Residue theorem

Template:Short description Template:Complex analysis sidebar 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

The statement is as follows:

File:Residue theorem illustration.svg

Illustration of the setting

Residue theorem: Let $U$ be a simply connected open subset of the complex plane containing a finite list of points $a_1,\dots ,a_n,$ $U_0=U\setminus \{a_1,\dots ,a_n\},$ and a function $f$ holomorphic on $U_0.$ Letting $\gamma$ be a closed rectifiable curve in $U_0,$ and denoting the residue of $f$ at each point $a_k$ by $\mathrm{Res}(f,a_k)$ and the winding number of $\gamma$ around $a_k$ by $\mathrm{I}(\gamma ,a_k),$ the line integral of $f$ around $\gamma$ is equal to $2\pi i$ times the sum of residues, each counted as many times as $\gamma$ winds around the respective point:

$${\oint }_{\gamma }f(z)dz=2\pi i\sum _{k=1}^n\mathrm{I}(\gamma ,a_k)\mathrm{Res}(f,a_k).$$

If $\gamma$ is a positively oriented simple closed curve, $\mathrm{I}(\gamma ,a_k)$ is $1$ if $a_k$ is in the interior of $\gamma$ and $0$ if not, therefore

$${\oint }_{\gamma }f(z)dz=2\pi i\sum \mathrm{Res}(f,a_k)$$

with the sum over those $a_k$ inside $\gamma .$

The relationship of the residue theorem to Stokes' theorem is given by the Jordan curve theorem. The general plane curve Template:Mvar must first be reduced to a set of simple closed curves $\{{\gamma }_i\}$ whose total is equivalent to $\gamma$ for integration purposes; this reduces the problem to finding the integral of $fdz$ along a Jordan curve ${\gamma }_i$ with interior $V.$ The requirement that $f$ be holomorphic on $U_0=U\setminus \{a_k\}$ is equivalent to the statement that the exterior derivative $d(fdz)=0$ on $U_0.$ Thus if two planar regions $V$ and $W$ of $U$ enclose the same subset $\{a_j\}$ of $\{a_k\},$ the regions $V\setminus W$ and $W\setminus V$ lie entirely in $U_0,$ hence

$$\underset{V\setminus W}{\int }d(fdz)-\underset{W\setminus V}{\int }d(fdz)$$

is well-defined and equal to zero. Consequently, the contour integral of $fdz$ along ${\gamma }_j=\partial V$ is equal to the sum of a set of integrals along paths ${\gamma }_j,$ each enclosing an arbitrarily small region around a single $a_j$ — the residues of $f$ (up to the conventional factor $2\pi i$ at $\{a_j\}.$ Summing over $\{{\gamma }_j\},$ we recover the final expression of the contour integral in terms of the winding numbers $\{\mathrm{I}(\gamma ,a_k)\}.$

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 $$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{e^{itx}}{x^2+1}dx$$

File:Contour example.svg

The contour Template:Mvar.

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 Template:Mvar that goes along the real line from −a to Template:Mvar and then counterclockwise along a semicircle centered at 0 from Template:Mvar to −a. Take Template:Mvar to be greater than 1, so that the imaginary unit Template:Mvar is enclosed within the curve. Now consider the contour integral $$\underset{C}{\int }f(z)dz=\underset{C}{\int }\frac{e^{itz}}{z^2+1}dz.$$

Since e^itz 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)(z − i), 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 $$\begin{array}{rl}\frac{e^{itz}}{z^2+1}& =\frac{e^{itz}}{2i}(\frac{1}{z-i}-\frac{1}{z+i})\\ & =\frac{e^{itz}}{2i(z-i)}-\frac{e^{itz}}{2i(z+i)},\end{array}$$ the residue of f(z) at z = i is $${\mathrm{Res}}_{z=i}f(z)=\frac{e^{-t}}{2i}.$$

According to the residue theorem, then, we have $$\underset{C}{\int }f(z)dz=2\pi i\cdot \underset{z=i}{\mathrm{Res}}f(z)=2\pi i\frac{e^{-t}}{2i}=\pi e^{-t}.$$

