Monthly Archives: April 2015

Applications of Residues: Evaluaton of Improper Integrals 3 (Indented Paths)

Suppose that a function $f(z)$ has a simple pole at a point $z=x_0$ on the real axis, with a Laurent series representation in a punctured disk $0<|z-x_0|<R_2$ and with residue $B_0$. Let $C_\rho$ denote the upper half of a circle $|z-x_0|=\rho$, where $\rho<R_2$ and with the clockwise direction.

indentedpath$f(z)$ can be written as
$$f(z)=g(z)+\frac{B_0}{z-x_0}\ (0<|z-x_0|<R_2),$$
where $g(z)=\sum_{n=0}^\infty a_n(z-x_0)^n$. So, we have
$$\int_{C_\rho}f(z)dz=\int_{C_\rho}g(z)dz+B_0\int_{C_\rho}\frac{dz}{z-x_0}.$$
Let $\rho<\rho_0<R_2$. Then $g(z)$ is bounded on $|z-x_0|\leq\rho_0$ i.e. there exists $M>0$ such that $|g(z)|\leq M$ whenever $|z-x_0|\leq\rho_0$. Thus, we get the estimate
$$\left|\int_{C_\rho}g(z)dz\right|\leq M\pi\rho.$$
Consequently, we have
$$\lim_{\rho\to 0}\int_{C_\rho} g(z)dz=0.$$
The semi-circle $-C_\rho$ has parametric representation
$$z=x_0+\rho e^{i\theta}\ (0\leq\theta\leq\pi).$$
Using this parametric representation, we calculate
\begin{align*}
\int_{C_\rho}\frac{dz}{z-x_0}&=-\int_{-C_{\rho}}\frac{dz}{z-x_0}\\
&=-\int_0^{\pi}\frac{1}{\rho e^{i\theta}}\rho i e^{i\theta}d\theta\\
&=-\pi i.
\end{align*}
Therefore, we obtain
\begin{equation}
\label{eq:indentpath}
\lim_{\rho\to 0}\int_{C_\rho}f(z)dz=-B_0\pi i.
\end{equation}

Example. [Singularity on Contour of Integration] Evaluate the improper Integral
$$I=\int_0^\infty\frac{\sin x}{x}dx.$$

Solution. The function $f(z)=\frac{e^{iz}}{z}$ has a simple pole at $z=0$.

indentedpath2Since $f(z)$ is analytic within and on the simple closed contour, we have
$$\int_{C_R} \frac{e^{iz}}{z}dz+\int_{-R}^{-\rho}\frac{e^{ix}}{x}dx+\int_{C_\rho}\frac{e^{iz}}{z}dz+\int_{\rho}^R\frac{e^{ix}}{x}dx=0.$$
Note that $\frac{1}{z}$ is analytic at all points $z$ in the upper half plane that are exterior to the circle $C_\rho$ and that for any $z$ on $C_R$, $\left|\frac{1}{z}\right|=\frac{1}{R}$
and $\lim_{R\to\infty}\frac{1}{R}=0$. Thus, by Jordan’s Lemma
$$\lim_{R\to\infty}\int_{C_R}f(z)dz=\lim_{R\to\infty}\int_{C_R}\frac{e^{iz}}{z}dz=0.$$
As $R\to\infty$ and $\rho\to 0$, we obtain
$$\int_{-\infty}^\infty\frac{e^{ix}}{x}dx=-\lim_{\rho\to 0}\int_{C_\rho}\frac{e^{iz}}{z}dz=\pi i.$$
Therefore,
$$\int_0^\infty\frac{\sin x}{x}dx=\frac{\pi}{2}.$$

Fresnel Integrals

In this lecture, we derive Fresnel integrals
$$\int_0^\infty\cos(x^2)dx=\int_0^\infty\sin(x^2)dx=\frac{1}{2}\sqrt{\frac{\pi}{2}},$$
which appear in optics and diffraction theory.

