# 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 can be seen as 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, U0 = U \ {a1, ..., an}, and a function f defined and holomorphic on U0. Let γ be a closed rectifiable curve in U0, 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:

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

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, therefore

${\displaystyle \oint _{\gamma }f(z)\,dz=2\pi i\sum \operatorname {Res} (f,a_{k})}$

with the sum over those ak inside γ. [1]

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 fdz 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(fdz) = 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

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

is well-defined and equal to zero. Consequently, the contour integral of fdz 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

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

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

${\displaystyle \int _{C}{f(z)}\,dz=\int _{C}{\frac {e^{itz}}{z^{2}+1}}\,dz.}$

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

{\displaystyle {\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

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

According to the residue theorem, then, we have

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

The contour C may be split into a straight part and a curved arc, so that

${\displaystyle \int _{\mathrm {straight} }f(z)\,dz+\int _{\mathrm {arc} }f(z)\,dz=\pi e^{-t}\,}$

and thus

${\displaystyle \int _{-a}^{a}f(z)\,dz=\pi e^{-t}-\int _{\mathrm {arc} }f(z)\,dz.}$

Using some estimations, we have

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

and

${\displaystyle \lim _{a\to \infty }{\frac {\pi a}{a^{2}-1}}=0.}$

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,

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

Therefore,

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

If t < 0 then a similar argument with an arc C that winds around i rather than i shows that

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

and finally we have

${\displaystyle \int _{-\infty }^{\infty }{\frac {e^{itz}}{z^{2}+1}}\,dz=\pi e^{-\left|t\right|}.}$

(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

${\displaystyle \displaystyle \sum _{n=-\infty }^{\infty }f(n).}$

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,

${\displaystyle {\frac {1}{2\pi i}}\int _{\Gamma _{N}}f(z)\pi \cot(\pi z)\,dz=\operatorname {Res} \limits _{z=0}+\sum _{n=-N \atop n\neq 0}^{N}n^{-2}.}$

The left-hand side goes to zero as N → ∞ since the integrand has order ${\displaystyle O(n^{-2})}$. On the other hand, [2]

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

(In fact, z/2 cot(z/2) = iz/1 − eiziz/2.) Thus, the residue Resz=0 is π2/3. We conclude:

${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}}}={\frac {\pi ^{2}}{6}}}$

which is a proof of the Basel problem.

The same trick can be used to establish the sum of the Eisenstein series:

${\displaystyle \pi \cot(\pi z)=\lim _{N\to \infty }\sum _{n=-N}^{N}(z-n)^{-1}.}$

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:

${\displaystyle \int _{\Gamma _{N}}{\frac {\pi \cot(\pi z)}{z}}\,dz=0}$

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,

${\displaystyle \int _{\Gamma _{N}}f(z)\pi \cot(\pi z)\,dz=\int _{\Gamma _{N}}\left({\frac {1}{w-z}}+{\frac {1}{z}}\right)\pi \cot(\pi z)\,dz}$

goes to zero as N → ∞.

## Notes

1. Whittaker & Watson 1920 , p. 112, §6.1.
2. Whittaker & Watson 1920 , p. 125, §7.2. Note that the Bernoulli number ${\displaystyle B_{2n}}$ is denoted by ${\displaystyle B_{n}}$ in Whittaker & Watson's book.

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## References

• Ahlfors, Lars (1979). Complex Analysis. McGraw Hill. ISBN   0-07-085008-9.
• Lindelöf, Ernst L. (1905). Le calcul des résidus et ses applications à la théorie des fonctions (in French). Editions Jacques Gabay (published 1989). ISBN   2-87647-060-8.
• Mitrinović, Dragoslav; Kečkić, Jovan (1984). The Cauchy method of residues: Theory and applications. D. Reidel Publishing Company. ISBN   90-277-1623-4.
• Whittaker, E. T.; Watson, G. N. (1920). A Course of Modern Analysis (3rd ed.). Cambridge University Press.