Consider the integral:
$$I=\int_0^\infty \frac{t\cos(t\cdot\ln(b))-a\sin(t\cdot\ln(b))}{(a^2+t^2)(e^{2\pi t}-1)}\,dt,\qquad a>0,\ b>1\tag{1}\label{intdef}$$
Define:
$$f(z)=-\frac{i}{2}\cdot\frac{1}{b^z(a-z)}$$
Easy to check:
$$f(it)-f(-it)=\frac{t\cos(t\cdot\ln(b))-a\sin(t\cdot\ln(b))}{a^2+t^2}$$
Use the Abel-Plana formula to calculate above integral. The Abel-Plana formula says:
$$\sum_{n=0}^\infty f(n)=\int_0^\infty f(x)dx+\frac{1}{2}f(0)+i\int_0^\infty \frac{f(it)-f(-it)}{e^{2\pi t}-1}dt\tag{2}$$
$$\sum_{n=0}^\infty -\frac{i}{2}\cdot\frac{1}{b^n(a-n)}=\int_0^\infty -\frac{i}{2}\cdot\frac{1}{b^x(a-x)}dx+\frac{1}{2}\left( -\frac{i}{2}\cdot\frac{1}{a} \right)+i\cdot I$$
$$I=\frac{1}{4a}+\frac{1}{2}\cdot\sum_{n=0}^\infty \frac{1}{b^n(n-a)}-\frac{1}{2}\cdot\int_0^\infty \frac{1}{b^x(x-a)}dx\tag{3}\label{inteqn}$$
Question:
If $a\in\mathbb{N^+}$, both series and integral diverge in Eq.\eqref{inteqn},
If $a\notin\mathbb{N^+}$, the integral diverges in Eq.\eqref{inteqn}.
But the integral \eqref{intdef} always exists for all $a\in \mathbb{R^+}$.
Why does the Abel-Plana formula fail here?
Best Answer
This formula requires the function $f(z)$ to be holomorphic on the right-half-plane with $\Re(z)>0$, but this function has a pole at $x=a,~(a>0)$, so it is not holomorphic. I think this is the reason it fails here.