Abel–Plana formula (original) (raw)

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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]

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

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

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

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,

ζ ( s , α ) = ∑ n = 0 ∞ 1 ( n + α ) s = α 1 − s s − 1 + 1 2 α s + 2 ∫ 0 ∞ sin ⁡ ( s arctan ⁡ t α ) ( α 2 + t 2 ) s 2 d t e 2 π t − 1 , {\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}},} $\zeta (s,\alpha )=\sum _{n=0}^{\infty }{\frac {1}{(n+\alpha )^{s}}}={\frac {\alpha ^{1-s}}{s-1}}+{\frac {1}{2\alpha ^{s}}}+2\int _{0}^{\infty }{\frac {\sin \left(s\arctan {\frac {t}{\alpha }}\right)}{(\alpha ^{2}+t^{2})^{\frac {s}{2}}}}{\frac {dt}{e^{2\pi t}-1}},$

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

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

Abel also gave the following variation for alternating sums:

∑ n = 0 ∞ ( − 1 ) n f ( n ) = 1 2 f ( 0 ) + i ∫ 0 ∞ f ( i t ) − f ( − i t ) 2 sinh ⁡ ( π t ) d t , {\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,} $\sum _{n=0}^{\infty }(-1)^{n}f(n)={\frac {1}{2}}f(0)+i\int _{0}^{\infty }{\frac {f(it)-f(-it)}{2\sinh(\pi t)}}\,dt,$

which is related to the Lindelöf summation formula [2]

∑ k = m ∞ ( − 1 ) k f ( k ) = ( − 1 ) m ∫ − ∞ ∞ f ( m − 1 / 2 + i x ) d x 2 cosh ⁡ ( π x ) . {\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)}}.} $\sum _{k=m}^{\infty }(-1)^{k}f(k)=(-1)^{m}\int _{-\infty }^{\infty }f(m-1/2+ix){\frac {dx}{2\cosh(\pi x)}}.$

Let f {\displaystyle f} $f$ be holomorphic on ℜ ( z ) ≥ 0 {\displaystyle \Re (z)\geq 0} $\Re (z)\geq 0$ , such that f ( 0 ) = 0 {\displaystyle f(0)=0} $f(0)=0$ , f ( z ) = O ( | z | k ) {\displaystyle f(z)=O(|z|^{k})} $f(z)=O(|z|^{k})$ and for arg ⁡ ( z ) ∈ ( − β , β ) {\displaystyle \operatorname {arg} (z)\in (-\beta ,\beta )} $\operatorname {arg} (z)\in (-\beta ,\beta )$ , f ( z ) = O ( | z | − 1 − δ ) {\displaystyle f(z)=O(|z|^{-1-\delta })} $f(z)=O(|z|^{-1-\delta })$ . Taking a = e i β / 2 {\displaystyle a=e^{i\beta /2}} $a=e^{i\beta /2}$ with the residue theorem ∫ a − 1 ∞ 0 + ∫ 0 a ∞ f ( z ) e − 2 i π z − 1 d z = − 2 i π ∑ n = 0 ∞ Res ⁡ ( f ( z ) e − 2 i π z − 1 ) = ∑ n = 0 ∞ f ( n ) . {\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).} $\int _{a^{-1}\infty }^{0}+\int _{0}^{a\infty }{\frac {f(z)}{e^{-2i\pi z}-1}}\,dz=-2i\pi \sum _{n=0}^{\infty }\operatorname {Res} \left({\frac {f(z)}{e^{-2i\pi z}-1}}\right)=\sum _{n=0}^{\infty }f(n).$

Then ∫ a − 1 ∞ 0 f ( z ) e − 2 i π z − 1 d z = − ∫ 0 a − 1 ∞ f ( z ) e − 2 i π z − 1 d z = ∫ 0 a − 1 ∞ f ( z ) e 2 i π z − 1 d z + ∫ 0 a − 1 ∞ f ( z ) d z = ∫ 0 ∞ f ( a − 1 t ) e 2 i π a − 1 t − 1 d ( a − 1 t ) + ∫ 0 ∞ f ( t ) d t . {\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}}} ${\begin{aligned}\int _{a^{-1}\infty }^{0}{\frac {f(z)}{e^{-2i\pi z}-1}}\,dz&=-\int _{0}^{a^{-1}\infty }{\frac {f(z)}{e^{-2i\pi z}-1}}\,dz\\&=\int _{0}^{a^{-1}\infty }{\frac {f(z)}{e^{2i\pi z}-1}}\,dz+\int _{0}^{a^{-1}\infty }f(z)\,dz\\&=\int _{0}^{\infty }{\frac {f(a^{-1}t)}{e^{2i\pi a^{-1}t}-1}}\,d(a^{-1}t)+\int _{0}^{\infty }f(t)\,dt.\end{aligned}}$

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

This identity stays true by analytic continuation everywhere the integral converges, letting a → i {\displaystyle a\to i} $a\to i$ we obtain the Abel–Plana formula ∑ n = 0 ∞ f ( n ) = ∫ 0 ∞ ( f ( t ) + i f ( i t ) − i f ( − i t ) e 2 π t − 1 ) d t . {\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.} $\sum _{n=0}^{\infty }f(n)=\int _{0}^{\infty }\left(f(t)+{\frac {i\,f(it)-i\,f(-it)}{e^{2\pi t}-1}}\right)\,dt.$

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

^ Hermite, C. (1901). "Extrait de quelques lettres de M. Ch. Hermite à M. S. Píncherle". Annali di Matematica Pura ed Applicata. Serie III. 5: 57–72.
^ "Summation Formulas of Euler-Maclaurin and Abel-Plana: Old and New Results and Applications" (PDF).

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