Editing Contour integration (section)

===Example 2 – Cauchy distribution===
The integral
<math display=block>\int_{-\infty}^\infty \frac{e^{itx}}{x^2+1}\,dx</math>

[[Image:contourDiagram.png|right|250px|the contour]]

(which arises in [[probability theory]] as a scalar multiple of the [[characteristic function (probability theory)|characteristic function]] of the [[Cauchy distribution]]) resists the techniques of elementary [[calculus]]. We will evaluate it by expressing it as a limit of contour integrals along 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. The contour integral is
<math display=block>\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. The [[residue (complex analysis)|residue]] of {{math|''f''(''z'')}} at {{math|1=''z'' = ''i''}} is
<math display=block>\lim_{z\to i}(z-i)f(z) = \lim_{z\to i}(z-i)\frac{e^{itz} }{ z^2+1} = \lim_{z\to i}(z-i)\frac{e^{itz} }{ (z-i)(z+i)} = \lim_{z\to i}\frac{e^{itz} }{ z+i} = \frac{e^{-t}}{2i}.</math>

According to the [[residue theorem]], then, we have
<math display=block>\int_C f(z)\,dz=2\pi i \operatorname{Res}_{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_\text{straight}+\int_\text{arc}=\pi e^{-t},</math>
and thus
<math display=block>\int_{-a}^a =\pi e^{-t}-\int_\text{arc}.</math>

According to [[Jordan's lemma]], '''if {{math|''t'' > 0}} then'''
<math display=block>\int_\text{arc}\frac{e^{itz} }{ z^2+1}\,dz \rightarrow 0 \mbox{ as } a\rightarrow\infty.</math>

Therefore, '''if {{math|''t'' > 0}} then'''
<math display=block>\int_{-\infty}^\infty \frac{e^{itx} }{ x^2+1}\,dx=\pi e^{-t}.</math>

A similar argument with an arc that winds around {{math|−''i''}} rather than {{mvar|i}} shows that '''if {{math|''t'' < 0}} then'''
<math display=block>\int_{-\infty}^\infty \frac{e^{itx} }{ x^2+1}\,dx=\pi e^t,</math>
and finally we have this:
<math display=block>\int_{-\infty}^\infty \frac{e^{itx} }{ x^2+1} \,dx=\pi e^{-|t|}.</math>

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