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Residue theorem
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==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|β''a''}} to {{mvar|a}} and then counterclockwise along a semicircle centered at 0 from {{mvar|a}} to {{math|β''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'' β ''i'')}}, that happens only where {{math|1=''z'' = ''i''}} or {{math|1=''z'' = β''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|Ο}} 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|β''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|''Ο'' cot(''Οz'')}} 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>β2</sup>}}. Let {{math|Ξ<sub>''N''</sub>}} be the rectangle that is the boundary of {{math|[β''N'' β {{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'' β β}} 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=Β§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 β ''e''<sup>β''iz''</sup>}} β {{sfrac|''iz''|2}}}}.) Thus, the residue {{math|Res{{sub|1=''z''=0}}}} is {{math|β{{sfrac|''Ο''<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> }}
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