# Abel–Plana formula

In mathematics, the Abel–Plana formula is a summation formula discovered independently by Niels Henrik Abel (1823) and Giovanni Antonio Amedeo Plana (1820). It states that

${\displaystyle \sum _{n=0}^{\infty }f(n)=\int _{0}^{\infty }f(x)\,dx+{\frac {1}{2}}f(0)+i\int _{0}^{\infty }{\frac {f(it)-f(-it)}{e^{2\pi t}-1}}\,dt.}$

It holds for functions f that are holomorphic in the region Re(z) ≥ 0, and satisfy a suitable growth condition in this region; for example it is enough to assume that |f| is bounded by C/|z|1+ε in this region for some constants C, ε > 0, though the formula also holds under much weaker bounds. (Olver 1997, p.290).

An example is provided by the Hurwitz zeta function,

${\displaystyle \zeta (s,\alpha )=\sum _{n=0}^{\infty }{\frac {1}{(n+\alpha )^{s}}}={\frac {\alpha ^{1-s}}{s-1}}+{\frac {1}{2\alpha ^{s}}}+2\int _{0}^{\infty }{\frac {\sin \left(s\arctan {\frac {t}{\alpha }}\right)}{(\alpha ^{2}+t^{2})^{\frac {s}{2}}}}{\frac {dt}{e^{2\pi t}-1}},}$

which holds for all s, s ≠ 1.

Abel also gave the following variation for alternating sums:

${\displaystyle \sum _{n=0}^{\infty }(-1)^{n}f(n)={\frac {1}{2}}f(0)+i\int _{0}^{\infty }{\frac {f(it)-f(-it)}{2\sinh(\pi t)}}\,dt.}$

## Proof

Let ${\displaystyle f}$ be holomorphic on ${\displaystyle \Re (z)\geq 0}$, such that ${\displaystyle f(0)=0}$, ${\displaystyle f(z)=O(|z|^{k})}$ and for ${\displaystyle {\text{arg}}(z)\in (-\beta ,\beta )}$, ${\displaystyle f(z)=O(|z|^{-1-\delta })}$. Taking ${\displaystyle a=e^{i\beta /2}}$ with the residue theorem

${\displaystyle \int _{a^{-1}\infty }^{0}+\int _{0}^{a\infty }{\frac {f(z)}{e^{-2i\pi z}-1}}dz=-2i\pi \sum _{n=0}^{\infty }Res\left({\frac {f(z)}{e^{-2i\pi z}-1}}\right)=\sum _{n=0}^{\infty }f(n).}$

Then

{\displaystyle {\begin{aligned}\int _{a^{-1}\infty }^{0}{\frac {f(z)}{e^{-2i\pi z}-1}}dz&=-\int _{0}^{a^{-1}\infty }{\frac {f(z)}{e^{-2i\pi z}-1}}dz=\int _{0}^{a^{-1}\infty }{\frac {f(z)}{e^{2i\pi z}-1}}dz+\int _{0}^{a^{-1}\infty }f(z)dz=\\&=\int _{0}^{\infty }{\frac {f(a^{-1}t)}{e^{2i\pi a^{-1}t}-1}}d(a^{-1}t)+\int _{0}^{\infty }f(t)dt.\end{aligned}}}

Using the Cauchy integral theorem for the last one. ${\displaystyle \int _{0}^{a\infty }{\frac {f(z)}{e^{-2i\pi z}-1}}dz=\int _{0}^{\infty }{\frac {f(at)}{e^{-2i\pi at}-1}}d(at)}$, thus obtaining

${\displaystyle \sum _{n=0}^{\infty }f(n)=\int _{0}^{\infty }\left(f(t)+{\frac {a\,f(at)}{e^{-2i\pi at}-1}}+{\frac {a^{-1}f(a^{-1}t)}{e^{2i\pi a^{-1}t}-1}}\right)dt.}$

This identity stays true by analytic continuation everywhere the integral converges, letting ${\displaystyle a\to i}$ we obtain Abel-Plana's formula

${\displaystyle \sum _{n=0}^{\infty }f(n)=\int _{0}^{\infty }\left(f(t)+{\frac {i\,f(it)-i\,f(-it)}{e^{2\pi t}-1}}\right)dt}$.

The case f(0) ≠ 0 is obtained similarly, replacing ${\displaystyle \int _{a^{-1}\infty }^{a\infty }{\frac {f(z)}{e^{-2i\pi z}-1}}dz}$ by two integrals following the same curves with a small indentation on the left and right of 0.