Abel–Plana formula

In mathematics, the Abel–Plana formula is a summation formula discovered independently by and. It states that

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

For the case $$a=0$$ we have
 * $$\sum_{n=0}^\infty f(n)=\frac 1 2 f(0)+ \int_0^\infty f(x) \, dx+ i \int_0^\infty \frac{f(i t)-f(-i t)}{e^{2\pi t}-1} \, dt.$$

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. .

An example is provided by the Hurwitz zeta function,
 * $$\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 \in \mathbb{C}$$, $s ≠ 1$. Another powerful example is applying the formula to the function $$e^{-n}n^{x}$$: we obtain

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

Abel also gave the following variation for alternating sums:
 * $$\sum_{n=0}^\infty (-1)^nf(n)= \frac {1}{2} f(0)+i \int_0^\infty \frac{f(i t)-f(-i t)}{2\sinh(\pi t)} \, dt, $$

which is related to the Lindelöf summation formula


 * $$\sum_{k=m}^\infty (-1)^kf(k)=(-1)^m\int_{-\infty}^\infty f(m-1/2+ix)\frac{dx}{2\cosh(\pi x)}. $$

Proof
Let $$f$$ be holomorphic on $$\Re(z) \ge 0$$, such that $$f(0) = 0$$, $$f(z) = O(|z|^k)$$ and for $$\operatorname{arg}(z)\in (-\beta,\beta)$$, $$f(z) = O(|z|^{-1-\delta})$$. Taking $$a=e^{i \beta/2}$$ with the residue theorem $$\int_{a^{-1}\infty}^0 + \int_0^{a\infty} \frac{f(z)}{e^{-2i\pi z}-1} \, dz = -2i\pi \sum_{n = 0}^\infty \operatorname{Res}\left(\frac{f(z)}{e^{-2i\pi z}-1}\right)=\sum_{n=0}^\infty f(n).$$

Then $$\begin{align} \int_{a^{-1}\infty}^0 \frac{f(z)}{e^{-2i\pi z}-1} \, dz&=-\int_0^{a^{-1}\infty} \frac{f(z)}{e^{-2i\pi z}-1} \, dz \\ &=\int_0^{a^{-1}\infty}\frac{f(z)}{e^{2i\pi z}-1} \, dz+\int_0^{a^{-1}\infty} f(z) \, dz\\ &= \int_0^\infty \frac{f(a^{-1}t)}{e^{2i\pi a^{-1} t}-1} \, d(a^{-1}t)+\int_0^\infty f(t) \, dt. \end{align}$$

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

This identity stays true by analytic continuation everywhere the integral converges, letting $$a\to i$$ we obtain the Abel–Plana formula $$\sum_{n=0}^\infty f(n)=\int_0^\infty \left(f(t)+\frac{i\, f(i t)-i\, f(-it)}{e^{2\pi t}-1}\right) \, dt.$$

The case ƒ(0) ≠ 0 is obtained similarly, replacing $\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.