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

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

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

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

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)

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 $$ \begin{split} \sum_{k=x}^{\infty} f (k) & = -2\pi\int_{-\infty}^{\infty} \frac {F(x-\frac {1}{2}+it)}{(e^{\pi t}+e^{-\pi t})^2}dt\\ & = 2\pi \int_{\infty}^{\infty} \frac {\int_{0}^{\infty}f(u)du-\int_{0}^{x-\frac {1}{2}+it}\sum_{k=0}^{\infty}\frac{f^{(k)}(0)(u)^k}{k!}du}{(e^{\pi t}+e^{-\pi t})^2}dt\\ & = 2\pi \int_{-\infty}^{\infty} \frac {\int_{0}^{\infty}f(u)du-\sum_{k=0}^{\infty}\frac{f^{(k)}(0)(x-\frac {1}{2}+it)^{k+1}}{k!(k+1)}du}{(e^{\pi t}+e^{-\pi t})^2}dt\\ & = \int_{0}^{\infty}f(u)du+\sum_{k=0}^{\infty}\frac{f^{(k)}(0)\zeta (-k,x)}{k!}\\ &=\sum_{k=-1}^{\infty}\frac{f^{(k)}(0)\zeta (-k,x)}{k!} \end{split} $$ The form above proves that Taylor series i truely defined at least as $f (n)=\sum_{k=a}^{\infty}\frac{f^{(k)}(0)n^k}{k!} \wedge a <0$ because for divergent summtion theorem $\sum_{n=x}^{\infty}f (n)=\sum_{k=-1}^{\infty}\frac{f^{(k)}(0)\zeta (x,-k)}{k!} $

Let's assume $x=1$: then we can use Riemann's zeta functional equation to derived some transformation of the above expression: $$ \begin{split} \sum_{k=1}^{\infty} f (k) & = \int_{0}^{\infty}f(t)dt+ \sum_{k=a}^{\infty}\frac{f^{(k)}(0)(2 \pi)^{-k-1} (i^{-k-1}+(-i)^{-k-1}) \Gamma (1+k)\zeta (1+k) }{k!} \\ & = \int_{0}^{\infty}f(t)dt+\sum_{n=1}^{\infty}\left (\sum_{k=0}^{\infty}+\sum_{a}^{-1} \frac { f^{(k)}(0)(2 \pi)^{-k-1} (i^{-k-1}+(-i)^{-k-1}) n^{-k-1}\Gamma (k+1)}{k!} \right) \\ & = \int_{0}^{\infty}f(t)dt+ \sum_{n=1}^{\infty}\int_{0}^{\infty}\sum_{k=0}^{\infty}\frac{f^{(k)}(0) t^{k}(e^{-2 \pi i nt}+e^{2 \pi i nt})}{k!}dt+\Omega (0) \\ & = \int_{0}^{\infty}f(t)dt+ \sum_{n=1}^{\infty}\int_{0}^{\infty}f(t)(e^{-2 \pi i nt}+e^{2 \pi i nt})dt+\Omega (0). \end{split} $$ This is enough to write down the formula as $$ \sum_{k=1}^{\infty} f (k)=\sum_{n=-\infty}^{\infty} \mathcal{L} \{ f \} (2 \pi i n)+\Omega (0)$$

From Poisson formula we know that $$\sum_{n=-\infty}^{\infty} f (n)= \sum_{n=-\infty}^{\infty} \hat { f} (n)$$, which is sufficent to identyfify $\Omega (0) $. In such way we finaly get

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

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

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

I obtained the very strange formula above (in the question title) and I'm starting to understand what does it mean. It's very similar to Poisson summation and indeed it is possible to get Poisson summation from that form (Almost because I lost one term in proving process but with correction it works). Natural question seems to be to ask if the equation above is stronger than Poisson summation or is it equivalent (Is it possible to prove my form from Poisson summation)
 

• 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.

