# 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

$\displaystyle{ \sum_{n=0}^\infty f(n)= \int_0^\infty f(x) \, dx+ \frac 1 2 f(0)+i \int_0^\infty \frac{f(i t)-f(-i t)}{e^{2\pi t}-1} \, dt. }$

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).

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.

Abel also gave the following variation for alternating sums:

$\displaystyle{ \sum_{n=0}^\infty (-1)^nf(n)= \frac {1}{2} f(0)+i \int_0^\infty \frac{f(i t)-f(-i t)}{2\sinh(\pi t)} \, dt. }$

## Proof

Let $\displaystyle{ f }$ be holomorphic on $\displaystyle{ \Re(z) \ge 0 }$, 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 $\displaystyle{ \int_{a^{-1}\infty}^0 + \int_0^{a\infty} \frac{f(z)}{e^{-2i\pi z}-1} dz = -2i\pi \sum_{n = 0}^\infty \mathrm{Res}\left(\frac{f(z)}{e^{-2i\pi z}-1}\right)=\sum_{n=0}^\infty f(n). }$

Then \displaystyle{ \begin{align} \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{align} }

Using the Cauchy integral theorem for the last one. $\displaystyle{ \int_0^{a\infty} \frac{f(z)}{e^{-2i\pi z}-1}dz = \int_0^\infty \frac{f(at)}{e^{-2i\pi a t}-1}d(at), }$ thus obtaining $\displaystyle{ \sum_{n=0}^\infty f(n)=\int_0^\infty \left(f(t)+\frac{a\, f(a t)}{e^{-2i\pi a t}-1}+\frac{a^{-1} f(a^{-1}t)}{e^{2i\pi a^{-1} t}-1}\right)dt. }$

This identity stays true by analytic continuation everywhere the integral converges, letting $\displaystyle{ a\to i }$ we obtain Abel-Plana's formula $\displaystyle{ \sum_{n=0}^\infty f(n)=\int_0^\infty \left(f(t)+\frac{i\, f(i t)-i\, f(-it)}{e^{2\pi t}-1}\right) dt. }$

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.