suffering from the problem of analytic continuation. We do not check that sum rule 3 is satisfied in the chiral and the weak coupling limits since this was already done in our previous paper [@Gubler:2016hnf]. Instead we just cite the order of magnitude of the three contributions: the transport peak contribution is of order ${C_{\text{em}}}T^2 \tau^2\sim {C_{\text{em}}}g^{-8}$ and is equal to the left-hand side of sum rule 3, while the continuum is much smaller, ${C_{\text{em}}}g^{-5}$. The UV tail contribution is the smallest and of order ${C_{\text{em}}}g^{-4}$.

Explicitly taking into account higher order terms in Eq. (\[eq:hydro-GR-T\]) and expanding $\delta\rho_T$ in $|{{\mathbf{p}}}|^2$ order by order, it should be possible to obtain a corresponding sum rule at finite momentum. We leave this task for future work.

Longitudinal channel {#ssc:sumrule-L}
--------------------

Before discussing the sum rules, let us remember that the retarded Green function in the longitudinal channel is exactly known at zero momentum [@Ding:2010ga] from the conservation law of the charge: $$\begin{aligned}
\label{eq:rho-L-p=0-exact}
\rho_{00}(\omega,{{\mathbf{0}}}) 
&= \pi \chi_q \omega\delta(\omega) ,\end{aligned}$$ where $\chi_q\equiv \int d^3{{\mathbf{x}}}\langle j^0({{\mathbf{x}}}) j^0({{\mathbf{0}}}) \rangle/T$ is the charge susceptibility. By matching this result with the hydro result of Eq.(\[eq:hydro-rho-L-p=0\]), we see that the hydro result is exact for all $\omega$ at 