Editing Residue theorem (section)

===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}}.)