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 theoremEdit
Template:See also The statement is as follows:
File:Residue theorem illustration.svgIllustration of the settingResidue theorem: Let <math>U</math> be a simply connected open subset of the complex plane containing a finite list of points <math>a_1, \ldots, a_n,</math> <math>U_0 = U \smallsetminus \{a_1, \ldots, a_n\},</math> and a function <math>f</math> holomorphic on <math>U_0.</math> Letting <math>\gamma</math> be a closed rectifiable curve in <math>U_0,</math> and denoting the residue of <math>f</math> at each point <math>a_k</math> by <math>\operatorname{Res}(f, a_k)</math> and the winding number of <math>\gamma</math> around <math>a_k</math> by <math>\operatorname{I}(\gamma, a_k),</math> the line integral of <math>f</math> around <math>\gamma</math> is equal to <math>2\pi i</math> times the sum of residues, each counted as many times as <math>\gamma</math> winds around the respective point:
<math display=block> \oint_\gamma f(z)\, dz = 2\pi i \sum_{k=1}^n \operatorname{I}(\gamma, a_k) \operatorname{Res}(f, a_k). </math>
If <math>\gamma</math> is a positively oriented simple closed curve, <math>\operatorname{I}(\gamma, a_k)</math> is <math>1</math> if <math>a_k</math> is in the interior of <math>\gamma</math> and <math>0</math> if not, therefore
<math display=block> \oint_\gamma f(z)\, dz = 2\pi i \sum \operatorname{Res}(f, a_k) </math>
with the sum over those <math>a_k</math> inside Template:Nobr
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 <math>\{\gamma_i\}</math> whose total is equivalent to <math>\gamma</math> for integration purposes; this reduces the problem to finding the integral of <math>f\, dz</math> along a Jordan curve <math>\gamma_i</math> with interior <math>V.</math> The requirement that <math>f</math> be holomorphic on <math>U_0 = U \smallsetminus \{a_k\}</math> is equivalent to the statement that the exterior derivative <math>d(f\, dz) = 0</math> on <math>U_0.</math> Thus if two planar regions <math>V</math> and <math>W</math> of <math>U</math> enclose the same subset <math>\{a_j\}</math> of <math>\{a_k\},</math> the regions <math>V \smallsetminus W</math> and <math>W \smallsetminus V</math> lie entirely in <math>U_0,</math> hence
<math display=block> \int_{V \smallsetminus W} d(f \, dz) - \int_{W \smallsetminus V} d(f \, dz) </math>
is well-defined and equal to zero. Consequently, the contour integral of <math>f\, dz</math> along <math>\gamma_j = \partial V</math> is equal to the sum of a set of integrals along paths <math>\gamma_j,</math> each enclosing an arbitrarily small region around a single <math>a_j</math> — the residues of <math>f</math> (up to the conventional factor <math>2\pi i</math> at <math>\{a_j\}.</math> Summing over <math>\{\gamma_j\},</math> we recover the final expression of the contour integral in terms of the winding numbers <math>\{\operatorname{I}(\gamma, a_k)\}.</math>
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.
