Extreme limit of sum of Faddeeva function I am looking for the proof of
$$
\lim_{t \to +0} \left[ \log t + 2\pi \sum_{n = 0}^{\infty} \left\{ \frac{1}{\omega_n} 
 - \sqrt{\pi} A t ~ \mathrm{W}(i\omega_n At)\right\} \right] = \log \frac{\mathrm{e}^{\gamma/2}}{\pi A}~,
$$
with $\omega_n = (2n + 1)\pi~$ and $~A>0$ . Here, $\mathrm{W}(z)$ is the Faddeeva function given by
$$
\mathrm{W}(z) = \mathrm{e}^{-z^2} \mathrm{erfc}(-iz)~.
$$
The proof of this equation is troubling to me because the equation is necessary for the Eilenberger equation of superconductivity to give the value of the upper critical field at absolute zero.
For example, using a function similar to the Fermi distribution function,
$$
F(z) = \frac{1}{{\mathrm e}^{z} + 1} ,
$$
the infinite sum can be rewritten as the sum of the residues of a complex function. After the substitution $i\omega_{n} \to z$ and the analytic continuation that move $z$ on the real axis, the equation can be written as
$$
2\pi \sum_{n = 0}^{\infty} \left\{ \frac{1}{\omega_n} 
 - \sqrt{\pi} A t ~ \mathrm{W}(i\omega_n At)\right\} = -\int_{-\infty}^{\infty} {\mathrm d}z ~ F(z) \left\{ \frac{1}{z} + i\sqrt{\pi} A t {\mathrm W}(zAt)\right\}.
$$
By using this, I was able to verify the limit value numerically. However, even in this form, it is difficult to evaluate the limit value of $t \to +0$ analytically.
How can I prove the equation? I am not quite sure how to proceed.
Thank you.
 A: $$
{\rm W}(iz)={\rm e}^{z^2}\frac{2}{\sqrt\pi}\int_z^\infty{\rm e}^{-t^2}\,{\rm d}t\underset{t=z+x}{\phantom{\big[}=\phantom{\big]}}\frac{2}{\sqrt\pi}\int_0^\infty{\rm e}^{-x^2-2zx}\,{\rm d}x,
$$
$$
\frac1\omega-\sqrt\pi At\,{\rm W}(i\omega At)=2At\int_0^\infty(1-{\rm e}^{-x^2}){\rm e}^{-2\omega Atx}\,{\rm d}x,
$$
$$
2\pi\sum_{n=0}^\infty\left(\frac1{\omega_n}-\sqrt\pi At\,{\rm W}(i\omega_n At)\right)=4\pi At\int_0^\infty(1-{\rm e}^{-x^2})\frac{{\rm e}^{-2\pi Atx}}{1-{\rm e}^{-4\pi Atx}}\,{\rm d}x=f(\pi At),
$$
$$
f(\lambda):=2\lambda\int_0^\infty\frac{1-{\rm e}^{-x^2}}{\sinh 2\lambda x}\,{\rm d}x\underset{\text{IBP}}{\phantom{\big[}=\phantom{\big]}}\int_0^\infty 2x{\rm e}^{-x^2}\log\coth\lambda x\,{\rm d}x,
$$
$$
\log\lambda+f(\lambda)\underset{x^2=y}{\phantom{\big[}=\phantom{\big]}}\int_0^\infty{\rm e}^{-y}\log(\lambda\coth\lambda\sqrt{y})\,{\rm d}y\underset{\lambda\to 0^+}{\longrightarrow}-\frac12\int_0^\infty{\rm e}^{-y}\log y\,{\rm d}y=\frac{\gamma}{2}
$$
(the limit/integral interchange on the last line is easy to justify using DCT).
