Residue Theorem - Example

Example

The integral

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

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

\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 f(z) at z = i is

According to the residue theorem, then, we have

The contour C may be split into a "straight" part and a curved arc, so that

and thus

Using some estimations, we have

\begin{align} & {} \quad \left|\int_{\mathrm{arc}}{e^{itz} \over z^2+1}\,dz\right| \leq \int_{\mathrm{arc}}\left|{e^{itz} \over z^2+1}\right| \, |dz| = \int_{\mathrm{arc}}{|e^{itz}| \over |z^2+1|}\,|dz| \\ \\ & \leq \int_{\mathrm{arc}}{1 \over |z^2+1|}\,|dz|\leq \int_{\mathrm{arc}}{1 \over a^2-1}\,|dz|=\frac{\pi a}{a^2-1} \rightarrow 0 \text{ as } a\rightarrow\infty. \end{align}

Note that, since t > 0 and for complex numbers in the upper halfplane the argument lies between 0 an, one can estimate

\left|e^{itz}\right|=\left|e^{it|z|(\cos\phi + j\sin\phi)}\right|=\left|e^{-t|z|\sin\phi + it|z|\cos\phi}\right|= e^{-t|z|\sin\phi} \le 1.

Therefore

If t < 0 then a similar argument with an arc C' that winds around −i rather than i shows that

and finally we have

(If t = 0 then the integral yields immediately to elementary calculus methods and its value is π.)

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