Mathematical Methods for Physics and Engineering : A Comprehensive Guide

25.4 SUMMATION OF SERIES

By considering

∮

C

πcotπz

(a+z)^2

dz,

whereais not an integer andCis a circle of large radius, evaluate

∑∞

n=−∞

1

(a+n)^2

.

The integrand has (i) simple poles atz= integern,for−∞<n<∞, due to the factor
cotπzand (ii) a double pole atz=−a.
(i) To find the residue of cotπz, putz=n+ξfor smallξ:

cotπz=

cos(nπ+ξπ)

sin(nπ+ξπ)

≈

cosnπ

(cosnπ)ξπ

=

1

ξπ

.

The residue of the integrand atz=nis thusπ(a+n)−^2 π−^1.
(ii) Puttingz=−a+ξfor smallξand determining the coefficient ofξ−^1 gives§

πcotπz

(a+z)^2

=

π

ξ^2

cot(−aπ+ξπ)

=

π

ξ^2

{

cot(−aπ)+ξ

[

d

dz

(cotπz)

]

z=−a

+···

}

,

so that the residue at the double polez=−ais given by

π[−πcosec^2 πz]z=−a=−π^2 cosec^2 πa.

Collecting together these results to express the residue theorem gives

I=

∮

C

πcotπz

(a+z)^2

dz=2πi

[ N

∑

n=−N

1

(a+n)^2

−π^2 cosec^2 πa

]

, (25.23)

whereNequals the integer part ofR. But as the radiusRofCtends to∞,cotπz→∓i
(depending on whether Imzis greater or less than zero, respectively). Thus

I<k

∫

dz

(a+z)^2

,

which tends to 0 asR→∞. ThusI→0asR(and henceN)→∞, and (25.23) establishes
the result

∑∞

n=−∞

1

(a+n)^2

=

π^2

sin^2 πa

.

Series with alternating signs in the terms, i.e. (−1)n, can also be attempted

in this way but using cosecπzrather than cotπz, since the former has residue

(−1)nπ−^1 atz=n(see exercise 25.11).

§This again illustrates one of the techniques for determining residues.