Residue theorem explained

In complex analysis, the residue theorem, sometimes called Cauchy's residue theorem, is a powerful tool to evaluate line integrals of analytic functions over closed curves; it can often be used to compute real integrals and infinite series as well. It generalizes the Cauchy integral theorem and Cauchy's integral formula. The residue theorem should not be confused with special cases of the generalized Stokes' theorem; however, the latter can be used as an ingredient of its proof.

Statement of Cauchy's residue theorem

See also: Residue (complex analysis). The statement is as follows:Let

U

be a simply connected open subset of the complex plane containing a finite list of points

a1,\ldots,an,

U0=U\smallsetminus\{a1,\ldots,an\},

and a function

f

holomorphic on

U0.

Letting

\gamma

be a closed rectifiable curve in

U0,

and denoting the residue of

f

at each point

ak

by

\operatorname{Res}(f,ak)

and the winding number of

\gamma

around

ak

by

\operatorname{I}(\gamma,ak),

the line integral of

f

around

\gamma

is equal to

2\pii

times the sum of residues, each counted as many times as

\gamma

winds around the respective point:

\oint_\gamma f(z)\, dz = 2\pi i \sum_^n \operatorname(\gamma, a_k) \operatorname(f, a_k).

If

\gamma

is a positively oriented simple closed curve,

\operatorname{I}(\gamma,ak)

is

1

if

ak

is in the interior of

\gamma

and

0

if not, therefore

\oint_\gamma f(z)\, dz = 2\pi i \sum \operatorname(f, a_k)

with the sum over those

ak

inside

The relationship of the residue theorem to Stokes' theorem is given by the Jordan curve theorem. The general plane curve must first be reduced to a set of simple closed curves

\{\gammai\}

whose total is equivalent to

\gamma

for integration purposes; this reduces the problem to finding the integral of

fdz

along a Jordan curve

\gammai

with interior

V.

The requirement that

f

be holomorphic on

U0=U\smallsetminus\{ak\}

is equivalent to the statement that the exterior derivative

d(fdz)=0

on

U0.

Thus if two planar regions

V

and

W

of

U

enclose the same subset

\{aj\}

of

\{ak\},

the regions

V\smallsetminusW

and

W\smallsetminusV

lie entirely in

U0,

hence

\int_ d(f \, dz) - \int_ d(f \, dz)

is well-defined and equal to zero. Consequently, the contour integral of

fdz

along

\gammaj=\partialV

is equal to the sum of a set of integrals along paths

\gammaj,

each enclosing an arbitrarily small region around a single

aj

— the residues of

f

(up to the conventional factor

2\pii

at

\{aj\}.

Summing over

\{\gammaj\},

we recover the final expression of the contour integral in terms of the winding numbers

\{\operatorname{I}(\gamma,ak)\}.

In order to evaluate real integrals, the residue theorem is used in the following manner: the integrand is extended to the complex plane and its residues are computed (which is usually easy), and a part of the real axis is extended to a closed curve by attaching a half-circle in the upper or lower half-plane, forming a semicircle. The integral over this curve can then be computed using the residue theorem. Often, the half-circle part of the integral will tend towards zero as the radius of the half-circle grows, leaving only the real-axis part of the integral, the one we were originally interested in.

Examples

An integral along the real axis

The integral\int_^\infty \frac\,dx

arises in probability theory when calculating the characteristic function of the Cauchy distribution. It resists the techniques of elementary calculus but can be evaluated by expressing it as a limit of contour integrals.

Suppose and define the contour that goes along the real line from to and then counterclockwise along a semicircle centered at 0 from to . Take to be greater than 1, so that the imaginary unit is enclosed within the curve. Now consider the contour integral\int_C \,dz = \int_C \frac\,dz.

Since is an entire function (having no singularities at any point in the complex plane), this function has singularities only where the denominator is zero. Since, that happens only where or . Only one of those points is in the region bounded by this contour. Because is\begin\frac & =\frac\left(\frac-\frac\right) \\& =\frac -\frac,\endthe residue of at is\operatorname_f(z)=\frac.

According to the residue theorem, then, we have\int_C f(z)\,dz=2\pi i\cdot\operatorname\limits_f(z)=2\pi i \frac = \pi e^.

The contour may be split into a straight part and a curved arc, so that\int_ f(z)\,dz+\int_ f(z)\,dz=\pi e^and thus\int_^a f(z)\,dz =\pi e^-\int_ f(z)\,dz.

Using some estimations, we have\left|\int_\frac\,dz\right| \leq \pi a \cdot \sup_ \left| \frac \right| \leq \pi a \cdot \sup_ \frac

\leq \frac,and\lim_ \frac = 0.

The estimate on the numerator follows since, and for complex numbers along the arc (which lies in the upper half-plane), the argument of lies between 0 and . So,\left|e^\right| = \left|e^\right|=\left|e^\right|=e^ \le 1.

Therefore,\int_^\infty \frac\,dz=\pi e^.

If then a similar argument with an arc that winds around rather than shows that

\int_^\infty\frac\,dz=\pi e^t,

and finally we have\int_^\infty\frac\,dz=\pi e^.

(If then the integral yields immediately to elementary calculus methods and its value is .)

Evaluating zeta functions

The fact that has simple poles with residue 1 at each integer can be used to compute the sum \sum_^\infty f(n).

Consider, for example, . Let be the rectangle that is the boundary of with positive orientation, with an integer . By the residue formula,

\frac \int_ f(z) \pi \cot(\pi z) \, dz = \operatorname\limits_ + \sum_^N n^.

The left-hand side goes to zero as since

|\cot(\piz)|

is uniformly bounded on the contour, thanks to using

x=\pm\left(

12
+

N\right)

on the left and right side of the contour, and so the integrand has order

O(N-2)

over the entire contour. On the other hand,[1]

B2=

1
6

.

(In fact, .) Thus, the residue is . We conclude:

\sum_^\infty \frac = \fracwhich is a proof of the Basel problem.

The same argument works for all

f(x)=x-2n

where

n

is a positive integer, giving us \zeta(2n) = \frac.The trick does not work when

f(x)=x-2n-1

, since in this case, the residue at zero vanishes, and we obtain the useless identity

0+\zeta(2n+1)-\zeta(2n+1)=0

.

Evaluating Eisenstein series

The same trick can be used to establish the sum of the Eisenstein series:\pi \cot(\pi z) = \lim_ \sum_^N (z - n)^.

See also

References

External links

Notes and References

  1. . Note that the Bernoulli number

    B2n

    is denoted by

    Bn

    in Whittaker & Watson's book.