{i\omega_{n}-k-2q}\right)\Bigg{)}$
$\displaystyle=-\frac{g^{2}C_{A}T^{2}}{32\pi^{2}}\times 0.00380658$ (D.61)

The divergent part then reads

$\displaystyle Z^{(c)}_{g\,k^{-2}\,G\,n\neq 0\,\text{div}}$
$\displaystyle=-\frac{g^{2}C_{A}T}{4\pi^{2}}\sum_{n\neq
0}\int_{k}\int_{0}^{\infty}dq\frac{n_{B}(q)}{k(\omega_{n}^{2}+k^{2})}\left(\ln\frac{i\omega_{n}+k-2q}{i\omega_{n}-k-2q}+\ln\frac{i\omega_{n}-k+2q}{i\omega_{n}+k+2q}\right).$
(D.62)

This can be integrated by rephrasing it back in Minkowskian signature and
separating $1/2+n_{\mathrm{B}}(|k^{0}|)$ into a vacuum part, $1/2$, and a
zero-mode subtracted thermal part $n_{\mathrm{B}}(|k^{0}|)-T/|k^{0}|$, so as
to avoid double-counting the zero mode. The integration proceeds in much the
same way as for the fermion case except here, the vacuum part is UV divergent.
The total reads

$\displaystyle Z^{(c)}_{g\,k^{-2}\,G\,n\neq 0\,\text{div}}$
$\displaystyle=\frac{g^{2}C_{A}T^{2}}{48\pi^{2}}\left(\frac{1}{\epsilon}-2\ln\frac{4\pi
T}{A^{12}\bar{\mu}}-2.44147-2.44091\right),$ (D.63)

where the last two numbers come from the $n_{\mathrm{B}}(|k^{0}|)-T/|k^{0}|$
contribution in the pole and cut channels respectively. Unlike the fermion
$k^{-2}$ piece, the gluonic one contains a divergence coming from both the
zero-mode and the non-zero modes.

Note that the gluonic “no log” equivalent (see Eq. (D.22)) is split here
between the zero-mode contribution, Eq. (6.49) and finite contribution from
the non-zero modes, Eq. (6.74)222It can nevertheless easily be obtained
analytically by making the replacement $2N_{F}T_{F}\rightarrow C_{A}$ in Eq.
(D.22)..

####  D.3.4 Gluonic Contribution: $\Pi_{L}-\Pi_{T}$ Piece

For the HTL part, the $q$ integrand is given by $q\,n_{\text{B}}(q)$.
Therefore, given that

$\int_{0}^{\infty}dqqn_{\text{B}}(q)=2\int_{0}^{\infty}dqqn_{\text{F}}(q)=\frac{\pi^{2}T}{6},$
(D.64)

we may simply make the replacement $2N_{F}T_{F}\rightarrow C_{A}$ in Eq.
(D.26) to obtain the gluonic equivalent. This yields

$\displaystyle Z_{g\,\text{G L-T HTL}}^{(c)}=$
$\displaystyle\frac{g^{2}C_{A}T^{2}}{48\pi^{2}}\bigg{[}\frac{1}{\epsilon^{2}}+\frac{2\ln\frac{\bar{\mu}\exp(\gamma_{E}+1/2)}{4\pi
T}}{\epsilon}+2\ln^{2}\frac{\bar{\mu}\exp(\gamma_{E}+1/2)}{4\pi
T}-4\gamma_{1}+\frac{\pi^{2}}{4}-2\gamma_{E}^{2}+\frac{3}{2}\bigg{]}$
$\displaystyle+\frac{g^{2}C_{A}T^{2}}{24\pi^{2}}(\ln(2)-1)\bigg{[}\frac{1}{\epsilon}+2\ln\frac{\bar{\mu}\exp(\gamma_{E}+1/2)}{4\pi
T}\bigg{]}.$ (D.65)

For the vacuum part of the non-HTL contribution we have instead

$\displaystyle Z_{g\,\text{G L-T non-HTL
vac}}^{(c)}=\frac{g^{2}C_{A}}{16\pi}\int_{0}^{\infty}dq\,n_{\mathrm{B}}(q)\int_{0}^{\infty}\frac{dk^{0}}{2\pi}\int_{k}\frac{k_{\perp}^{2}}{k^{5}}\bigg{[}\frac{1}{(k^{0}-k^{z}-i\varepsilon)^{2}}+\frac{1}{(k^{0}+k^{z}+i\varepsilon)^{2}}\bigg{]}$
$\displaystyle\times\bigg{\\{}\Theta(k-2q)\bigg{[}\left(3(2q-k^{0})^{2}-k^{2}\right)\Theta(k+2q-k^{0})\Theta(k^{0}-k+2q)-24qk^{0}\Theta(k-2q-k^{0})\bigg{]}$
$\displaystyle+\Theta(2q-k)\left(3(2q-k^{0})^{2}-k^{2}\right)\Theta(2q+k-k^{0})\Theta(k^{0}-2q+k)\bigg{\\}}.$
(D.66)

