Residue theorem

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In complex analysis, a discipline within mathematics, 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. From a geometrical perspective, it is a special case of the generalized Stokes' theorem.

Statement

The statement is as follows:

Let U be a simply connected open subset of the complex plane containing a finite list of points a1, ..., an, and f a function defined and holomorphic on U \{a1, ..., an}. Let γ be a closed rectifiable curve in U which does not meet any of the ak, and denote the winding number of γ around ak by I(γ, ak). The line integral of f around γ is equal to 2πi times the sum of residues of f at the points, each counted as many times as γ winds around the point:

$\oint _{\gamma }f(z)\,dz=2\pi i\sum _{k=1}^{n}\operatorname {I} (\gamma ,a_{k})\operatorname {Res} (f,a_{k}).$ 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 ak inside γ.

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

$\int _{V\smallsetminus W}d(f\,dz)-\int _{W\smallsetminus V}d(f\,dz)$ 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.

Examples

An integral along the real axis

The integral

$\int _{-\infty }^{\infty }{\frac {e^{itx}}{x^{2}+1}}\,dx$ 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 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. Now consider the contour integral

$\int _{C}{f(z)}\,dz=\int _{C}{\frac {e^{itz}}{z^{2}+1}}\,dz.$ 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{aligned}{\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{aligned}} the residue of f(z) at z = i is

$\operatorname {Res} \limits _{z=i}f(z)={\frac {e^{-t}}{2i}}.$ According to the residue theorem, then, we have

$\int _{C}f(z)\,dz=2\pi i\cdot \operatorname {Res} \limits _{z=i}f(z)=2\pi i{\frac {e^{-t}}{2i}}=\pi e^{-t}.$ 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

$\left|\int _{\mathrm {arc} }{\frac {e^{itz}}{z^{2}+1}}\,dz\right|\leq \pi a\cdot \sup _{\text{arc}}\left|{\frac {e^{itz}}{z^{2}+1}}\right|\leq \pi a\cdot \sup _{\text{arc}}{\frac {1}{|z^{2}+1|}}\leq {\frac {\pi a}{a^{2}-1}},$ and

$\lim _{a\to \infty }{\frac {\pi a}{a^{2}-1}}=0.$ The estimate on the numerator follows since t > 0, and for complex numbers z along the arc (which lies in the upper halfplane), the argument φ of z lies between 0 and π. So,

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

$\int _{-\infty }^{\infty }{\frac {e^{itz}}{z^{2}+1}}\,dz=\pi e^{-t}.$ If t < 0 then a similar argument with an arc C that winds around i rather than i shows that

$\int _{-\infty }^{\infty }{\frac {e^{itz}}{z^{2}+1}}\,dz=\pi e^{t},$ and finally we have

$\int _{-\infty }^{\infty }{\frac {e^{itz}}{z^{2}+1}}\,dz=\pi e^{-\left|t\right|}.$ (If t = 0 then the integral yields immediately to elementary calculus methods and its value is π.)

An infinite sum

The fact that π cot(πz) has simple poles with residue 1 at each integer can be used to compute the sum

$\displaystyle \sum _{n=-\infty }^{\infty }f(n).$ Consider, for example, f(z) = z−2. Let ΓN be the rectangle that is the boundary of [−N1/2, N + 1/2]2 with positive orientation, with an integer N. By the residue formula,

${\frac {1}{2\pi i}}\int _{\Gamma _{N}}f(z)\pi \cot(\pi z)\,dz=\operatorname {Res} \limits _{z=0}+\sum _{n=-N \atop n\neq 0}^{N}n^{-2}.$ The left-hand side goes to zero as N → ∞ since the integrand has order O(N−2). On the other hand,

${\frac {z}{2}}\cot \left({\frac {z}{2}}\right)=1-B_{2}{\frac {z^{2}}{2!}}+\cdots ;\qquad B_{2}={\frac {1}{6}}.$ (In fact, z/2 cot(z/2) = iz/1 − eiziz/2.) Thus, the residue is π2/3. We conclude:

$\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}={\frac {\pi ^{2}}{6}}$ which is a proof of the Basel problem.

The same trick can be used to establish

$\pi \cot(\pi z)=\lim _{N\to \infty }\sum _{n=-N}^{N}(z-n)^{-1},$ that is, the Eisenstein series.

We take f(z) = (wz)−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:

$\int _{\Gamma _{N}}{\frac {\pi \cot(\pi z)}{z}}\,dz=0$ since the integrand is an even function and so the contributions from the contour in the left-half plane and the contour in the right cancel each other out. Thus,

$\int _{\Gamma _{N}}f(z)\pi \cot(\pi z)\,dz=\int _{\Gamma _{N}}\left({\frac {1}{w-z}}+{\frac {1}{z}}\right)\pi \cot(\pi z)\,dz$ goes to zero as N → ∞.