# Residue theorem

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.

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 closed rectifiable curve in U which does not meet any of the ak,

$\oint _{\gamma }f(z)\,dz=2\pi i\sum _{k=1}^{n}\operatorname {I} (\gamma ,a_{k})\operatorname {Res} (f,a_{k}).$ 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 $\int _{V\backslash W}d(f\,dz)-\int _{W\backslash V}d(f\,dz)$ 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

The integral

$\int _{-\infty }^{\infty }{e^{itx} \over x^{2}+1}\,dx$ 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. 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{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

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

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

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

$\int _{-\infty }^{\infty }{e^{itz} \over z^{2}+1}\,dz=\pi e^{t},$ 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

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

${1 \over 2\pi i}\int _{\Gamma _{N}}f(z)\pi \operatorname {cot} (\pi z)\,dz=\operatorname {Res} _{z=0}+\sum _{n=-N,n\neq 0}^{N}n^{-2}$ .
${z/2}\operatorname {cot} (z/2)=1-B_{2}{z^{2} \over 2!}+\cdots ,\,B_{2}={1 \over 6}$ .
$\sum _{n=1}^{\infty }{1 \over n^{2}}={\pi ^{2} \over 6}$ (cf. Basel problem.)

The same trick can be used to establish

$\pi \operatorname {cot} (\pi z)=\lim _{N\to \infty }\sum _{-N}^{N}(z-n)^{-1}$ (cf. Eisenstein series.) We take $f(z)=(w-z)^{-1}$ 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 contour in the left-half plane and the contour in the right cancel each other out. Thus,

$\int _{\Gamma _{N}}f(z)\pi \operatorname {cot} (\pi z)\,dz=\int _{\Gamma _{N}}\left({1 \over w-z}+{1 \over z}\right)\pi \operatorname {cot} (\pi z)\,dz$ See the corresponding article in French Wikipedia for further examples.