The contour Template:Mvar may be split into a straight part and a curved arc, so that $$\underset{\mathrm{s}\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}}{\int }f(z)dz+\underset{\mathrm{a}\mathrm{r}\mathrm{c}}{\int }f(z)dz=\pi e^{-t}$$ and thus $$\underset{-a}{\overset{a}{\int }}f(z)dz=\pi e^{-t}-\underset{\mathrm{a}\mathrm{r}\mathrm{c}}{\int }f(z)dz.$$

Using some estimations, we have $$\left|\underset{\mathrm{a}\mathrm{r}\mathrm{c}}{\int }\frac{e^{itz}}{z^2+1}dz\right|\le \pi a\cdot \underset{\text{arc}}{\sup }\left|\frac{e^{itz}}{z^2+1}\right|\le \pi a\cdot \underset{\text{arc}}{\sup }\frac{1}{|z^2+1|}\le \frac{\pi a}{a^2-1},$$ and $$\underset{a\to \mathrm{\infty }}{\lim }\frac{\pi a}{a^2-1}=0.$$

The estimate on the numerator follows since t > 0, and for complex numbers Template:Mvar along the arc (which lies in the upper half-plane), the argument Template:Mvar of Template:Mvar lies between 0 and Template:Pi. So, $$\left|e^{itz}\right|=\left|e^{it|z|(\cos \phi +i\sin \phi )}\right|=\left|e^{-t|z|\sin \phi +it|z|\cos \phi }\right|=e^{-t|z|\sin \phi }\le 1.$$

Therefore, $$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{e^{itz}}{z^2+1}dz=\pi e^{-t}.$$

If t < 0 then a similar argument with an arc Template:Prime that winds around −i rather than i shows that

File:Contour example 2.svg

The contour Template:Prime.

$$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{e^{itz}}{z^2+1}dz=\pi e^t,$$

and finally we have $$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{e^{itz}}{z^2+1}dz=\pi e^{-\left|t\right|}.$$

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

Evaluating zeta functions

The fact that π cot(πz) has simple poles with residue 1 at each integer can be used to compute the sum $$\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}f(n).$$

Consider, for example, f(z) = z⁻². Let Γ_N be the rectangle that is the boundary of [−N − Template:Sfrac, N + Template:Sfrac]² with positive orientation, with an integer Template:Mvar. By the residue formula,

$$\frac{1}{2\pi i}\underset{{\mathrm{\Gamma }}_N}{\int }f(z)\pi \cot (\pi z)dz=\underset{z=0}{\mathrm{Res}}+\sum _{\genfrac{}{}{0ex}{}{n=-N}{n\ne 0}}^N{n}^{-2}.$$

The left-hand side goes to zero as N → ∞ since $|\cot (\pi z)|$ is uniformly bounded on the contour, thanks to using $x=\pm (\frac{1}{2}+N)$ 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,

$$\frac{z}{2}\cot \left(\frac{z}{2}\right)=1-B_2\frac{z^2}{2!}+\cdots$$ where the Bernoulli number $B_2=\frac{1}{6}.$

(In fact, Template:Sfrac cot(Template:Sfrac) = Template:Sfrac − Template:Sfrac.) Thus, the residue Res_z=0 is −Template:Sfrac. We conclude:

$$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}=\frac{{\pi }^2}{6}$$ which 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$$\zeta (2n)=\frac{(-1{)}^{n+1}{B}_{2n}(2\pi {)}^{2n}}{2(2n)!}.$$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:$$\pi \cot (\pi z)=\underset{N\to \mathrm{\infty }}{\lim }\sum _{n=-N}^N(z-n{)}^{-1}.$$

See also

• Residue (complex analysis)
• Cauchy's integral formula
• Glasser's master theorem
• Jordan's lemma
• Methods of contour integration
• Morera's theorem
• Nachbin's theorem
• Residue at infinity
• Logarithmic form

External links

• Template:Springer
• Residue theorem in MathWorld