# 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 [1]

${\displaystyle \sum _{n=0}^{\infty }f\left(a+n\right)=\int _{a}^{\infty }f\left(x\right)dx+{\frac {f\left(a\right)}{2}}+\int _{0}^{\infty }{\frac {f\left(a-ix\right)-f\left(a+ix\right)}{i\left(e^{2\pi x}-1\right)}}dx}$

For the case ${\displaystyle a=0}$ we have

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

It holds for functions ƒ 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 |ƒ| 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 ${\displaystyle s\in \mathbb {C} }$, s ≠ 1. Another powerful example is applying the formula to the function ${\displaystyle e^{-n}n^{x}}$: we obtain

${\displaystyle \Gamma (x+1)=\operatorname {Li} _{-x}\left(e^{-1}\right)+\theta (x)}$ where ${\displaystyle \Gamma (x)}$ is the gamma function, ${\displaystyle \operatorname {Li} _{s}\left(z\right)}$ is the polylogarithm and ${\displaystyle \theta (x)=\int _{0}^{\infty }{\frac {2t^{x}}{e^{2\pi t}-1}}\sin \left({\frac {\pi x}{2}}-t\right)dt}$.

Abel also gave the following variation for alternating sums:

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

which is related to the Lindelöf summation formula [2]

${\displaystyle \sum _{k=m}^{\infty }(-1)^{k}f(k)=(-1)^{m}\int _{-\infty }^{\infty }f(m-1/2+ix){\frac {dx}{2\cosh(\pi x)}}.}$

## 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 ${\displaystyle \operatorname {arg} (z)\in (-\beta ,\beta )}$, ${\displaystyle f(z)=O(|z|^{-1-\delta })}$. Taking ${\displaystyle a=e^{i\beta /2}}$ with the residue theorem

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

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