I get some very stange equation 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 not periodic functions which infinite series defined as analytic continuation. Equation in title (it can't be use for numerical calulations because of divergence). If you know anything about it or what does it mean please get me know.


Let's f (x) will be some analytic function with Taylor series. By using theorem of divergent summation derived by Ramanujan we can associate  $\int_x^{\infty}f (t) dt $ as being convergent so we can write summation in terms of Euler-Maclaurin formula as follow $\sum_{k=x}^{\infty}f (k)=- \sum_{n=0}^{\infty} \frac {(-1)^nf^{(n-1)}(x)B_n}{n!}$.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 can derive $ \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 $ and by using Taylor series get =$-2\pi\int_{-\infty}^{\infty} \frac {F(x-\frac {1}{2}+it)}{(e^{\pi t}+e^{-\pi t})^2}dt $


Let's use consequances of theorem derived by Ramanujan another one

$\displaystyle \sum_{k=x}^{\infty} f (k) =-2\pi\int_{-\infty}^{\infty} \frac {F(x-\frac {1}{2}+it)}{(e^{\pi t}+e^{-\pi t})^2}dt =\displaystyle 2\pi \int_{\infty}^{\infty} \frac {\int_{0}^{\infty}f(u)du-\int_{0}^{x-\frac {1}{2}+it}\sum_{k=0}^{\infty}\frac{f^{(k)}(0)(u)^k}{k!}du}{(e^{\pi t}+e^{-\pi t})^2}dt$ 


$=\displaystyle 2\pi \int_{-\infty}^{\infty} \frac {\int_{0}^{\infty}f(u)du-\sum_{k=0}^{\infty}\frac{f^{(k)}(0)(x-\frac {1}{2}+it)^{k+1}}{k!(k+1)}du}{(e^{\pi t}+e^{-\pi t})^2}dt=\displaystyle \int_{0}^{\infty}f(u)du+\sum_{k=0}^{\infty}\frac{f^{(k)}(0)\zeta (-k,x)}{k!}$ 



Let's assume $x=1$. We can use Riemann zeta functional equation to derived some transformation. 





$\displaystyle \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!}= $ 

$\displaystyle   \int_{0}^{\infty}f(t)dt+\sum_{n=1}^{\infty}\sum_{k=0}^{\infty} \frac { f^{(k)}(0)(2 \pi)^{-k-1} (i^{-k-1}+(-i)^{-k-1}) n^{-k-1}\Gamma (k+1)}{k!}  =$ 



$\displaystyle  \int_{0}^{\infty}f(t)dt+ \sum_{n=1}^{\infty}\int_{0}^{\infty}\sum_{k=0}^{\infty}\frac{f^{(k)}(0) t^{k}(e^{-2 \pi i nt}+e^{2 \pi i nt})}{k!}dt=$

$\displaystyle  \int_{0}^{\infty}f(t)dt+ \sum_{n=1}^{\infty}\int_{0}^{\infty}f(t)(e^{-2 \pi i nt}+e^{2 \pi i nt})dt$ 


This is enough to write down solution as 

$\displaystyle \sum_{k=1}^{\infty} f (k)=\sum_{n=-\infty}^{\infty}\mathscr {L} \{ f \} (2 \pi i n)$


PS:if someone is good at editing I world be very greatfull for some editing. I know it looks terible