Summation of Series via the Residue Theorem - University of North ...

Summation of Series via the Residue Theorem

R. Herman

Let f (z) to be a meromorphic function with a finite set of poles that are not integers. Furthermore, let's assume that

| f (z)|

<

M |z|k

for k > 1 and M a constant. We seek to show that

f (n) = - Res [ cot z f (z); zk]

n=-

where the second sum is over the poles of f (z). We consider the integral

I = cot z f (z) dz,

(1)

CN

where the contour CN is shown in Figure 1.

N

+

1 2

(-1 + i)

y

N

+

1 2

(1 + i)

Figure 1: Contour for integration.

CN

-N

-2 -1

12

Nx

N

+

1 2

(-1 - i)

N

+

1 2

(1 - i)

There are simple poles of g(z) = cot z f (z) at z = 0, ?1, ?2, . . . .

The residues are found as

Res[g(z), z = n] = lim (z - n) cos z f (z) = f (n).

zn

sin z

Thus, for the above contour, the Residue theorem gives

N

cot z f (z) dz = 2i f (n) + Res [ cot z f (z); zk] ,

CN

n=-N

k

(2)

where the second sum is over the poles of f (z).

We eventually will let N . We need to consider the value of

the contour integral around the rectangle and equate it to this result.

We will show that

lim cot z f (z) dz = 0,

N CN

leading to the desired result.

N

+

1 2

(-1 + i)

C2 y

summation of series via the residue theorem 2

N

+

1 2

(1 + i)

Figure 2: Contour for integration.

-N C3

-2 -1

12

C1 Nx

N

+

1 2

(-1 - i)

C4

N

+

1 2

(1 - i)

The contour CN can be broken into four pieces, as noted in Figure 2.

| cot z| =

eiz + e-iz eiz - e-iz

=

eix-y + e-ix+y eix-y - e-ix+y

|eix-y| + |e-ix+y| |eix+y| - |e-ix-y|

=

e-y + e+y ey - e-y

= coth y.

(3)

Along path C2, y

>

1 2

.

So,

we

have

|

coth

z|

coth /2. Along

path

C4,

y

>

1 2

and

we

also

have

| cot z|

coth /2.

For

paths

C1

and

C3

we

also

have

-

1 2

y

1 2

.

In

these

cases

z

=

N

+

1 2

+

iy.

Then,

| cot z| =

cot

N

+

1 2

+

iy

= | tanh y|

tanh

2

<

coth

2

.

(4)

Therefore,

cot z f (z) dz | cot z|| f (z)| dz

CN

CN

M

coth

2

dz CN |z|k

M Nk

coth

2

dz

CN

=

M Nk

coth

2

4(2N

+

1).

(5)

summation of series via the residue theorem 3

Therefore, we have shown that

lim cot z f (z) dz = 0.

N CN

From Equation (2) we have

N

2i f (n) + Res [ cot z f (z); zk] = 0.

n=-N

k

This gives

N

f (n) = - Res [ cot z f (z); zk] .

n=-N

k

Example 1. Prove

N

n=-N

1 n2 + a2

=

a

coth a

for a>0.

Here

f (z)

=

z2

1 +a2

.

This

function

has

simple

poles

at

z

=

?ia.

The residues of cot z f (z) are then

lim (z

z?ia

ia)

cot z2 +

z a2

=

lim (z

z?ia

ia)

(z

cot z + ia)(z -

ia)

= cot ia 2ia

= cos ia 2ia sin ia

= - cosh a 2a sinh a

=

-

2a

coth

a.

(6)

Adding these residues gives the result. A similar series can be obtained under the same hypotheses:

(-1)n f (n) = - Res [ csc z f (z); zk]

n=-

where the second sum is over the poles of f (z).

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