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 

$$
\begin {split}
\sum_{k=x}^{\infty} f (k) & = -F (x)+\frac {f (x)}{2}+ \sum_{k=1}^{\infty}\frac{f^{(k)}(x)(2 \pi)^{-k-1} (i^{-k-1}+(-i)^{-k-1}) \Gamma (1+k)\zeta (1+k) }{k!} \\ 
& =-F (x)+\frac {f (x)}{2}+ \sum_{k=1}^{\infty}\frac{f^{(k)}(x)(2 \pi)^{-k-1} (i^{-k-1}+(-i)^{-k-1}) \int_0^{\infty}\frac {t^k}{e^t-1}dt}{k !}\\ 
& =-F (x)+\frac {f (x)}{2}+ \sum_{k=0}^{\infty}\frac{f^{(k)}(x)(2 \pi)^{-k-1} (i^{-k-1}+(-i)^{-k-1}) \int_0^{\infty}\frac {t^k}{e^t-1}dt}{k !}\\ 
& =-F (x)+\frac {f (x)}{2}+ \int_0^{\infty}\frac {\frac {f (x+\frac {t}{2\pi i})}{2\pi i}+\frac {f (x+\frac {t}{-2\pi i})}{-2\pi i}}{e^t-1} dt \\ 
&=-F (x)+\frac {f (x)}{2}+i\int_{0}^{\infty} \frac{f (x+it)-f (x-it)}{e^{2\pi t}-1} dt \\
\end {split}
$$