Calculation of residuesEdit
ExamplesEdit
An integral along the real axisEdit
The integral <math display="block">\int_{-\infty}^\infty \frac{e^{itx}}{x^2+1}\,dx</math>
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 Template:Math and define the contour Template:Mvar that goes along the real line from Template:Math to Template:Mvar and then counterclockwise along a semicircle centered at 0 from Template:Mvar to Template:Math. 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 <math display="block">\int_C {f(z)}\,dz = \int_C \frac{e^{itz}}{z^2+1}\,dz.</math>
Since Template:Math is an entire function (having no singularities at any point in the complex plane), this function has singularities only where the denominator Template:Math is zero. Since Template:Math, that happens only where Template:Math or Template:Math. Only one of those points is in the region bounded by this contour. Because Template:Math is <math display="block">\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}</math> the residue of Template:Math at Template:Math is <math display="block">\operatorname{Res}_{z=i}f(z)=\frac{e^{-t}}{2i}.</math>
According to the residue theorem, then, we have <math display="block">\int_C f(z)\,dz=2\pi i\cdot\operatorname{Res}\limits_{z=i}f(z)=2\pi i \frac{e^{-t}}{2i} = \pi e^{-t}.</math>
The contour Template:Mvar may be split into a straight part and a curved arc, so that <math display="block">\int_{\mathrm{straight}} f(z)\,dz+\int_{\mathrm{arc}} f(z)\,dz=\pi e^{-t}</math> and thus <math display="block">\int_{-a}^a f(z)\,dz =\pi e^{-t}-\int_{\mathrm{arc}} f(z)\,dz.</math>
Using some estimations, we have <math display="block">\left|\int_{\mathrm{arc}}\frac{e^{itz}}{z^2+1}\,dz\right| \leq \pi a \cdot \sup_{\text{arc}} \left| \frac{e^{itz}}{z^2+1} \right| \leq \pi a \cdot \sup_{\text{arc}} \frac{1}{|z^2+1|} \leq \frac{\pi a}{a^2 - 1},</math> and <math display="block">\lim_{a \to \infty} \frac{\pi a}{a^2-1} = 0.</math>
The estimate on the numerator follows since Template:Math, 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, <math display="block">\left|e^{itz}\right| = \left|e^{it|z|(\cos\varphi + i\sin\varphi)}\right|=\left|e^{-t|z|\sin\varphi + it|z|\cos\varphi}\right|=e^{-t|z| \sin\varphi} \le 1.</math>
Therefore, <math display="block">\int_{-\infty}^\infty \frac{e^{itz}}{z^2+1}\,dz=\pi e^{-t}.</math>
If Template:Math then a similar argument with an arc Template:Math that winds around Template:Math rather than Template:Math shows that
<math display="block">\int_{-\infty}^\infty\frac{e^{itz}}{z^2+1}\,dz=\pi e^t,</math>
and finally we have <math display="block">\int_{-\infty}^\infty\frac{e^{itz}}{z^2+1}\,dz=\pi e^{-\left|t\right|}.</math>
(If Template:Math then the integral yields immediately to elementary calculus methods and its value is Template:Pi.)
Evaluating zeta functionsEdit
The fact that Template:Math has simple poles with residue 1 at each integer can be used to compute the sum <math display="block"> \sum_{n=-\infty}^\infty f(n).</math>
Consider, for example, Template:Math. Let Template:Math be the rectangle that is the boundary of Template:Math with positive orientation, with an integer Template:Mvar. By the residue formula,
<math display="block">\frac{1}{2 \pi i} \int_{\Gamma_N} f(z) \pi \cot(\pi z) \, dz = \operatorname{Res}\limits_{z = 0} + \sum_{n = -N \atop n\ne 0}^N n^{-2}.</math>
The left-hand side goes to zero as Template:Math since <math>|\cot(\pi z)|</math> is uniformly bounded on the contour, thanks to using <math>x = \pm \left(\frac 12 + N\right)</math> on the left and right side of the contour, and so the integrand has order <math>O(N^{-2})</math> over the entire contour. On the other hand,<ref>Template:Harvnb. Note that the Bernoulli number <math>B_{2n}</math> is denoted by <math>B_{n}</math> in Whittaker & Watson's book.</ref>
<math display="block">\frac{z}{2} \cot\left(\frac{z}{2}\right) = 1 - B_2 \frac{z^2}{2!} + \cdots </math> where the Bernoulli number <math>B_2 = \frac{1}{6}.</math>
(In fact, Template:Math.) Thus, the residue Template:Math is Template:Math. We conclude:
<math display="block">\sum_{n = 1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}</math> which is a proof of the Basel problem.
The same argument works for all <math>f(x) = x^{-2n}</math> where <math>n</math> is a positive integer, giving us<math display="block"> \zeta(2n) = \frac{(-1)^{n+1}B_{2n}(2\pi)^{2n}}{2(2n)!}.</math>The trick does not work when <math>f(x) = x^{-2n-1}</math>, since in this case, the residue at zero vanishes, and we obtain the useless identity <math>0 + \zeta(2n+1) - \zeta(2n+1) = 0</math>.
Evaluating Eisenstein seriesEdit
The same trick can be used to establish the sum of the Eisenstein series:<math display="block">\pi \cot(\pi z) = \lim_{N \to \infty} \sum_{n=-N}^N (z - n)^{-1}.</math>
See alsoEdit
- 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