Residue theorem

{{See also|Residue (complex analysis)}}

The statement is as follows:

File:Residue theorem illustration.svg

Residue theorem: Let $U$ be a simply connected open subset of the complex plane containing a finite list of points $a\_1,\; \backslash ldots,\; a\_n,$ $U\_0\; =\; U\; \backslash smallsetminus\; \backslash \{a\_1,\; \backslash ldots,\; a\_n\backslash \},$ and a function $f$ holomorphic on $U\_0.$ Letting $\backslash gamma$ be a closed rectifiable curve in $U\_0,$ and denoting the residue of $f$ at each point $a\_k$ by $\backslash operatorname\{Res\}(f,\; a\_k)$ and the winding number of $\backslash gamma$ around $a\_k$ by $\backslash operatorname\{I\}(\backslash gamma,\; a\_k),$ the line integral of $f$ around $\backslash gamma$ is equal to $2\backslash pi\; i$ times the sum of residues, each counted as many times as $\backslash gamma$ winds around the respective point:

$$$$

\oint_\gamma f(z)\, dz = 2\pi i \sum_{k=1}^n \operatorname{I}(\gamma, a_k) \operatorname{Res}(f, a_k).

If $\backslash gamma$ is a positively oriented simple closed curve, $\backslash operatorname\{I\}(\backslash gamma,\; a\_k)$ is $1$ if $a\_k$ is in the interior of $\backslash gamma$ 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 {{nobr|$\backslash gamma.${{harvnb|Whittaker|Watson|1920|loc=§6.1|page=112}}.}}

The relationship of the residue theorem to Stokes' theorem is given by the Jordan curve theorem. The general plane curve {{mvar|γ}} must first be reduced to a set of simple closed curves $\backslash \{\backslash gamma\_i\backslash \}$ whose total is equivalent to $\backslash gamma$ for integration purposes; this reduces the problem to finding the integral of $f\backslash ,\; dz$ along a Jordan curve $\backslash gamma\_i$ with interior $V.$ The requirement that $f$ be holomorphic on $U\_0\; =\; U\; \backslash smallsetminus\; \backslash \{a\_k\backslash \}$ is equivalent to the statement that the exterior derivative $d(f\backslash ,\; dz)\; =\; 0$ on $U\_0.$ Thus if two planar regions $V$ and $W$ of $U$ enclose the same subset $\backslash \{a\_j\backslash \}$ of $\backslash \{a\_k\backslash \},$ the regions $V\; \backslash smallsetminus\; W$ and $W\; \backslash smallsetminus\; V$ lie entirely in $U\_0,$ 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 $f\backslash ,\; dz$ along $\backslash gamma\_j\; =\; \backslash partial\; V$ is equal to the sum of a set of integrals along paths $\backslash gamma\_j,$ each enclosing an arbitrarily small region around a single $a\_j$ — the residues of $f$ (up to the conventional factor $2\backslash pi\; i$ at $\backslash \{a\_j\backslash \}.$ Summing over $\backslash \{\backslash gamma\_j\backslash \},$ we recover the final expression of the contour integral in terms of the winding numbers $\backslash \{\backslash operatorname\{I\}(\backslash gamma,\; a\_k)\backslash \}.$

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.

{{Excerpt|Residue (complex analysis)|Calculation of residues|subsections=yes}}

The integral

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

Image:Contour example.svg

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 {{math|t > 0}} and define the contour {{mvar|C}} that goes along the real line from {{math|−a}} to {{mvar|a}} and then counterclockwise along a semicircle centered at 0 from {{mvar|a}} to {{math|−a}}. Take {{mvar|a}} to be greater than 1, so that the imaginary unit {{mvar|i}} is enclosed within the curve. Now consider the contour integral

$$\backslash int\_C\; \{f(z)\}\backslash ,dz\; =\; \backslash int\_C\; \backslash frac\{e^\{itz\}\}\{z^2+1\}\backslash ,dz.$$

Since {{math|e^itz}} is an entire function (having no singularities at any point in the complex plane), this function has singularities only where the denominator {{math|z² + 1}} is zero. Since {{math|1=z² + 1 = (z + i)(z − i)}}, that happens only where {{math|1=z = i}} or {{math|1=z = −i}}. Only one of those points is in the region bounded by this contour. Because {{math|f(z)}} is

$$\backslash begin\{align\}$$

\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{align}

the residue of {{math|f(z)}} at {{math|1=z = i}} is

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

According to the residue theorem, then, we have

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

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

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

and thus

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

Using some estimations, we have

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

and

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

The estimate on the numerator follows since {{math|t > 0}}, and for complex numbers {{mvar|z}} along the arc (which lies in the upper half-plane), the argument {{mvar|φ}} of {{mvar|z}} lies between 0 and {{pi}}. So,

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

Therefore,

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

If {{math|t < 0}} then a similar argument with an arc {{math|{{prime|C}}}} that winds around {{math|−i}} rather than {{math|i}} shows that

Image:Contour example 2.svg

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

and finally we have

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

(If {{math|1=t = 0}} then the integral yields immediately to elementary calculus methods and its value is {{pi}}.)

