Residue theorem

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In complex analysis, a field in mathematics, the residue theorem, sometimes called Cauchy's residue theorem (one of many things named after Augustin-Louis Cauchy), is a powerful tool to evaluate line integrals of analytic functions over closed curves; it can often be used to compute real integrals as well. It generalizes the Cauchy integral theorem and Cauchy's integral formula. From a geometrical perspective, it is a special case of the generalized Stokes' theorem.

Illustration of the setting.

The statement is as follows:

Suppose U is a simply connected open subset of the complex plane, and a1,...,an are finitely many points of U and f is a function which is defined and holomorphic on U \ {a1,...,an}. If γ is a rectifiable curve in U which does not meet any of the ak, and whose start point equals its endpoint, then

\oint_\gamma f(z)\, dz =
2\pi i \sum_{k=1}^n \operatorname{I}(\gamma, a_k) 
\operatorname{Res}( f, a_k ).

If γ is a positively oriented simple closed curve, I(γ, ak) = 1 if ak is in the interior of γ, and 0 if not, so

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

with the sum over those k for which ak is inside γ.

Here, Res(f, ak) denotes the residue of f at ak, and I(γ, ak) is the winding number of the curve γ about the point ak. This winding number is an integer which intuitively measures how many times the curve γ winds around the point ak; it is positive if γ moves in a counter clockwise ("mathematically positive") manner around ak and 0 if γ doesn't move around ak at all.

The relationship of the residue theorem to Stokes' theorem is given by the Jordan curve theorem. The general plane curve γ must first be reduced to a set of simple closed curves {γi} whose total is equivalent to γ for integration purposes; this reduces the problem to finding the integral of f dz along a Jordan curve γi with interior V. The requirement that f be holomorphic on U0 = U \ {ak} is equivalent to the statement that the exterior derivative d(f dz) = 0 on U0. Thus if two planar regions V and W of U enclose the same subset {aj} of {ak}, the regions V\W and W\V lie entirely in U0, and hence \scriptstyle\int_{V \backslash W} d(f \, dz) - \int_{W \backslash V} d(f \, dz) is well-defined and equal to zero. Consequently, the contour integral of f dz along γj = ∂V is equal to the sum of a set of integrals along paths λj, each enclosing an arbitrarily small region around a single aj—the residues of f (up to the conventional factor 2πi) at {aj}. Summing over {γj}, we recover the final expression of the contour integral in terms of the winding numbers {I(γ, 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.

Example[edit]

The integral

\int_{-\infty}^\infty {e^{itx} \over x^2+1}\,dx
The contour C.

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 t > 0 and define the contour C that goes along the real line from −a to a and then counterclockwise along a semicircle centered at 0 from a to −a. Take a to be greater than 1, so that the imaginary unit i is enclosed within the curve. The contour integral is

\int_C {f(z)}\,dz =\int_C {e^{itz} \over z^2+1}\,dz.

Since eitz is an entire function (having no singularities at any point in the complex plane), this function has singularities only where the denominator z2 + 1 is zero. Since z2 + 1 = (z + i)(zi), that happens only where z = i or z = −i. Only one of those points is in the region bounded by this contour. Because f(z) is

\begin{align}
\frac{e^{itz}}{z^2+1} & =\frac{e^{itz}}{2i}\left(\frac{1}{z-i}-\frac{1}{z+i}\right) \\
& =\frac{e^{itz}}{2i(z-i)} -\frac{e^{itz}}{2i(z+i)} ,
\end{align}

the residue of f(z) at z = i is

\operatorname{Res}\limits_{z=i}f(z)={e^{-t}\over 2i}.

According to the residue theorem, then, we have

\int_C f(z)\,dz=2\pi i\cdot\operatorname{Res}\limits_{z=i}f(z)=2\pi i{e^{-t} \over 2i}=\pi e^{-t}.

The contour C may be split into a "straight" part and a curved arc, so that

\int_{\mathrm{straight}} f(z)\,dz+\int_{\mathrm{arc}} f(z)\,dz=\pi e^{-t}\,

and thus

\int_{-a}^a f(z)\,dz =\pi e^{-t}-\int_{\mathrm{arc}} f(z)\,dz.

Using some estimations, we have

\left|\int_{\mathrm{arc}}{e^{itz} \over z^2+1}\,dz\right| \leq \int_{\mathrm{arc}}\left|{e^{itz} \over z^2+1}\right| dz \le \int_{\mathrm{arc}}{1 \over |z^2+1|}dz\leq \int_{\mathrm{arc}}{1 \over a^2-1}dz =  \frac{\pi a}{a^2-1}.

and

\lim_{a \to \infty} \frac{\pi a}{a^2-1} =  0.

Note that, since t > 0 and for complex numbers in the upper halfplane the argument lies between 0 and π, one can estimate

\left|e^{itz}\right|=\left|e^{it|z|(\cos\phi + i\sin\phi)}\right|=\left|e^{-t|z|\sin\phi + it|z|\cos\phi}\right|=e^{-t|z|\sin\phi} \le 1.

Therefore

\int_{-\infty}^\infty {e^{itz} \over z^2+1}\,dz=\pi e^{-t}.

If t < 0 then a similar argument with an arc C' that winds around −i rather than i shows that

The contour C'.
\int_{-\infty}^\infty{e^{itz} \over z^2+1}\,dz=\pi e^t,

and finally we have

\int_{-\infty}^\infty{e^{itz} \over z^2+1}\,dz=\pi e^{-\left|t\right|}.

(If t = 0 then the integral yields immediately to elementary calculus methods and its value is π.)

Example 2[edit]

The fact that \pi \operatorname{cot}(\pi z) has simple poles with residue one at each integer can be used to compute the sum \displaystyle \sum_{-\infty}^\infty f(n). Consider, for example, f(z) = z^{-2}. Let \Gamma_N be the rectangle with sides |z| = N + 1/2 and with positive orientation. By the residue formula,

{1 \over 2 \pi i} \int_{\Gamma_N} f(z) \pi \operatorname{cot}(\pi z) \, dz = \operatorname{Res}_{z = 0} + \sum_{n = -N, n\ne 0}^N n^{-2}.

The left-hand side goes to zero as N \to \infty since the integrand has order O(N^{-2}). On the other hand,

{z / 2} \operatorname{cot}(z/2) = 1 - B_1 {z^2 \over 2!} + \cdots, \, B_1 = {1 \over 6}.[1]

Thus, the residue \operatorname{Res}_{z = 0} at z = 0 is -{\pi^2 \over 3}. We conclude:

\sum_{n = 1}^\infty {1 \over n^2} = {\pi^2 \over 6}

(cf. Basel problem.)

The same trick can be used to establish

\pi \operatorname{cot}(\pi z) = \lim_{N \to \infty} \sum_{-N}^N (z - n)^{-1}.

We take f(z) = (z - w)^{-1} with w a non-integer and we shall show the above for w. The difficulty in this case is to show the vanishing of the contour integral at infinity. We have: \int_{\Gamma_N} {\pi \operatorname{cot}(\pi z) \over z} \, dz = 0 since the integrad is an even function and so the contributions from the left contour and the right contour cancel each other out. Thus,

\int_{\Gamma_N} f(z) \pi \operatorname{cot}(\pi z) \, dz = \int_{\Gamma_N} ({1 \over z - w} - {1 \over z}) \pi \operatorname{cot}(\pi z) \, dz

goes to zero as N \to \infty.

See also[edit]

References[edit]

  1. ^ Whittaker-Watson, a course of modern analysis, § 7.2.

External links[edit]