I obtained the very strange formula above (in the question title) and I can't understand what does it mean (or if it does mean anything). It's very similar to Poisson summation but it should work for non-periodic functions for which infinite series are defined as analytic continuation.

• First of all, I point out that the above formula cannot be used for numerical calculations because of its divergence.
• If I understand it correctly, to sum a series from $-\infty $ to $\infty $ means to broke it in two parts at some point $a$ and to treat each of them as an analytic continuation: if my understanding is not correct and/or you know anything about it or what does it mean, please let me know.

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 $$ \sum_{k=x}^{\infty}f (k)=- \sum_{n=0}^{\infty} \frac {(-1)^nf^{(n-1)}(x)B_n}{n!}. $$ We know also that $$ (-1)^nB_n=2 \pi \int_{-\infty}^{\infty} \frac {(-\frac {1}{2}+it) ^n}{(e^{\pi t}+e^{-\pi t})^2}dt, $$ so we get $$ \begin{split} \sum_{k=x}^{\infty} f (k) &=-2 \pi \sum_{n=0}^{\infty}\frac{f^{(n-1)} (x)}{n!}\int_{-\infty}^{\infty} \frac {(-\frac {1}{2}+it) ^n}{(e^{\pi t}+e^{-\pi t})^2}dt\\ & \text{ and by using Taylor series }\\ &=-2\pi\int_{-\infty}^{\infty} \frac {F(x-\frac {1}{2}+it)}{(e^{\pi t}+e^{-\pi t})^2}dt \end{split}$$

Let's use the consequences of Ramanujan's theorem another time $$ \begin{split} \sum_{k=x}^{\infty} f (k) & = -2\pi\int_{-\infty}^{\infty} \frac {F(x-\frac {1}{2}+it)}{(e^{\pi t}+e^{-\pi t})^2}dt\\ & = 2\pi \int_{\infty}^{\infty} \frac {\int_{0}^{\infty}f(u)du-\int_{0}^{x-\frac {1}{2}+it}\sum_{k=0}^{\infty}\frac{f^{(k)}(0)(u)^k}{k!}du}{(e^{\pi t}+e^{-\pi t})^2}dt\\ & = 2\pi \int_{-\infty}^{\infty} \frac {\int_{0}^{\infty}f(u)du-\sum_{k=0}^{\infty}\frac{f^{(k)}(0)(x-\frac {1}{2}+it)^{k+1}}{k!(k+1)}du}{(e^{\pi t}+e^{-\pi t})^2}dt\\ & = \int_{0}^{\infty}f(u)du+\sum_{k=0}^{\infty}\frac{f^{(k)}(0)\zeta (-k,x)}{k!} \end{split} $$ Now let's assume $x=1$: then we can use Riemann's zeta functional equation to derived some transformation of the above expression: $$ \begin{split} \sum_{k=1}^{\infty} f (k) & = \int_{0}^{\infty}f(t)dt+ \sum_{k=0}^{\infty}\frac{f^{(k)}(0)(2 \pi)^{-k-1} (i^{-k-1}+(-i)^{-k-1}) \Gamma (1+k)\zeta (1+k) }{k!} \\ & = \int_{0}^{\infty}f(t)dt+\sum_{n=1}^{\infty}\sum_{k=0}^{\infty} \frac { f^{(k)}(0)(2 \pi)^{-k-1} (i^{-k-1}+(-i)^{-k-1}) n^{-k-1}\Gamma (k+1)}{k!} \\ & = \int_{0}^{\infty}f(t)dt+ \sum_{n=1}^{\infty}\int_{0}^{\infty}\sum_{k=0}^{\infty}\frac{f^{(k)}(0) t^{k}(e^{-2 \pi i nt}+e^{2 \pi i nt})}{k!}dt\\ & = \int_{0}^{\infty}f(t)dt+ \sum_{n=1}^{\infty}\int_{0}^{\infty}f(t)(e^{-2 \pi i nt}+e^{2 \pi i nt})dt\\ \end{split} $$ This is enough to write down the formula as $$ \sum_{k=1}^{\infty} f (k)=\sum_{n=-\infty}^{\infty} \mathcal{L} \{ f \} (2 \pi i n)$$

To contrast there is formula of Poisson summation, where $\hat {f}(x)=\int_{-\infty}^{\infty}f (t) e^{-2 \pi i xt}dt$

$$\sum_{n=-\infty}^{\infty} f (n)= \sum_{n=-\infty}^{\infty} \hat { f} (n)$$

PS:Now when I wrote it next to each other I thought there could exsist some conectiom like $\sum_{n=x+1}^{\infty} f (n)= \sum_{n=-\infty}^{\infty} \int_{x}^{\infty}f(t)e^{-2 \pi i nt}dt $ or something in type of generalisation of Abel-Plana formula as follow $ \sum_{k=x}^{\infty}f (k)=-F(x)+\frac {1}{2}f (x)+\int_{0}^{\infty} \frac{f (x+it)-f (x-it)}{e^{2\pi t}-1} dt. $