Why Abel-Plana formula does not work for exponent? Abel-Plana formula:
$$\sum _{n=0}^{\infty }f(n)=\int _{0}^{\infty }f(x)\,dx+{\frac {1}{2}}f(0)+i\int _{0}^{\infty }{\frac {f(it)-f(-it)}{e^{2\pi t}-1}}\,dt$$
If we take $f(x)=e^{-x}$, the right-hand side is $3/2$, the left-hand side is $\frac{e}{e-1}$.
 A: 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$$\frac32+\frac{3-e}{2e-2}=\frac{e}{e-1}.$$
A: Take $f(x)=e^{-x}$, and compute them term by term:
$$\begin{align}
\sum _{n=0}^{\infty }f(n)&=\sum _{n=0}^{\infty} e^{-n}=\sum _{n=0}^{\infty} \left(\frac{1}{e}\right)^{n}=\frac{1}{1-\frac{1}{e}}=\frac{e}{e-1}\\
\\
\int _{0}^{\infty}f(x)~dx&=\int _{0}^{\infty}e^{-x}~dx=1\\
\\
\frac {1}{2}f(0)&=\frac {1}{2}\\
\\
f(it)-f(-it)&=e^{-it}-e^{it}=-2i\cdot\sin(t)\\
\\
i\int _{0}^{\infty }{\frac {f(it)-f(-it)}{e^{2\pi t}-1}}~dt&=\int _{0}^{\infty }{\frac {2\sin(t)}{e^{2\pi t}-1}}~dt
\end{align}$$
Plug into the Abel-Plana formula:
$$\frac{e}{e-1}=1+\frac {1}{2}+\int _{0}^{\infty }{\frac {2\sin(t)}{e^{2\pi t}-1}}~dt$$
Simplify:
$$\int _{0}^{\infty }{\frac {2\sin(t)}{e^{2\pi t}-1}}~dt=\frac{1}{e-1}-\frac{1}{2}=\frac{1}{2}\coth\left(\frac{1}{2}\right)-1$$
Divide $2$ on both sides:
$$\int _{0}^{\infty }{\frac {\sin(t)}{e^{2\pi t}-1}}~dt=\frac{1}{2e-2}-\frac{1}{4}=\frac{1}{4}\coth\left(\frac{1}{2}\right)-\frac{1}{2}$$