Almost Proof because I missed $\frac{f (0)}{2} $ somewhere: 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 $$ \begin{split} \sum_{k=x}^{\infty} f (k) & = -2\pi\int_{-\infty}^{\infty} \frac {F(x-\frac {1}{2}+it)}{(e^{\pi t}+e^{-\pi t})^2}dt\\ & = 2\pi \int_{\infty}^{\infty} \frac {\int_{0}^{\infty}f(u)du-\int_{0}^{x-\frac {1}{2}+it}\sum_{k=0}^{\infty}\frac{f^{(k)}(0)(u)^k}{k!}du}{(e^{\pi t}+e^{-\pi t})^2}dt\\ &\text {Here I think i lost that $\frac {f (0)}{2}$ because I do not use limits} \\ & = 2\pi \int_{-\infty}^{\infty} \frac {\int_{0}^{\infty}f(u)du-\sum_{k=0}^{\infty}\frac{f^{(k)}(0)(x-\frac {1}{2}+it)^{k+1}}{k!(k+1)}du}{(e^{\pi t}+e^{-\pi t})^2}dt\\ & = \int_{0}^{\infty}f(u)du+\sum_{k=0}^{\infty}\frac{f^{(k)}(0)\zeta (-k,x)}{k!}. \end{split} $$ 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 $$ \sum_{k=1}^{\infty} f (k)=\sum_{n=-\infty}^{\infty} \mathcal{L} \{ f \} (2 \pi i n).$$

Making correction with observation of @Carlo Beenakker about losted $\frac {f (0)}{2} $ somewhere in proving proces we get

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

Now indeed we can get Poisson summation equation by adding corrected formula for $f(k) $ with $f (-k)$.

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

PS:If someone find losted $\frac {f (0)}{2} $ I world be very greatful

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)

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 $$ \begin{split} \sum_{k=x}^{\infty} f (k) & = -2\pi\int_{-\infty}^{\infty} \frac {F(x-\frac {1}{2}+it)}{(e^{\pi t}+e^{-\pi t})^2}dt\\ & = 2\pi \int_{\infty}^{\infty} \frac {\int_{0}^{\infty}f(u)du-\int_{0}^{x-\frac {1}{2}+it}\sum_{k=0}^{\infty}\frac{f^{(k)}(0)(u)^k}{k!}du}{(e^{\pi t}+e^{-\pi t})^2}dt\\ & = 2\pi \int_{-\infty}^{\infty} \frac {\int_{0}^{\infty}f(u)du-\sum_{k=0}^{\infty}\frac{f^{(k)}(0)(x-\frac {1}{2}+it)^{k+1}}{k!(k+1)}du}{(e^{\pi t}+e^{-\pi t})^2}dt\\ & = \int_{0}^{\infty}f(u)du+\sum_{k=0}^{\infty}\frac{f^{(k)}(0)\zeta (-k,x)}{k!}\\ &=\sum_{k=-1}^{\infty}\frac{f^{(k)}(0)\zeta (-k,x)}{k!} \end{split} $$ The form above proves that Taylor series i truely defined at least as $f (n)=\sum_{k=a}^{\infty}\frac{f^{(k)}(0)n^k}{k!} \wedge a <0$ because for divergent summtion theorem $\sum_{n=x}^{\infty}f (n)=\sum_{k=-1}^{\infty}\frac{f^{(k)}(0)\zeta (x,-k)}{k!} $

Let's assume $x=1$: then we can use Riemann's zeta functional equation to derived some transformation of the above expression: $$ \begin{split} \sum_{k=1}^{\infty} f (k) & = \int_{0}^{\infty}f(t)dt+ \sum_{k=a}^{\infty}\frac{f^{(k)}(0)(2 \pi)^{-k-1} (i^{-k-1}+(-i)^{-k-1}) \Gamma (1+k)\zeta (1+k) }{k!} \\ & = \int_{0}^{\infty}f(t)dt+\sum_{n=1}^{\infty}\left (\sum_{k=0}^{\infty}+\sum_{a}^{-1} \frac { f^{(k)}(0)(2 \pi)^{-k-1} (i^{-k-1}+(-i)^{-k-1}) n^{-k-1}\Gamma (k+1)}{k!} \right) \\ & = \int_{0}^{\infty}f(t)dt+ \sum_{n=1}^{\infty}\int_{0}^{\infty}\sum_{k=0}^{\infty}\frac{f^{(k)}(0) t^{k}(e^{-2 \pi i nt}+e^{2 \pi i nt})}{k!}dt+\Omega (0) \\ & = \int_{0}^{\infty}f(t)dt+ \sum_{n=1}^{\infty}\int_{0}^{\infty}f(t)(e^{-2 \pi i nt}+e^{2 \pi i nt})dt+\Omega (0). \end{split} $$ This is enough to write down the formula as $$ \sum_{k=1}^{\infty} f (k)=\sum_{n=-\infty}^{\infty} \mathcal{L} \{ f \} (2 \pi i n)+\Omega (0)$$

From Poisson formula we know that $$\sum_{n=-\infty}^{\infty} f (n)= \sum_{n=-\infty}^{\infty} \hat { f} (n)$$, which is sufficent to identyfify $\Omega (0) $. In such way we finaly get

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