I obtained the very strange formula above and at begining I was just wanted know how to interpretate it. But now when I know what is this with help of @Carlo Beenakker, I am leaving it as a proof. BTW I found some [article](https://www.google.com/url?q=https://www.peertechzpublications.com/articles/AMP-5-141.pdf&sa=U&ved=2ahUKEwjDor-tpNb9AhXX_yoKHYGHAC4QFnoECAQQAg&usg=AOvVaw2L5byXDO356phO20CxHXRS) about non formal take of that theorem (I hope my take on this could be colled formal). I made some changes because I am ritarded and I realised that half of calculation was unnecesary.


**proof** Let $f(x)$ be some an analytic function with a given Taylor series. By using the theorem on divergent summation derived by Ramanujan, we can associate to it the convergent integral $\int\_x^{\infty}f (t) dt $ so we can write the summation in terms of Euler–Maclaurin formula as follows
$$
\sum\_{k=x}^{\infty}f (k)=\sum\_{n=-1}^{\infty} \frac {f^{(n)}(x)\zeta (-n)}{n!}.
$$


We can use Riemann's zeta functional equation to derived some transformation of the above expression:
$$
\begin{split} 
\sum\_{k=x}^{\infty} f (k) & = -F (x)+\frac {f (x)}{2}+ \sum\_{k=1}^{\infty}\frac{f^{(k)}(x)(2 \pi)^{-k-1} (i^{-k-1}+(-i)^{-k-1}) \Gamma (1+k)\zeta (1+k) }{k!} \\ 
& = -F (x)+\frac {f (x)}{2} +\sum\_{n=1}^{\infty}\sum\_{k=1}^{\infty} \frac { f^{(k)}(x)(2 \pi)^{-k-1} (i^{-k-1}+(-i)^{-k-1}) n^{-k-1}\Gamma (k+1)}{k!} \\ 
& = -F (x)+\frac {f (x)}{2} +\sum\_{n=1}^{\infty}\sum\_{k=0}^{\infty} \frac { f^{(k)}(x)(2 \pi)^{-k-1} (i^{-k-1}+(-i)^{-k-1}) n^{-k-1}\Gamma (k+1)}{k!} \\ 
& = -F (x)+\frac {f (x)}{2} +\sum\_{n=1}^{\infty}\int\_{0}^{\infty}\sum\_{k=0}^{\infty}\frac{f^{(k)}(x) t^{k}(e^{-2 \pi i nt}+e^{2 \pi i nt})}{k!}dt \\
& = -F (x)+\frac {f (x)}{2} + \sum\_{n=1}^{\infty}\int\_{0}^{\infty}f(t+x)(e^{-2 \pi i nt}+e^{2 \pi i nt})dt
\end{split}
$$
And now, for x=0, without any strange equations it is possible to write down formula as
$$ \frac {f (0)}{2}+\sum\_{k=1}^{\infty} f (k)=\sum\_{n=-\infty}^{\infty} \mathcal{L} \{ f \} (2 \pi i n)$$


PS: I feel like. Is it all Euler-Maclaurin formula? Always has been. Watch this trivial proof of Abel-Plana formula


$$
\begin {split}
\sum\_{k=x}^{\infty} f (k) & = -F (x)+\frac {f (x)}{2}+ \sum\_{k=1}^{\infty}\frac{f^{(k)}(x)(2 \pi)^{-k-1} (i^{-k-1}+(-i)^{-k-1}) \Gamma (1+k)\zeta (1+k) }{k!} \\ 
& =-F (x)+\frac {f (x)}{2}+ \sum\_{k=1}^{\infty}\frac{f^{(k)}(x)(2 \pi)^{-k-1} (i^{-k-1}+(-i)^{-k-1}) \int\_0^{\infty}\frac {t^k}{e^t-1}dt}{k !}\\ 
& =-F (x)+\frac {f (x)}{2}+ \sum\_{k=0}^{\infty}\frac{f^{(k)}(x)(2 \pi)^{-k-1} (i^{-k-1}+(-i)^{-k-1}) \int\_0^{\infty}\frac {t^k}{e^t-1}dt}{k !}\\ 
& =-F (x)+\frac {f (x)}{2}+ \int\_0^{\infty}\frac {\frac {f (x+\frac {t}{2\pi i})}{2\pi i}+\frac {f (x+\frac {t}{-2\pi i})}{-2\pi i}}{e^t-1} dt \\ 
&=-F (x)+\frac {f (x)}{2}+i\int\_{0}^{\infty} \frac{f (x+it)-f (x-it)}{e^{2\pi t}-1} dt \\
\end {split}
$$



$\frac {f (0)}{2}+ \sum_{k=1}^{\infty}f (k)=\sum_{n=-\infty}^{\infty} \mathcal{L} \{ f \} (2 \pi i n)$
*This is an answer to the question as it was originally formulated, it has now been heavily edited.*




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I consider this formula in the OP,
$$\sum\_{k=1}^{\infty} f (k) = \int\_{0}^{\infty}f(t)dt+ \sum\_{n=1}^{\infty}\int\_{0}^{\infty}f(t)(e^{-2 \pi i nt}+e^{2 \pi i nt})dt.\qquad\qquad(\ast)
$$
This is *almost* the [Extension of the Poisson Summation Formula](https://doi.org/10.1006/jmaa.1997.5767) using [Cesàro summation](https://en.wikipedia.org/wiki/Ces%C3%A0ro_summation), see Eq. 3.4 of the cited paper:
$$\sum\_{k=-\infty}^\infty f(k)-\int\_{-\infty}^\infty f(t)\,dt=\sum\_{n=1}^\infty \int\_{-\infty}^\infty f(t)\left(e^{2\pi int}+e^{-2\pi int}\right)\,dt.\qquad\qquad(\ast\ast)$$
The expression on the left-hand side is to be evaluated in the Cesàro sense, when the sum over $f(k)$ diverges (for example, when $f(k)$ is a polynomial). The formula $(\ast\ast)$ follows from the identification of the distributions
$$\sum\_{k=-\infty}^\infty \delta(x-k)-1=\sum\_{n=1}^\infty\left(e^{2\pi inx}+e^{-2\pi inx}\right).$$


Now I can take an even function $f(t)=f(-t)$, and then I almost find from $(\ast\ast)$ the expression $(\ast)$, up to an extra term $f(0)$ on the left-hand-side.




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Here is a check that a term $\tfrac{1}{2}f(0)$ is missing from the left-hand-side of equation $(\ast)$ from the OP: take the case $f(t)=e^{-t}$; then the left-hand-side of equation $(\ast)$ evaluates to $\frac{1}{e-1}-1$, while the right-hand-side evaluates to $\frac{1}{e-1}-\tfrac{1}{2}$, so there is a $1/2$ difference.