Let us consider a contour shown in the following figure.

fresnel

As seen in the figure, $C_R$ is a part of the circle $z=Re^{i\theta}$, where $0\leq\theta\leq\frac{\pi}{4}$. Let $f(z)=e^{iz^2}$. Then $f(z)$ is analytic on and within the positively oriented simple closed contour $C$ shown in the figure. So, we have $\int_Cf(z)dz=0$ which amounts to the following expression:
$$\int_0^Re^{ix^2}dx+\int_{C_R}e^{iz^2}dz-\int_0^Re^{-r^2}\left(\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}\right)dr=0.$$
Separating this expression into the real and the imaginary parts, we obtain
\begin{align}
\label{eq:cos}
\int_0^R\cos(x^2)dx&=\frac{1}{\sqrt{2}}\int_0^Re^{-r^2}dr-\mathrm{Re}\int_{C_R}e^{iz^2}dz,\\
\label{eq:sin}
\int_0^R\sin(x^2)dx&=\frac{1}{\sqrt{2}}\int_0^Re^{-r^2}dr-\mathrm{Im}\int_{C_R}e^{iz^2}dz.
\end{align}
\begin{align*}
\left|\int_{C_R}e^{iz^2}dz\right|&=\left|\int_0^{\frac{\pi}{4}}e^{iR^2e^{2i\theta}}Rie^{i\theta}d\theta\right|\\
&\leq R\int_0^{\frac{\pi}{4}}|e^{iR^2e^{2i\theta}}|d\theta\\
&=R\int_0^{\frac{\pi}{4}}e^{-R^2\sin 2\theta}d\theta\\
&=\frac{R}{2}\int_0^{\frac{\pi}{2}}e^{-R^2\sin\phi}d\phi\ (\mbox{by subsitution}\ \phi=2\theta)\\
&=\frac{R}{4}\int_0^\pi e^{-R^2\sin\phi}d\phi\\
&<\frac{R}{4}\frac{\pi}{R^2}\\
&=\frac{\pi}{4R}\to 0
\end{align*}
as $R\to\infty$. The inequality in the second line to the last was obtained by Jordan’s Inequality. Hence, as $R\to\infty$ \eqref{eq:cos} and \eqref{eq:sin} become
\begin{align*}
\int_0^\infty\cos(x^2)dx&=\int_0^\infty\sin(x^2)dx\\
&=\frac{1}{\sqrt{2}}\int_0^\infty e^{-r^2}dr\\
&=\frac{1}{2}\sqrt{\frac{\pi}{2}}.
\end{align*}

Jordan’s Lemma

Suppose that

  1. a function $f(z)$ is analytic at all points $z$ in the upper half plane $y\geq 0$ that are exterior to a circle $|z|=R_0$.
  2. For any $z$ on $C_R: |z|=R>R_0$, there exists a positive real number $M_R>0$ such that $|f(z)|\leq M_R$ and $\lim_{R\to\infty}M_R=0$.

Then for any positive real number $a$,
$$\lim_{R\to\infty}\int_{C_R}f(z)e^{iaz}dz=0.$$

Proof. We first show Jordan’s Inequality
\begin{equation}\label{eq:jordan}\int_0^\pi e^{-R\sin\theta}d\theta<\frac{\pi}{R}\ (R>0).\end{equation}

The graphs of y=sin(theta) (in red) and y=(2/pi)theta (in blue)

The graphs of y=sin(theta) (in red) and y=(2/pi)theta (in blue)

As shown in the figure, $\frac{2}{\pi}\theta\leq\sin\theta$ for $0\leq\theta\leq\frac{\pi}{2}$. If $R>0$, then $$e^{-R\sin\theta}\leq e^{-2R\theta/\pi},\ 0\leq\theta\leq\frac{\pi}{2}.$$So, we have
\begin{align*}
\int_0^{\frac{\pi}{2}}e^{-R\sin\theta}d\theta&\leq\int_0^{\frac{\pi}{2}}e^{-2R\theta/\pi}d\theta\\
&=\frac{\pi}{2R}(1-e^{-R})\\
&<\frac{\pi}{2R}.
\end{align*}Since the graph of $y=\sin\theta$ is symmetric about $\theta=\frac{\pi}{2}$ on the interval $0\leq\theta\leq\pi$,
$$\int_0^\pi e^{-R\sin\theta}d\theta=2\int_0^{\frac{\pi}{2}}e^{-R\sin\theta}d\theta<\frac{\pi}{R}.$$

Let $C_R$ denote the positively oriented circle $z=Re^{i\theta}$ where $0\leq\theta\leq\pi$. Then
$$\int_{C_R}f(z)e^{iaz}dz=\int_0^\pi f(Re^{i\theta})\exp(iaRe^{i\theta})iRe^{i\theta}d\theta$$
and so
\begin{align*}
\left|\int_{C_R}f(z)e^{iaz}dz\right|&\leq M_RR\int_0^\pi e^{-aR\sin\theta}d\theta\\
&<\frac{M_R\pi}{a}\ (\mbox{by Jordan Inequality \eqref{eq:jordan}})\\
&\to 0
\end{align*}
as $R\to\infty$ since by assumption $\lim_{R\to\infty}M_R=0$.