Using the methods of Sec. D.3.2 this gives

$\displaystyle Z_{g\,\text{G L-T non-HTL
vac}}^{(c)}=\frac{g^{2}C_{A}T^{2}}{48\pi^{2}}\bigg{\\{}\frac{1}{\epsilon^{2}}+\frac{2}{\epsilon}\ln\frac{\bar{\mu}A^{12}e^{-1/2}}{8\pi
T}+2\ln^{2}\frac{\bar{\mu}A^{12}e^{-1/2}}{8\pi T}$
$\displaystyle+48\ln(A)(\gamma_{E}+\ln(2\pi)-6\ln(A))+\frac{12\zeta^{\prime\prime}(2)}{\pi^{2}}+\frac{1}{2}\text{Li}_{2}\left(\frac{2}{3}\right)+\frac{1}{4}\text{Li}_{2}\left(\frac{3}{4}\right)$
$\displaystyle-\frac{1}{2}-\frac{3}{2}\ln^{2}(2)-2\ln(\pi)\ln(4\pi)-2\gamma_{E}\left(\gamma_{E}+\ln\left(4\pi^{2}\right)\right)-\frac{1}{2}\ln(3)\coth^{-1}(7)\bigg{\\}}.$
(D.67)

For the thermal part we have instead to subtract the zero mode, which is
evaluated in Eq. (6.53). No zero-mode subtraction is necessary in the HTL
part, since the zero mode is scale-free there. This yields

$\displaystyle Z_{g\,\text{G L-T non-HTL
th}}^{(c)}=\frac{g^{2}C_{A}}{4\pi}\int_{0}^{\infty}dq\,n_{\mathrm{B}}(q)\int_{0}^{\infty}\frac{dk^{0}}{2\pi}\int_{0}^{\infty}\frac{dk}{4\pi^{2}}\frac{1}{k}\bigg{[}n_{\mathrm{B}}(k^{0})-\frac{T}{k^{0}}\bigg{]}$
$\displaystyle\times\bigg{[}\frac{2k^{0}\tanh^{-1}\left(\frac{2kk^{0}}{k^{2}+k_{0}^{2}}\right)}{k^{3}}-\frac{4}{k^{2}}\bigg{]}\bigg{\\{}\Theta(k-2q)\bigg{[}\left(3(2q-k^{0})^{2}-k^{2}\right)\Theta(2q+k-k^{0})$
$\displaystyle-\left(3(2q+k^{0})^{2}-k^{2}\right)\Theta(k-2q-k^{0})\bigg{]}$
$\displaystyle+\Theta(2q-k)\left(3(2q-k^{0})^{2}-k^{2}\right)\Theta(2q+k-k^{0}))\Theta(k^{0}+k-2q)\bigg{\\}}.$
(D.68)

Here we instead perform $k$ first analytically, followed by $k^{0}$ and $q$
numerically. We find

$Z_{g\,\text{G L-T non-HTL
th}}^{(c)}=\frac{g^{2}C_{A}T^{2}}{48\pi^{2}}\bigg{[}\underbrace{0.145277}_{k<2q}\underbrace{-5.13832}_{k>2q}\bigg{]}.$
(D.69)

The total is then

$\displaystyle Z_{g\,\text{G L-T
}}^{(c)}=\frac{g^{2}C_{A}T^{2}}{48\pi^{2}}\bigg{\\{}\frac{2}{\epsilon^{2}}+\frac{2\ln\frac{\bar{\mu}\exp(\gamma_{E}+1/2)}{4\pi
T}+2(\ln(2)-1)+2\ln\frac{\bar{\mu}A^{12}e^{-1/2}}{8\pi T}}{\epsilon}$
$\displaystyle+2\ln^{2}\frac{\bar{\mu}\exp(\gamma_{E}+1/2)}{4\pi
T}-4\gamma_{1}+\frac{\pi^{2}}{4}-2\gamma_{E}^{2}+\frac{3}{2}$
$\displaystyle+2(\ln(2)-1)\bigg{[}2\ln\frac{\bar{\mu}\exp(\gamma_{E}+1/2)}{4\pi
T}\bigg{]}$ $\displaystyle+2\ln^{2}\frac{\bar{\mu}A^{12}e^{-1/2}}{8\pi T}$
$\displaystyle+48\ln(A)(\gamma_{E}+\ln(2\pi)-6\ln(A))+\frac{12\zeta^{\prime\prime}(2)}{\pi^{2}}+\frac{1}{2}\text{Li}_{2}\left(\frac{2}{3}\right)+\frac{1}{4}\text{Li}_{2}\left(\frac{3}{4}\right)$
$\displaystyle-\frac{1}{2}-\frac{3}{2}\ln^{2}(2)-2\ln(\pi)\ln(4\pi)-2\gamma_{E}\left(\gamma_{E}+\ln\left(4\pi^{2}\right)\right)-\frac{1}{2}\ln(3)\coth^{-1}(7)$
$\displaystyle+\underbrace{0.145277}_{k<2q}\underbrace{-5.13832}_{k>2q}\bigg{\\}}.$
(D.70)

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See pages 2 of Cover.pdf