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.
See also: Residue (complex analysis). The statement is as follows:Let
U
a1,\ldots,an,
U0=U\smallsetminus\{a1,\ldots,an\},
f
U0.
\gamma
U0,
f
ak
\operatorname{Res}(f,ak)
\gamma
ak
\operatorname{I}(\gamma,ak),
f
\gamma
2\pii
\gamma
If
\gamma
\operatorname{I}(\gamma,ak)
1
ak
\gamma
0
with the sum over those
ak
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\}
\gamma
fdz
\gammai
V.
f
U0=U\smallsetminus\{ak\}
d(fdz)=0
U0.
V
W
U
\{aj\}
\{ak\},
V\smallsetminusW
W\smallsetminusV
U0,
is well-defined and equal to zero. Consequently, the contour integral of
fdz
\gammaj=\partialV
\gammaj,
aj
f
2\pii
\{aj\}.
\{\gammaj\},
\{\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.
The integral
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
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 isthe residue of at is
According to the residue theorem, then, we have
The contour may be split into a straight part and a curved arc, so thatand thus
Using some estimations, we have
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,
Therefore,
If then a similar argument with an arc that winds around rather than shows that
and finally we have
(If then the integral yields immediately to elementary calculus methods and its value is .)
The fact that has simple poles with residue 1 at each integer can be used to compute the sum
Consider, for example, . Let be the rectangle that is the boundary of with positive orientation, with an integer . By the residue formula,
The left-hand side goes to zero as since
|\cot(\piz)|
x=\pm\left(
12 | |
+ |
N\right)
O(N-2)
B2=
1 | |
6 |
.
(In fact, .) Thus, the residue is . We conclude:
which is a proof of the Basel problem.
The same argument works for all
f(x)=x-2n
n
f(x)=x-2n-1
0+\zeta(2n+1)-\zeta(2n+1)=0
The same trick can be used to establish the sum of the Eisenstein series: