# Residue theorem

The residue theorem in complex analysis is a powerful tool to evaluate path integrals of meromorphic functions over closed curves and can often be used to compute real integrals as well. It generalizes the Cauchy integral theorem and Cauchy's integral formula.

The statement is as follows. Suppose U is a simply connected open subset of the complex plane C, 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 doesn't meet any of the points ak and whose start point equals its endpoint, then

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

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. 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 if it is large enough, leaving only the real-axis part of the integral, the one we were originally interested in.

## Example

The integral

$\int _{-\infty }^{\infty }{e^{itx} \over x^{2}+1}\,dx$ (which arises in probability theory as (a scalar multiple of) the characteristic function of the Cauchy distribution) resists the techniques of elementary calculus. We will evaluate it by expressing it as a limit of contour integrals along 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

$\int _{C}{e^{itz} \over z^{2}+1}\,dz.$ 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

${\frac {e^{itz}}{z^{2}+1}}={\frac {e^{itz}}{2i}}\left({\frac {1}{z-i}}-{\frac {1}{z+i}}\right)$ $={\frac {e^{-t}}{2i}}{\frac {1}{z-i}}+{\frac {e^{-t}}{2i}}{\frac {e^{it(z-i)}-1}{z-i}}-{\frac {e^{itz}}{2i(z+i)}}$ ,

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

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

$\int _{C}f(z)\,dz=2\pi i\cdot \operatorname {Res} _{z=i}f(z)=2\pi i{e^{-t} \over 2i}=\pi e^{-t}.$ The contour C may be split into a "straight" part and a curved arc, so that

$\int _{\mbox{straight}}+\int _{\mbox{arc}}=\pi e^{-t},$ and thus

$\int _{-a}^{a}=\pi e^{-t}-\int _{\mbox{arc}}.$ It can be shown that if t > 0 then

$\int _{\mbox{arc}}{e^{itz} \over z^{2}+1}\,dz\rightarrow 0\ {\mbox{as}}\ a\rightarrow \infty .$ Therefore if t > 0 then

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

$\int _{-\infty }^{\infty }{e^{itz} \over z^{2}+1}\,dz=\pi e^{t},$ 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 first-year calculus methods and its value is π.)

## Humor

Q: What's the value of a contour integral around Western Europe?

A: Zero, because all the Poles are in Eastern Europe. de:Residuensatz fr:Théorème des résidus he:משפט השאריות