# 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.

$f(z)$ 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

\label{eq:indentpath}
\lim_{\rho\to 0}\int_{C_\rho}f(z)dz=-B_0\pi i.

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 $z=0$.

Since $f(z)$ 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.

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
$$\label{eq:jordan}\int_0^\pi e^{-R\sin\theta}d\theta<\frac{\pi}{R}\ (R>0).$$

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 $\lim_{R\to\infty}M_R=0$.

# Harmonic Functions

Throughout this course, a connected open subset of $\mathbb{C}$ is called a domain. Suppose that a function $f(z)=u(x,y)+iv(x,y)$ is analytic in a domain $\mathcal{D}$. Then $f(z)$ satisfies the Cauchy-Riemann equations i.e.

\label{eq:c-r}
u_x=v_y,\ u_y=-v_x.

Differentiating the Cauchy-Riemann equations \eqref{eq:c-r} with respect to $x$, we obtain

\label{eq:c-r2}
u_{xx}=v_{yx},\ u_{yx}=-v_{xx}.

Differentiating the Cauchy-Riemann equations \eqref{eq:c-r} with respect to $y$, we obtain

\label{eq:c-r3}
u_{xy}=v_{yy},\ u_{yy}=-v_{xy}.

By the continuity of the partial derivatives of $u(x,y)$ and $v(x,y)$, we have

\label{eq:cont}
u_{xy}=u_{yx},\ v_{xy}=v_{yx}.

Applying \eqref{eq:cont} to \eqref{eq:c-r2} and \eqref{eq:c-r3}, we obtain the Laplace equations:
$$u_{xx}+u_{yy}=0,\ v_{xx}+v_{yy}=0.$$
That is to say, $u(x,y)$ and $v(x,y)$ are harmonic maps in $\mathcal{D}$.

Example. The function $f(z)=e^{-y}\sin x-ie^{-y}\cos x$ is entire (i.e analytic on the complex plane $\mathbb{C}$), so both $e^{-y}\sin x$ and $-e^{-y}\cos x$ are harmonic. (You can of course check it for yourself!)

If two functions $u(x,y)$, $v(x,y)$ are harmonic in a domain $\mathcal{D}$ and their first-order partial derivatives satisfy the Cauchy-Riemann equations \eqref{eq:c-r} throughout $\mathcal{D}$, $v(x,y)$ is said to be a harmonic conjugate of $u(x,y)$.

Theorem. A function $f(z)=u(x,y)+iv(x,y)$ is analytic in a domain $\mathcal{D}$ if and only if $v(x,y)$ is a harmonic conjugate of $u(x,y)$.

Remark. If $v(x,y)$ is a harmonic conjugate of $u(x,y)$ in some domain, it is not in general true that $u$ is a harmonic conjugate of $v$ there.

Example. Let $u(x,y)=x^2-y^2$ and $v(x,y)=2xy$. Since $f(z)=z^2=(x^2-y^2)+i2xy$ is entire, $v(x,y)$ is a harmonic conjugate of $u(x,y)$. However, $u(x,y)$ cannot be a harmonic conjugate of $v(x,y)$ since $2xy+i(x^2-y^2)$ is not analytic anywhere.

Example. [Finding a harmonic conjugate of a harmonic function] Let $u(x,y)=y^3-3x^2y$ and $v(x,y)$ be a harmonic conjugate of $u(x,y)$. Then it follows from the Cauchy-Riemann equations \eqref{eq:c-r} that $v_y=-6xy$. Integrating this with respect to $y$, we obtain
$$v(x,y)=-3xy^2+\phi(x),$$
where $\phi(x)$ is some unknown function. We determine $\phi(x)$ using $u_y=-v_x$:
$$v_x=-3y^2+\phi'(x).$$
Comparing this with $-u_y=-3y^2+3x^2$, we get $\phi'(x)=3x^2$ and so, $\phi(x)=x^3+C$ where $C$ is a constant. Hence, we find a harmonic conjugate of $u(x,y)$:
$$v(x,y)=-3xy^2+x^3+C,$$
where $C$ is a constant. The corresponding analytic function is
$$f(z)=(y^3-3x^2y)+i(-3xy^2+x^3+C),$$
where $C$ is a constant.

# Analytic Continuation

The function $f(z)=\displaystyle\frac{1}{1+z}$ has an isolated singularity at $z=-1$. It has the Maclaurin series representation

$$f(z)=\sum_{n=0}^\infty(-1)^nz^n$$
for $|z|<1$. The power series $f_1(z)=\displaystyle\sum_{n=0}^\infty(-1)^nz^n$ converges only on the open unit disk $D_1:\ |z|<1$. For instance, the series diverges at $z=\frac{3}{2}i$ i.e. $f_1\left(\frac{3}{2}i\right)$ is not defined. The first 25 partial sums of the series $f_1\left(\frac{3}{2}i\right)$ are listed below and they do not appear to be approaching somewhere.

