$\frac {f (0)}{2}+ \sum_{k=1}^{\infty}f (k)=\sum_{n=-\infty}^{\infty} \mathcal{L} \{ f \} (2 \pi i n)$

Question

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 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

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

Answer 1

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 using Cesàro 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.