If we accept this as valid Proof of Poisson summation
$\displaystyle \sum_{n=-\infty}^{\infty}\hat {f}(t) dt= \sum_{n=-\infty}^{\infty}\int_{-\infty}^{\infty}f (t)e^{-2 i \pi n t} dt= \int_{-\infty}^{\infty}f (t)\sum_{n=-\infty}^{\infty}e^{-2 i \pi n t} dt = \int_{-\infty}^{\infty}f (t)\sum_{n=-\infty}^{\infty}\delta (t-n) dt=\sum_{n=-\infty}^{\infty}\int_{-\infty}^{\infty}f (t)\delta (t-n) dt=\sum_{n=-\infty}^{\infty}f (n)$
Let's use fact that $ \int_{a}^{\infty}f (t)\delta (t-n)=f (n)$, if $a <n $ and $ \int_{a}^{\infty}f (t)\delta (t-n)=0$, if $a>n $
$\displaystyle \sum_{n=-\infty}^{\infty}\int_{0^+}^{\infty}f (t)e^{-2 i \pi n t} dt= \int_{0^+}^{\infty}f (t)\sum_{n=-\infty}^{\infty}e^{-2 i \pi n t} dt = \int_{0^+}^{\infty}f (t)\sum_{n=-\infty}^{\infty}\delta (t-n) dt=\sum_{n=-\infty}^{\infty}\int_{0^+}^{\infty}f (t)\delta (t-n) dt=\sum_{n=1}^{\infty}f (n) \wedge \displaystyle \sum_{n=-\infty}^{\infty}\int_{0^-}^{\infty}f (t)e^{-2 i \pi n t} dt= \left[...\right] =\sum_{n=0}^{\infty}f (n)$
Ergo
$\displaystyle \sum_{n=-\infty}^{\infty} \mathcal{L} \{ f \} (2 \pi i n)= \sum_{n=-\infty}^{\infty}\int_{0}^{\infty}f (t)e^{-2 i \pi n t} dt=\frac {f (0)}{2}+\sum_{n=1}^{\infty}f (n)$
And the question is. Is this so trivial that no one is talking about it or people just missed it. I mean at this article author is confused to about the status of equation (he get a little different form)