Category Archives: Functions of a Complex Variable

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,
$$\int_0^\infty\frac{\sin x}{x}dx=\frac{\pi}{2}.$$

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

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.
\begin{equation}
\label{eq:c-r}
u_x=v_y,\ u_y=-v_x.
\end{equation}
Differentiating the Cauchy-Riemann equations \eqref{eq:c-r} with respect to $x$, we obtain
\begin{equation}
\label{eq:c-r2}
u_{xx}=v_{yx},\ u_{yx}=-v_{xx}.
\end{equation}
Differentiating the Cauchy-Riemann equations \eqref{eq:c-r} with respect to $y$, we obtain
\begin{equation}
\label{eq:c-r3}
u_{xy}=v_{yy},\ u_{yy}=-v_{xy}.
\end{equation}
By the continuity of the partial derivatives of $u(x,y)$ and $v(x,y)$, we have
\begin{equation}
\label{eq:cont}
u_{xy}=u_{yx},\ v_{xy}=v_{yx}.
\end{equation}
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 10 partial sums

The first 20 partial sums

The first 20 partial sums

The first 30 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 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

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

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

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.