Residue theorem

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. 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 $a 1, ..., a n$ , $U0 = U \ {a1, …, an}$ , and a function $f$ defined and holomorphic on $U 0$ . Let $γ$ be a closed rectifiable curve in $U 0$ , and denote the winding number of $γ$ around $a k$ by $I(γ, a k)$ . The line integral of $f$ around $γ$ is equal to $2 πi$ 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, $I(γ, a k) = 1$ if $a k$ is in the interior of $γ$ , and 0 if not, therefore $\oint _{\gamma }f(z)\,dz=2\pi i\sum \operatorname {Res} (f,a_{k})$ with the sum over those $a k$ 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 ${γ i}$ whose total is equivalent to $γ$ for integration purposes; this reduces the problem to finding the integral of $f dz$ along a Jordan curve $γ i$ 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 $U 0$ . Thus if two planar regions $V$ and $W$ of $U$ enclose the same subset ${a j}$ of ${a k}$ , the regions $V \ W$ and $W \ V$ lie entirely in $U 0$ , and hence $\int _{V\setminus W}d(f\,dz)-\int _{W\setminus V}d(f\,dz)$ is well-defined and equal to zero. Consequently, the contour integral of $f dz$ along $γ j = \partial V$ is equal to the sum of a set of integrals along paths $λ j$ , each enclosing an arbitrarily small region around a single $a j$ — the residues of $f$ (up to the conventional factor $2 πi$ ) at ${a j}$ . Summing over ${γ j}$ , we recover the final expression of the contour integral in terms of the winding numbers ${I(γ, a k)}$ .

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 $e itz$ is an entire function (having no singularities at any point in the complex plane), this function has singularities only where the denominator $z 2 + 1$ is zero. Since $z 2 + 1 = (z + i)(z - i)$ , 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 $\operatorname {Res} _{z=i}f(z)={\frac {e^{-t}}{2i}}.$

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 half-plane), the argument $φ$ of $z$ lies between 0 and $π$ . So, $\left|e^{itz}\right|=\left|e^{it|z|(\cos \varphi +i\sin \varphi )}\right|=\left|e^{-t|z|\sin \varphi +it|z|\cos \varphi }\right|=e^{-t|z|\sin \varphi }\leq 1.$

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

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

Consider, for example, $f (z) = z -2$ . Let $Γ N$ be the rectangle that is the boundary of $[−N − .mw-parser-output .sfrac{white-space:nowrap}.mw-parser-output .sfrac.tion,.mw-parser-output .sfrac .tion{display:inline-block;vertical-align:-0.5em;font-size:85%;text-align:center}.mw-parser-output .sfrac .num{display:block;line-height:1em;margin:0.0em 0.1em;border-bottom:1px solid}.mw-parser-output .sfrac .den{display:block;line-height:1em;margin:0.1em 0.1em}.mw-parser-output .sr-only{border:0;clip:rect(0,0,0,0);clip-path:polygon(0px 0px,0px 0px,0px 0px);height:1px;margin:-1px;overflow:hidden;padding:0;position:absolute;width:1px}⁠1/2⁠, N + ⁠1/2⁠]2$ 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 $N \to \infty$ since the integrand has order $O(n^{-2})$ . On the other hand,^[2]

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

(In fact, $⁠ z / 2 ⁠ cot(⁠ z / 2 ⁠) = ⁠ iz / 1 - e - iz ⁠ - ⁠ iz / 2 ⁠$ .) Thus, the residue $Res z =0$ is $- ⁠ π 2 / 3 ⁠$ . 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 the sum of the Eisenstein series: $\pi \cot(\pi z)=\lim _{N\to \infty }\sum _{n=-N}^{N}(z-n)^{-1}.$

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: $\int _{\Gamma _{N}}{\frac {\pi \cot(\pi z)}{z}}\,dz=0$ 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 $N \to \infty$ .

Notes

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

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.

External links

"Cauchy integral theorem", Encyclopedia of Mathematics, EMS Press, 2001 [1994]
Residue theorem in MathWorld

[1] Whittaker & Watson 1920, p. 112, §6.1.

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

[1]

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