Editing Residue theorem

{{short description|Concept of complex analysis}}
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In [[complex analysis]], the '''residue theorem''', sometimes called '''Cauchy's residue theorem''', is a powerful tool to evaluate [[line integral]]s of [[analytic function]]s 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]].  The '''residue theorem''' should not be confused with special cases of the [[generalized Stokes' theorem]]; however, the latter can be used as an ingredient of its proof.

==Statement of Cauchy's residue theorem==
{{See also|Residue (complex analysis)}}
The statement is as follows:

<blockquote>
[[File:Residue theorem illustration.svg|thumb|Illustration of the setting]]
'''Residue theorem''': Let <math>U</math> be a [[simply connected]] [[open subset]] of the [[complex plane]] containing a finite list of points <math>a_1, \ldots, a_n,</math> <math>U_0 = U \smallsetminus \{a_1, \ldots, a_n\},</math> and a function <math>f</math> [[holomorphic function|holomorphic]] on <math>U_0.</math> Letting <math>\gamma</math> be a closed [[rectifiable curve]] in <math>U_0,</math> and denoting the [[residue (complex analysis)|residue]] of <math>f</math> at each point <math>a_k</math> by <math>\operatorname{Res}(f, a_k)</math> and the [[winding number]] of <math>\gamma</math> around <math>a_k</math> by <math>\operatorname{I}(\gamma, a_k),</math> the line integral of <math>f</math> around <math>\gamma</math> is equal to <math>2\pi i</math> times the sum of residues, each counted as many times as <math>\gamma</math> winds around the respective point:

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

If <math>\gamma</math> is a [[Curve orientation|positively oriented]] [[Jordan curve|simple closed curve]], <math>\operatorname{I}(\gamma, a_k)</math> is <math>1</math> if <math>a_k</math> is in the interior of <math>\gamma</math> and <math>0</math> if not, therefore

<math display=block>
\oint_\gamma f(z)\, dz = 2\pi i \sum \operatorname{Res}(f, a_k)
</math>

with the sum over those <math>a_k</math> inside {{nobr|<math>\gamma.</math><ref>{{harvnb|Whittaker|Watson|1920|loc=§6.1|page=112}}.</ref>}}
</blockquote>

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 <math>\{\gamma_i\}</math> whose total is equivalent to <math>\gamma</math> for integration purposes; this reduces the problem to finding the integral of <math>f\, dz</math> along a Jordan curve <math>\gamma_i</math> with interior <math>V.</math> The requirement that <math>f</math> be holomorphic on <math>U_0 = U \smallsetminus \{a_k\}</math> is equivalent to the statement that the [[exterior derivative]] <math>d(f\, dz) = 0</math> on <math>U_0.</math>  Thus if two planar regions <math>V</math> and <math>W</math> of <math>U</math> enclose the same subset <math>\{a_j\}</math> of <math>\{a_k\},</math> the regions <math>V \smallsetminus W</math> and <math>W \smallsetminus V</math> lie entirely in <math>U_0,</math> hence

<math display=block>
\int_{V \smallsetminus W} d(f \, dz) - \int_{W \smallsetminus V} d(f \, dz)
</math>

is well-defined and equal to zero. Consequently, the contour integral of <math>f\, dz</math> along <math>\gamma_j = \partial V</math> is equal to the sum of a set of integrals along paths <math>\gamma_j,</math> each enclosing an arbitrarily small region around a single <math>a_j</math> — the residues of <math>f</math> (up to the conventional factor <math>2\pi i</math> at <math>\{a_j\}.</math> Summing over <math>\{\gamma_j\},</math> we recover the final expression of the contour integral in terms of the winding numbers <math>\{\operatorname{I}(\gamma, a_k)\}.</math>

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.

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

==Examples==
===An integral along the real axis===
The integral
<math display="block">\int_{-\infty}^\infty \frac{e^{itx}}{x^2+1}\,dx</math>

[[Image:Contour example.svg|class=skin-invert-image|right|300px|thumb|The contour {{mvar|C}}.]]

arises in [[probability theory]] when calculating the [[characteristic function (probability theory)|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 integral]]s.

Suppose {{math|''t'' > 0}} and define the contour {{mvar|C}} that goes along the [[real number|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 number|imaginary]] unit {{mvar|i}} is enclosed within the curve.  Now consider the contour integral
<math display="block">\int_C {f(z)}\,dz = \int_C \frac{e^{itz}}{z^2+1}\,dz.</math>

Since {{math|''e''<sup>''itz''</sup>}} is an [[entire function]] (having no [[mathematical singularity|singularities]] at any point in the complex plane), this function has singularities only where the denominator {{math|''z''<sup>2</sup> + 1}} is zero. Since {{math|1=''z''<sup>2</sup> + 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
<math display="block">\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}</math>
the [[residue (complex analysis)|residue]] of {{math|''f''(''z'')}} at {{math|1=''z'' = ''i''}} is
<math display="block">\operatorname{Res}_{z=i}f(z)=\frac{e^{-t}}{2i}.</math>

