Abel–Plana formula

Template:Short description Template:Distinguish In mathematics, the Abel–Plana formula is a summation formula discovered independently by Template:Harvs and Template:Harvs. It states that

$\sum _{n=0}^{\mathrm{\infty }}f(a+n)=\frac{f\left(a\right)}{2}+{\int }_a^{\mathrm{\infty }}f\left(x\right)dx+i{\int }_0^{\mathrm{\infty }}\frac{f(a+it)-f(a-it)}{e^{2\pi t}-1}dt$

For the case $a=0$ we have

$\sum _{n=0}^{\mathrm{\infty }}f(n)=\frac{f(0)}{2}+{\int }_0^{\mathrm{\infty }}f(x)dx+i{\int }_0^{\mathrm{\infty }}\frac{f(it)-f(-it)}{e^{2\pi t}-1}dt.$

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. Template:Harv.

An example is provided by the Hurwitz zeta function,

$\zeta (s,\alpha )=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(n+\alpha {)}^s}=\frac{{\alpha }^{1-s}}{s-1}+\frac{1}{2{\alpha }^s}+2{\int }_0^{\mathrm{\infty }}\frac{\sin (s\arctan \frac{t}{\alpha })}{({\alpha }^2+t^2{)}^{\frac{s}{2}}}\frac{dt}{e^{2\pi t}-1},$

which holds for all $s\in ℂ$, Template:Math. Another powerful example is applying the formula to the function $e^{-n}{n}^x$: we obtain

$\mathrm{\Gamma }(x+1)={\mathrm{Li}}_{-x}\left(e^{-1}\right)+\theta (x)$ where $\mathrm{\Gamma }(x)$ is the gamma function, ${\mathrm{Li}}_s\left(z\right)$ is the polylogarithm and $\theta (x)={\int }_0^{\mathrm{\infty }}\frac{2t^x}{e^{2\pi t}-1}\sin (\frac{\pi x}{2}-t)dt$.

Abel also gave the following variation for alternating sums:

$\sum _{n=0}^{\mathrm{\infty }}(-1{)}^{n}f(n)=\frac{1}{2}f(0)+i{\int }_0^{\mathrm{\infty }}\frac{f(it)-f(-it)}{2\sinh (\pi t)}dt,$

which is related to the Lindelöf summation formula

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

Proof

Let $f$ be holomorphic on $\mathrm{\Re }(z)\ge 0$, such that $f(0)=0$, $f(z)=O(|z{|}^k)$ and for $\arg (z)\in (-\beta ,\beta )$, $f(z)=O(|z{|}^{-1-\delta })$. Taking $a=e^{i\beta /2}$ with the residue theorem $${\int }_{a^{-1}\mathrm{\infty }}^0+{\int }_0^{a\mathrm{\infty }}\frac{f(z)}{e^{-2i\pi z}-1}dz=-2i\pi \sum _{n=0}^{\mathrm{\infty }}\mathrm{Res}\left(\frac{f(z)}{e^{-2i\pi z}-1}\right)=\sum _{n=0}^{\mathrm{\infty }}f(n).$$

Then $$\begin{array}{cc}{\int }_{a^{-1}\mathrm{\infty }}^0\frac{f(z)}{e^{-2i\pi z}-1}dz& =-{\int }_0^{a^{-1}\mathrm{\infty }}\frac{f(z)}{e^{-2i\pi z}-1}dz\\ & ={\int }_0^{a^{-1}\mathrm{\infty }}\frac{f(z)}{e^{2i\pi z}-1}dz+{\int }_0^{a^{-1}\mathrm{\infty }}f(z)dz\\ & ={\int }_0^{\mathrm{\infty }}\frac{f(a^{-1}t)}{e^{2i\pi a^{-1}t}-1}d(a^{-1}t)+{\int }_0^{\mathrm{\infty }}f(t)dt.\end{array}$$

Using the Cauchy integral theorem for the last one. $${\int }_0^{a\mathrm{\infty }}\frac{f(z)}{e^{-2i\pi z}-1}dz={\int }_0^{\mathrm{\infty }}\frac{f(at)}{e^{-2i\pi at}-1}d(at),$$ thus obtaining $$\sum _{n=0}^{\mathrm{\infty }}f(n)={\int }_0^{\mathrm{\infty }}(f(t)+\frac{af(at)}{e^{-2i\pi at}-1}+\frac{a^{-1}f(a^{-1}t)}{e^{2i\pi a^{-1}t}-1})dt.$$

This identity stays true by analytic continuation everywhere the integral converges, letting $a\to i$ we obtain the Abel–Plana formula $$\sum _{n=0}^{\mathrm{\infty }}f(n)={\int }_0^{\mathrm{\infty }}(f(t)+\frac{if(it)-if(-it)}{e^{2\pi t}-1})dt.$$

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

See also

• Euler–Maclaurin summation formula
• Euler–Boole summation
• Ramanujan summation

References

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External links

• Template:Mathworld