[@Brandt:2013faa] with methods that do not suffer from the problem of analytic continuation, so this sum rule can be used to constrain the spectral function. At $|{{\mathbf{p}}}|=0$, it reduces to $$\begin{aligned}
\label{eq:sumrule-1-p0}
\begin{split}
0&= \frac{2}{\pi}\int^\infty_{0} d\omega
\frac{\delta \rho_{T}(\omega,{{\mathbf{0}}})}{\omega},
\end{split}\end{aligned}$$ which was already obtained in Ref. [@Gubler:2016hnf], and also in Ref. [@Bernecker:2011gh] by using the conservation of the EM current.

To get a feeling of how this sum rule is satisfied, let us check the respective contributions from the transport peak, the continuum, and the UV tail below. This also gives us an indication about the sensitivity of the sum rule integral to these three structures. The transport peak in the spectral function at small momentum is given by Eq. (\[eq:transport-T\]), and its contribution to sum rule 1 reads $$\begin{aligned}
\label{eq:T-sumrule1-transport}
\begin{split}
 \frac{2}{\pi}\int^\infty_{0} d\omega
\frac{\delta \rho_{T}(\omega,{{\mathbf{p}}})}{\omega}
&\simeq  {C_{\text{em}}}{N_c}\chi \frac{2}{3} \frac{2}{\pi}
\int^\infty_{0} d\omega \frac{\tau^{-1} }{\omega^2+\tau^{-2}} \\
&~~~\times \left[1 +\frac{{{\mathbf{p}}}^2}{5} \frac{(3\omega^2-\tau^{-2})}{(\omega^2+\tau^{-2})^2}\right]\\
&= {C_{\text{em}}}{N_c}\chi \frac{2}{3},
\end{split}\end{aligned}$$ where ${C_{\text{em}}}\equiv e^2\sum_f q^2_f$, $\tau\sim (g^4T)^{-1}$ is the relaxation time introduced in the Boltzmann equation, and $\chi\equiv T^2/6$. For the leading term we have used $\int^\infty_{0} d\omega \tau^{-1}/(\omega^2+\tau^{-2})=\pi/2$, while the momentum dependent term vanishes because of $\int^\infty_{0} d\omega (3\omega^2-\tau^{-2})/(\omega^2+\tau^{-2})^3=-[\omega/(\omega^2+\tau^{-2})^2]^\infty_0=0$.

The contribution from the continuum (\[eq:continuum-T\]) to sum rule 