According to the residue theorem, then, we have
<math display="block">\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}.</math>

The contour {{mvar|C}} may be split into a straight part and a curved arc, so that
<math display="block">\int_{\mathrm{straight}} f(z)\,dz+\int_{\mathrm{arc}} f(z)\,dz=\pi e^{-t}</math>
and thus
<math display="block">\int_{-a}^a f(z)\,dz =\pi e^{-t}-\int_{\mathrm{arc}} f(z)\,dz.</math>

Using some [[Estimation lemma|estimations]], we have
<math display="block">\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},</math>
and
<math display="block">\lim_{a \to \infty} \frac{\pi a}{a^2-1} = 0.</math>

The estimate on the numerator follows since {{math|''t'' > 0}}, and for [[complex number]]s {{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,
<math display="block">\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} \le 1.</math>

Therefore,
<math display="block">\int_{-\infty}^\infty \frac{e^{itz}}{z^2+1}\,dz=\pi e^{-t}.</math>

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|class=skin-invert-image|right|300px|thumb|The contour {{math|{{prime|''C''}}}}.]]

<math display="block">\int_{-\infty}^\infty\frac{e^{itz}}{z^2+1}\,dz=\pi e^t,</math>

and finally we have
<math display="block">\int_{-\infty}^\infty\frac{e^{itz}}{z^2+1}\,dz=\pi e^{-\left|t\right|}.</math>

(If {{math|1=''t'' = 0}} then the integral yields immediately to elementary calculus methods and its value is {{pi}}.)

===Evaluating zeta functions===
The fact that {{math|''π'' cot(''πz'')}} has simple poles with residue 1 at each integer can be used to compute the sum
<math display="block"> \sum_{n=-\infty}^\infty f(n).</math>

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

<math display="block">\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\ne 0}^N n^{-2}.</math>

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

<math display="block">\frac{z}{2} \cot\left(\frac{z}{2}\right) = 1 - B_2 \frac{z^2}{2!} + \cdots </math> where the [[Bernoulli number]] <math>B_2 = \frac{1}{6}.</math>

(In fact, {{math|1={{sfrac|''z''|2}} cot({{sfrac|''z''|2}}) = {{sfrac|''iz''|1 − ''e''<sup>−''iz''</sup>}} − {{sfrac|''iz''|2}}}}.) Thus, the residue {{math|Res{{sub|1=''z''=0}}}} is {{math|−{{sfrac|''π''<sup>2</sup>|3}}}}. We conclude:

<math display="block">\sum_{n = 1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}</math>
which is a proof of the [[Basel problem]].

The same argument works for all <math>f(x) = x^{-2n}</math> where <math>n</math> is a positive integer, [[Particular values of the Riemann zeta function|giving us]]<math display="block"> \zeta(2n) = \frac{(-1)^{n+1}B_{2n}(2\pi)^{2n}}{2(2n)!}.</math>The trick does not work when <math>f(x) = x^{-2n-1}</math>, since in this case, the residue at zero vanishes, and we obtain the useless identity <math>0 + \zeta(2n+1) - \zeta(2n+1) = 0</math>.

=== Evaluating Eisenstein series ===
The same trick can be used to establish the sum of the [[Eisenstein series]]:<math display="block">\pi \cot(\pi z) = \lim_{N \to \infty} \sum_{n=-N}^N (z - n)^{-1}.</math>

{{Math proof|title=Proof|proof=  

Pick an arbitrary <math>w \in \mathbb C\setminus \Z</math>. As above, define

<math display="block">g(z) := \frac{1}{w-z} \pi \cot(\pi z)</math>

By the Cauchy residue theorem, for all <math>N</math> large enough such that <math>\Gamma_N</math> encircles <math>w</math>,

<math display="block"> \frac{1}{2 \pi i} \oint_{\Gamma_N} g(z) dz = -\pi \cot(\pi z) + \sum_{n=-N}^N \frac{1}{z-n}</math>

It remains to prove the integral converges to zero. Since <math>\pi\cot(\pi z) /z</math> is an even function, and <math>\Gamma_N</math> is symmetric about the origin, we have <math>\oint_{\Gamma_N} \pi\cot(\pi z) /z dz = 0</math>, and so

<math display="block">\oint_{\Gamma_N} g(z) dz = \oint_{\Gamma_N} \left(\frac 1z + \frac{1}{w-z}\right) \pi\cot(\pi z)dz = -w \oint_{\Gamma_N} \frac{1}{z(z-w)} \pi\cot(\pi z) dz = O(1/N)</math>
}}

==See also==
* [[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]]

==Notes==
{{reflist}}

==References==
* {{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
 | last = Lindelöf
 | 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
}}

==External links==
* {{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]]