line, which is obtained by using the residual theorem for the integral $\oint_C d\omega' 
[\delta G_{T}(\omega',{{\mathbf{p}}})-\delta G_{T}(\infty,{{\mathbf{p}}})]/(\omega'+i\omega)$. By subtracting Eq. (\[eq:derivation\]) and $i\omega \delta G_T{}'=i\omega^2\delta G_T{}'2\int^\infty_0 d\omega'/[\pi(\omega^2+\omega'{}^2)]$, which is necessary to regularize the IR singularity in the integral, we get $$\begin{aligned}
\begin{split}
&\delta G_{T}(i\omega,{{\mathbf{p}}})-\delta G_{T}(0,{{\mathbf{p}}}) -i\omega \delta G_T{}'(0,{{\mathbf{p}}}) \\
&= \frac{2}{\pi }\omega^2 \int^\infty_{0} d\omega' \frac{1}{\omega^2+\omega'{}^2}
\left[ \delta \rho_{T}(\omega',{{\mathbf{p}}})\frac{-1}{\omega'} 
+\delta \rho'_T{}(0,{{\mathbf{p}}})\right],
\end{split}\end{aligned}$$ where $'$ stands for the derivative in terms of energy ($\omega$, $\omega'$). Taking the $\omega \to 0$ limit, this reduces to $$\begin{aligned}
\label{eq:derivation-sumrule3-T}
\begin{split}
\frac{1}{2}\delta G_{T}{}''(0,{{\mathbf{p}}})
= \frac{2}{\pi } \int^\infty_{0} d\omega \frac{1}{\omega^3}
\left[ \delta \rho_{T}(\omega,{{\mathbf{p}}})
-\omega \delta \rho'_T{}(0,{{\mathbf{p}}})\right]. 
\end{split}\end{aligned}$$ Here, we have changed the integration variable from $\omega'$ to $\omega$ for simplicity. To get an explicit form of this sum rule, one needs to evaluate $\delta G_{T}{}''(0,{{\mathbf{p}}})$ and $\delta \rho'_T{}(0,{{\mathbf{p}}})$. In the expansion of Eq.(\[eq:hydro-GR-T\]), we get only the $|{{\mathbf{p}}}|=0$ terms as $\delta G_{T}{}''(0,{{\mathbf{p}}})=-2\sigma\tau_J+{\cal O}({{\mathbf{p}}}^2)$ and $\delta \rho'_T{}(0,{{\mathbf{p}}})=\sigma+{\cal O}({{\mathbf{p}}}^2)$. Therefore, we can obtain a sum rule for the $|{{\mathbf{p}}}|=0$ case, which reads $$\begin{aligned}
\label{eq:sumrule-3-T}
-\sigma\tau_J
&= \frac{2}{\pi} \int^\infty_{0} \frac{d\omega}{\omega^3} 
\left[\delta \rho_T(\omega,{{\mathbf{0}}})-\sigma\omega\right].\end{aligned}$$ We call this equation sum rule 3 for the transverse channel. We note that the transport coefficients in the left-hand side cannot be computed by lattice QCD without 