Английская Википедия:Abel–Plana formula
Шаблон:Distinguish In mathematics, the Abel–Plana formula is a summation formula discovered independently by Шаблон:Harvs and Шаблон:Harvs. It states that [1]
<math>\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</math>
For the case <math>a=0</math> we have
- <math>\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.</math>
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. Шаблон:Harv.
An example is provided by the Hurwitz zeta function,
- <math>\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},</math> which holds for all <math>s \in \mathbb{C}</math>, Шаблон:Math. Another powerful example is applying the formula to the function <math>e^{-n}n^{x}</math>: we obtain
<math>\Gamma(x+1)=\operatorname{Li}_{-x}\left(e^{-1}\right)+\theta(x)</math> where <math>\Gamma(x)</math> is the gamma function, <math>\operatorname{Li}_{s}\left(z\right)</math> is the polylogarithm and <math>\theta(x)=\int_{0}^{\infty}\frac{2t^{x}}{e^{2\pi t}-1}\sin\left(\frac{\pi x}{2}-t\right)dt</math>.
Abel also gave the following variation for alternating sums:
- <math>\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, </math>
which is related to the Lindelöf summation formula [2]
- <math>\sum_{k=m}^\infty (-1)^kf(k)=(-1)^m\int_{-\infty}^\infty f(m-1/2+ix)\frac{dx}{2\cosh(\pi x)}. </math>
Proof
Let <math>f</math> be holomorphic on <math>\Re(z) \ge 0</math>, such that <math>f(0) = 0</math>, <math>f(z) = O(|z|^k)</math> and for <math>\operatorname{arg}(z)\in (-\beta,\beta)</math>, <math>f(z) = O(|z|^{-1-\delta})</math>. Taking <math>a=e^{i \beta/2}</math> with the residue theorem <math display="block">\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).</math>
Then <math display="block">\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}</math>
Using the Cauchy integral theorem for the last one. <math display="block">\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),</math> thus obtaining <math display="block">\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.</math>
This identity stays true by analytic continuation everywhere the integral converges, letting <math>a\to i</math> we obtain the Abel–Plana formula <math display="block">\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.</math>
The case ƒ(0) ≠ 0 is obtained similarly, replacing <math display="inline">\int_{a^{-1}\infty}^{a\infty} \frac{f(z)}{e^{-2i\pi z}-1} \, dz </math> by two integrals following the same curves with a small indentation on the left and right of 0.
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