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

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

Consider, for example, {{math|1=f(z) = z⁻²}}. Let {{math|Γ_N}} be the rectangle that is the boundary of {{math|[−N − {{sfrac|1|2}}, N + {{sfrac|1|2}}]²}} with positive orientation, with an integer {{mvar|N}}. By the residue formula,

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

The left-hand side goes to zero as {{math|N → ∞}} since $|\backslash cot(\backslash pi\; z)|$ is uniformly bounded on the contour, thanks to using $x\; =\; \backslash pm\; \backslash left(\backslash frac\; 12\; +\; N\backslash right)$ on the left and right side of the contour, and so the integrand has order $O(N^\{-2\})$ over the entire contour. On the other hand,{{harvnb|Whittaker|Watson|1920|loc=§7.2|page=125}}. Note that the Bernoulli number $B\_\{2n\}$ is denoted by $B\_\{n\}$ in Whittaker & Watson's book.

$$\backslash frac\{z\}\{2\}\; \backslash cot\backslash left(\backslash frac\{z\}\{2\}\backslash right)\; =\; 1\; -\; B\_2\; \backslash frac\{z^2\}\{2!\}\; +\; \backslash cdots$$ where the Bernoulli number $B\_2\; =\; \backslash frac\{1\}\{6\}.$

(In fact, {{math|1={{sfrac|z|2}} cot({{sfrac|z|2}}) = {{sfrac|iz|1 − e^−iz}} − {{sfrac|iz|2}}}}.) Thus, the residue {{math|Res{{sub|1=z=0}}}} is {{math|−{{sfrac|π²|3}}}}. We conclude:

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

which is a proof of the Basel problem.

The same argument works for all $f(x)\; =\; x^\{-2n\}$ where $n$ is a positive integer, giving us$$\backslash zeta(2n)\; =\; \backslash frac\{(-1)^\{n+1\}B\_\{2n\}(2\backslash pi)^\{2n\}\}\{2(2n)!\}.$$The trick does not work when $f(x)\; =\; x^\{-2n-1\}$, since in this case, the residue at zero vanishes, and we obtain the useless identity $0\; +\; \backslash zeta(2n+1)\; -\; \backslash zeta(2n+1)\; =\; 0$.

The same trick can be used to establish the sum of the Eisenstein series:$$\backslash pi\; \backslash cot(\backslash pi\; z)\; =\; \backslash lim\_\{N\; \backslash to\; \backslash infty\}\; \backslash sum\_\{n=-N\}^N\; (z\; -\; n)^\{-1\}.$$

{{Math proof|title=Proof|proof=

Pick an arbitrary $w\; \backslash in\; \backslash mathbb\; C\backslash setminus\; \backslash Z$. As above, define

$$g(z)\; :=\; \backslash frac\{1\}\{w-z\}\; \backslash pi\; \backslash cot(\backslash pi\; z)$$

By the Cauchy residue theorem, for all $N$ large enough such that $\backslash Gamma\_N$ encircles $w$,

$$\backslash frac\{1\}\{2\; \backslash pi\; i\}\; \backslash oint\_\{\backslash Gamma\_N\}\; g(z)\; dz\; =\; -\backslash pi\; \backslash cot(\backslash pi\; z)\; +\; \backslash sum\_\{n=-N\}^N\; \backslash frac\{1\}\{z-n\}$$

It remains to prove the integral converges to zero. Since $\backslash pi\backslash cot(\backslash pi\; z)\; /z$ is an even function, and $\backslash Gamma\_N$ is symmetric about the origin, we have $\backslash oint\_\{\backslash Gamma\_N\}\; \backslash pi\backslash cot(\backslash pi\; z)\; /z\; dz\; =\; 0$, and so

$$\backslash oint\_\{\backslash Gamma\_N\}\; g(z)\; dz\; =\; \backslash oint\_\{\backslash Gamma\_N\}\; \backslash left(\backslash frac\; 1z\; +\; \backslash frac\{1\}\{w-z\}\backslash right)\; \backslash pi\backslash cot(\backslash pi\; z)dz\; =\; -w\; \backslash oint\_\{\backslash Gamma\_N\}\; \backslash frac\{1\}\{z(z-w)\}\; \backslash pi\backslash cot(\backslash pi\; z)\; dz\; =\; O(1/N)$$

}}

• Residue (complex analysis)
• Cauchy's integral formula
• Glasser's master theorem
• Jordan's lemma
• Methods of contour integration
• Morera's theorem
• Nachbin's theorem
• Residue at infinity
• Logarithmic form

{{reflist}}

• {{cite book

| last = Ahlfors

| first = Lars

| author-link = Lars Ahlfors

| title = Complex Analysis

| publisher = McGraw Hill

| year = 1979

| isbn = 0-07-085008-9

}}

• {{cite book

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| first = Ernst L.

| author-link = Ernst Leonard Lindelöf

| title = Le calcul des résidus et ses applications à la théorie des fonctions

| publisher = Editions Jacques Gabay

| year = 1905

| language = fr

| publication-date = 1989

| isbn = 2-87647-060-8

}}

• {{cite book

| last1 = Mitrinović

| first1 = Dragoslav

| last2 = Kečkić

| first2 = Jovan

| title = The Cauchy method of residues: Theory and applications

| publisher = D. Reidel Publishing Company

| year = 1984

| isbn = 90-277-1623-4

}}

• {{cite book

| last1 = Whittaker

| first1 = E. T.

| author1-link = E. T. Whittaker

| last2 = Watson

| first2 = G. N.

| author2-link = G. N. Watson

| title = A Course of Modern Analysis

| publisher = Cambridge University Press

| edition = 3rd

| year = 1920

}}

• {{springer|title=Cauchy integral theorem|id=p/c020900}}
• [http://mathworld.wolfram.com/ResidueTheorem.html Residue theorem] in MathWorld

Category:Theorems in complex analysis

Category:Analytic functions