1 is given as $$\begin{aligned}
\label{eq:T-sumrule1-cont}
\begin{split}
& \frac{2}{\pi}\int^\infty_{0} d\omega
\frac{\delta \rho_{T}(\omega,{{\mathbf{p}}})}{\omega} \\
&\simeq -C_{\text{em}}{N_c}\frac{1}{3\pi^2}
\int^\infty_{0} d\omega \omega
 \Biggl[{n_F}\left(\frac{\omega}{2}\right) \\
&~~~ +{{\mathbf{p}}}^2\left\{\frac{1}{20}{n_F}''\left(\frac{\omega}{2}\right)
 -\frac{1}{\omega^2}{n_F}\left(\frac{\omega}{2}\right)\right\}\Biggr] \\
 &\simeq -C_{\text{em}}{N_c}\frac{1}{3}
 \Biggl[\frac{T^2}{3} 
 -\frac{{{\mathbf{p}}}^2}{2\pi^2}\ln\frac{T}{\mu}\Biggr] ,
\end{split}\end{aligned}$$ where ${n_F}(k^0)\equiv[e^{k^0/T}+1]^{-1}$ is the Fermi distribution function. We have used $\int^\infty_{0} d\omega \omega {n_F}(\omega/2)=\pi^2T^2/3$, and introduced the IR cutoff $\mu$ for the continuum contribution because of a logarithmic IR divergence. The nonsingular parts of the ${{\mathbf{p}}}^2$ terms have been omitted. The result for the continuum (\[eq:continuum-T\]) is obtained from a one-loop calculation, and becomes unreliable when $\omega\lesssim gT$, where the hard-thermal loop resummation becomes necessary [@Moore:2006qn; @Braaten:1990wp]. Therefore, we see that $\mu\sim gT$.

The contribution from the UV tail (\[eq:UVtail-T\]) is estimated as $$\begin{aligned}
\label{eq:T-sumrule1-UVtail}
\begin{split}
 \frac{2}{\pi}\int^\infty_{0} d\omega
\frac{\delta \rho_{T}(\omega,{{\mathbf{p}}})}{\omega}
&\sim {C_{\text{em}}}g^2T^2, {C_{\text{em}}}g^2{{\mathbf{p}}}^2.
\end{split} \end{aligned}$$ Here we have used the fact that the IR cutoff of the UV tail is of order $\sim T$, because the derivation of the UV tail is based on the OPE, which is valid when $\omega\gg T$.

Now let us check that sum rule 1 (\[eq:sumrule-1-T\]) is satisfied. Because $\kappa_B=0$ at weak coupling (see Appendix \[app:transport-peak\]), the integral in Eq. (\[eq:sumrule-1-T\]) needs to vanish in order to satisfy the sum rule. We first see that 