$$\sum _{n=0}^{\infty }f(n)=\int _{0}^{\infty }f(x)\,dx+{\frac {1}{2}}f(0)+i\int _{0}^{\infty }{\frac {f(it)-f(-it)}{e^{2\pi t}-1}}\,dt$$
If we take $f(x)=e^{-x}$, the right-hand side is $3/2$, the left-hand side is $\frac{e}{e-1}$.
Why Abel-Plana formula does not work for exponent
complex-analysisintegrationsequences-and-series
Related Solutions
EDIT: this "answer" contains errors, as explained by Achille Hui. I have tried to amend it in the next "answer".
You are very kind: thank you a lot! So we have $\int_{\pi/2}^{3\pi/2}f(\varepsilon e^{it})\pi\cot(\pi\varepsilon e^{it})i\varepsilon e^{it}dt+\int_{0}^{\infty}f(i\varepsilon+t)\pi\cot(\pi(i\varepsilon+t))dt-\int_{0}^{\infty}f(-i\varepsilon+t)\pi\cot(\pi(-i\varepsilon+t))dt=2\pi i\sum_{n=0}^{\infty}f(n)$ [EDIT: I wrote wrong sings!] and therefore, as $\varepsilon\to 0$, $\int_{0}^{\infty}f(i\varepsilon+t)\pi\cot(\pi(i\varepsilon+t))dt-\int_{0}^{\infty}f(-i\varepsilon+t)\pi\cot(\pi(-i\varepsilon+t))dt=2\pi i\sum_{n=0}^{\infty}f(n)-\pi i\text{Res}_{z=0}f(z)\pi\cot(\pi z)\to2\pi i\sum_{n=0}^{\infty}f(n)-\pi i f(0)$.
Rewriting $\pi\cot(\pi z)$ and dividing by $2\pi i$ we get $\int_{0}^{\infty}\frac{f(i\varepsilon+t)}{e^{2\pi i(i\varepsilon+t)}-1}+\frac{1}{2}f(i\varepsilon+t)dt+\int_{0}^{\infty}\frac{f(-i\varepsilon+t)}{e^{-2\pi i(-i\varepsilon+t)}-1}+\frac{1}{2}f(-i\varepsilon+t)dt\to\sum_{n=0}^{\infty}f(n)-\frac{1}{2}f(0)$
If we knew that $\int_{0}^{\infty}f(\pm i\varepsilon+t)$ converges we could be allowed to write $\int_{0}^{\infty}\frac{f(i\varepsilon+t)}{e^{2\pi i(i\varepsilon+t)}-1}dt+\int_{0}^{\infty}\frac{f(-i\varepsilon+t)}{e^{-2\pi i(-i\varepsilon+t)}-1}dt+\int_{0}^{\infty}\frac{1}{2}f(i\varepsilon+t)+\frac{1}{2}f(-i\varepsilon+t)dt\to\sum_{n=0}^{\infty}f(n)-\frac{1}{2}f(0)$ but can we be sure that such a step is allowed?
Then it would be easy to change variables [EDIT: I had convinced myself that it is possible to change variable as in real integration, which is not true in general] to write $\int_{0}^{\infty}\frac{f(-i\varepsilon+s)}{e^{-2\pi i(-i\varepsilon+s)}-1}ds+\int_{0}^{\infty}\frac{f(i\varepsilon+s)}{e^{2\pi i(i\varepsilon+s)}-1}ds+\int_{0}^{\infty}\frac{1}{2}f(i\varepsilon+t)+\frac{1}{2}f(-i\varepsilon+t)dt$ $=i\int_{0}^{\infty}\frac{f(-i\varepsilon+it)}{e^{2\pi(\varepsilon+t)}-1}dt-i\int_{0}^{\infty}\frac{f(i\varepsilon-it)}{e^{2\pi(t-\varepsilon)}-1}dt+\int_{0}^{\infty}\frac{1}{2}f(i\varepsilon+t)+\frac{1}{2}f(-i\varepsilon+t)dt$ which would be the searched result if we could substitute $\varepsilon$ with 0, but I am not sure how we can pass to the limit under the integral sign: can we and, if we can, why can we? I heartily thank you again!!!
Thank you to @Gary for mentioning this term I missed. And here is the full solution.
Use Abel-Plana formula, Eq.(2.10.2) $$\sum_{n=a}^N f(n)=\int_a^N f(x)dx+\frac{1}{2}f(a)+\frac{1}{2}f(N)+i\int_0^\infty \frac{f(it)-f(-it)}{e^{2\pi t}-1}dt+\hat{R}$$
where $\hat{R}$ is the remainder representing the last two terms in the second line of Eq.(2.10.2).
Let $$f(z)=\frac{1}{2}\ln(1+z)$$
then the remainder $\hat{R}$ vanishes when $N\rightarrow\infty$ for our choice of $f(z)$, namely,
$$\lim_{N\to \infty} \hat{R}=0$$
Let
$$I=\int_0^\infty \frac{\arctan(t)}{e^{2\pi t}-1} dt$$ Simplify the following terms $$f(it)-f(-it)=\frac{\ln(1+it)}{2}-\frac{\ln(1-it)}{2}=\mathrm{artanh}(it)=i\cdot \arctan(t)$$
Let $~a=0$, and we get
$$\sum_{n=0}^N \frac{\ln(1+n)}{2}=\int_0^N \frac{\ln(1+x)}{2}dx+0+\frac{1}{2}\cdot\frac{\ln(1+N)}{2}-\int_0^\infty \frac{\arctan(t)}{e^{2\pi t}-1}dt+\hat{R}$$
Organize terms and take the limit on both sides,
$$I=\int_0^\infty \frac{\arctan(t)}{e^{2\pi t}-1}dt=\lim_{N\rightarrow \infty} \left( \int_0^N \frac{\ln(1+x)}{2}dx+\frac{\ln(1+N)}{4}-\sum_{n=0}^N \frac{\ln(1+n)}{2} \right) $$ Further, we get $$I=\frac{1}{2}\lim_{N\rightarrow \infty} \left( ~(N+1)\ln(N+1)-N+\frac{1}{2}\cdot\ln(1+N)- \ln[(N+1)!] ~\right) $$
let $n=N+1$:
$$\begin{align}I&=\frac{1}{2}\lim_{n\rightarrow \infty} \left( ~n\ln(n)-n+1+\frac{1}{2}\cdot\ln(n)- \ln (n!) ~\right)\\ \\ &=\frac{1}{2}\lim_{n\rightarrow \infty} \left(1+\ln\left(\frac{n^{n+\frac{1}2}}{n!e^n}\right)\right) \end{align}$$
Use Stirling's formula, we get the limit:
$$\lim_{n\rightarrow \infty} \left(\frac{n^{n+\frac{1}2}}{n!e^n}\right)=\frac{1}{\sqrt{2\pi}}$$
Finally,
$$\int_0^\infty \frac{\arctan(t)}{e^{2\pi t}-1} dt =\frac{1}{2}-\frac{1}{4}\cdot\ln(2\pi)$$
Best Answer
Since$$i\int_0^\infty\frac{e^{-(2\pi+i)t}-e^{(2\pi-i)t}}{1-e^{-2\pi t}}dt=2\sum_{n\ge1}\frac{1}{4\pi^2n^2+1}=\frac12\coth\frac12-1=\frac{3-e}{2e-2},$$the right-hand side is$$\frac32+\frac{3-e}{2e-2}=\frac{e}{e-1}.$$