S[1] = 1.
S[2] = 1. – 1.500000000 I
S[3] = -1.250000000 – 1.500000000 I
S[4] = -1.250000000 + 1.875000000 I
S[5] = 3.812500000 + 1.875000000 I
S[6] = 3.812500000 – 5.718750000 I
S[7] = -7.578125000 – 5.718750000 I
S[8] = -7.578125000 + 11.36718750 I
S[9] = 18.05078125 + 11.36718750 I
S[10] = 18.05078125 – 27.07617188 I
S[11] = -39.61425781 – 27.07617188 I
S[12] = -39.61425781 + 59.42138672 I
S[13] = 90.13208008 + 59.42138672 I
S[14] = 90.13208008 – 135.1981201 I
S[15] = -201.7971802 – 135.1981201 I
S[16] = -201.7971802 + 302.6957703 I
S[17] = 455.0436554 + 302.6957703 I
S[18] = 455.0436554 – 682.5654831 I
S[19] = -1022.848225 – 682.5654831 I
S[20] = -1022.848225 + 1534.272337 I
S[21] = 2302.408505 + 1534.272337 I
S[22] = 2302.408505 – 3453.612758 I
S[23] = -5179.419137 – 3453.612758 I
S[24] = -5179.419137 + 7769.128706 I
S[25] = 11654.69306 + 7769.128706 I

Also shown below are the graphics of partial sums of the series $f_1\left(\frac{3}{2}i\right)$.

The first 10 partial sums

The first 20 partial sums

The first 30 partial sums

Let us expand $f(z)=\displaystyle\frac{1}{1+z}$ at $z=i$. Then we obtain
\begin{align*}
f(z)&=\frac{1}{1+z}\\
&=\frac{1}{1+i}\cdot\frac{1}{1+\frac{z-i}{1+i}}\\
&=\sum_{n=0}^\infty (-1)^n\frac{(z-i)^n}{(1+i)^{n+1}}
\end{align*}
for $|z-i|<\sqrt{2}$. Let $f_2(z)=\displaystyle\sum_{n=0}^\infty (-1)^n\frac{(z-i)^n}{(1+i)^{n+1}}$. This series converges only on the open disk $D_2:\ |z-i|<\sqrt{2}$, in particular at $z=\frac{3}{2}i$ and $f_2\left(\frac{3}{2}i\right)=f\left(\frac{3}{2}i\right)=\frac{4}{13}-\frac{6}{13}i$. The first 25 partial sums of the series $f_2\left(\frac{3}{2}i\right)$ are listed below and it appears that they are approaching a number. In fact, they are approaching the complex number $f\left(\frac{3}{2}i\right)=\frac{4}{13}-\frac{6}{13}i$.

S[1] = 0.5000000000 – 0.5000000000 I
S[2] = 0.2500000000 – 0.5000000000 I
S[3] = 0.3125000000 – 0.4375000000 I
S[4] = 0.3125000000 – 0.4687500000 I
S[5] = 0.3046875000 – 0.4609375000 I
S[6] = 0.3085937500 – 0.4609375000 I
S[7] = 0.3076171875 – 0.4619140625 I
S[8] = 0.3076171875 – 0.4614257812 I
S[9] = 0.3077392578 – 0.4615478516 I
S[10] = 0.3076782227 – 0.4615478516 I
S[11] = 0.3076934814 – 0.4615325928 I
S[12] = 0.3076934814 – 0.4615402222 I
S[13] = 0.3076915741 – 0.4615383148 I
S[14] = 0.3076925278 – 0.4615383148 I
S[15] = 0.3076922894 – 0.4615385532 I
S[16] = 0.3076922894 – 0.4615384340 I
S[17] = 0.3076923192 – 0.4615384638 I
S[18] = 0.3076923043 – 0.4615384638 I
S[19] = 0.3076923080 – 0.4615384601 I
S[20] = 0.3076923080 – 0.4615384620 I
S[21] = 0.3076923075 – 0.4615384615 I
S[22] = 0.3076923077 – 0.4615384615 I
S[23] = 0.3076923077 – 0.4615384616 I
S[24] = 0.3076923077 – 0.4615384615 I
S[25] = 0.3076923077 – 0.4615384615 I

The following graphics shows that the real parts of the partial sums of the series $f_2\left(\frac{3}{2}i\right)$ are approaching $\frac{3}{14}$ (blue line).

The real parts of the first 25 partial sums

The next graphics shows that the imaginary parts of the partial sums of the series $f_2\left(\frac{3}{2}i\right)$  are approaching $-\frac{6}{13}$ (blue line).

The imaginary parts of the first 25 partial sums

Also shown below is the graphics of the first 25 partial sums of the series $f_2\left(\frac{3}{2}i\right)$. They are approaching the complex number $f\left(\frac{3}{2}i\right)=\frac{4}{13}-\frac{6}{13}i$ (the intersection of horizontal and vertical blue lines).

The first 25 partial sums

Note that $f_1(z)=f_2(z)$ on $D_1\cap D_2$. Define $F(z)$ as

$$F(z)=\left\{\begin{array}{ccc} f_1(z) & \mbox{if} & z\in D_1,\\ f_2(z) & \mbox{if} & z\in D_2. \end{array}\right.$$

Analytic continuation

Then $F(z)$ is analytic in $D_1\cup D_2$. The function $F(z)$ is called the analytic continuation into $D_1\cup D_2$ of either $f_1$ or $f_2$, and $f_1$ and $f_2$ are called elements of $F$. The function $f_1(z)$ can be continued analytically to the punctured plane $\mathbb{C}\setminus\{-1\}$ and the function $f(z)=\frac{1}{1+z}$ is indeed the analytic continuation into $\mathbb{C}\setminus\{-1\}$ of $f_1$. In general, whenever analytic continuation exists it is unique.