In this lecture, we study improper integrals of the form $\int_{-\infty}^\infty f(x)\sin axdx$ or $\int_{-\infty}^\infty f(x)\cos axdx$, where $a$ denotes a positive constant. These integrals appear in Fourier analysis. Assume that $f(x)=\frac{p(x)}{q(x)}$, where $p(x)$ and $q(x)$ are polynomials with real coefficients and no factors in common. Also, $q(z)$ has no real zeros. We discuss how to evaluate improper integrals of the above type through the following example.

Example. Evaluate $\int_{-\infty}^\infty\frac{\cos 3x}{(x^2+1)^2}dx$.

Solution. Let $f(z)=\frac{1}{(z^2+1)^2}$. Then $f(z)e^{3iz}$ is analytic everywhere on and above the real axis except at $z=i$. Let $C_R$ be the upper semi-circle centered at the origin with radius $R>1$. Then by Cauchy’s Residue Theorem,
$$\int_{-R}^R\frac{e^{i3x}}{(x^2+1)^2}dx=2\pi i B_1-\int_{C_R}f(z)e^{i3z}dz,$$
where $B_1=\mathrm{Res}_{z=i}[f(z)e^{i3z}]$. $f(z)e^{i3z}$ can be written as
$$f(z)e^{i3z}=\frac{\phi(z)}{(z-i)^2},$$
where $\phi(z)=\frac{e^{i3z}}{(z+i)^2}$. Since $z=i$ is a pole of order $2$ of $f(z)$,
$$B_1=\phi'(i)=\frac{1}{ie^3}.$$
On $C_R$, $|z|=R$ and so by triangle inequality we obtain
$$|(z+i)^2|\geq (R^2-1)^2$$
and thereby
$$|f(z)|\leq\frac{1}{(R^2-1)^2}.$$
$|e^{i3z}|=e^{-3y}\leq 1$ for all $y\geq 0$. Hence, we find that
$$\left|\mathrm{Re}\int_{C_R}f(z)e^{i3z}dz\right|\leq\left|\int_{C_R}f(z)e^{i3z}dz\right|\leq\frac{\pi R}{(R^2-1)^2}\to 0$$
as $R\to\infty$. Therefore,
$$\int_{-\infty}^\infty\frac{e^{i3x}}{(x^2+1)^2}dx=\frac{2\pi}{e^3}.$$

Recall the definition of improper integrals in calculus:
\begin{align*}
\int_0^\infty f(x)dx&=\lim_{R\to\infty}\int_0^R f(x)dx,\\
\int_{-\infty}^\infty f(x)dx&=\lim_{R_1\to\infty}\int_{-R_1}^0 f(x)dx+\lim_{R_2\to\infty}\int_0^{R_2}f(x)dx.
\end{align*}
The Cauchy Principal Value (P.V.) is given by
$$\mathrm{P.V.}\int_{-\infty}^\infty f(x)dx=\lim_{R\to\infty}\int_{-R}^R f(x)dx.$$
The Cauchy principal value of an improper integral is not necessarily the same as the improper integral. For example,
$$\mathrm{P.V}\int_{-\infty}^\infty xdx=\lim_{R\to\infty}\int_{-R}^R xdx=0,$$
while
$$\int_{-\infty}^\infty xdx=\lim_{R_1\to\infty}\int_{-R_1}^0xdx+\lim_{R_2\to\infty}\int_0^{R_2}xdx=-\infty+\infty$$
is undefined. In general, if $\int_{-\infty}^\infty f(x)dx<\infty$ then $\mathrm{R.V.}\int_{-\infty}^\infty f(x)dx<\infty$, but the converse need not be true. Suppose that $f(x)$ is an even function. Then
\begin{align*}
\int_0^R f(x)dx&=\frac{1}{2}\int_{-R}^R f(x)dx,\\
\int_{-R_1}^0 f(x)dx&=\int_0^{R_1}f(x) dx.
\end{align*}
So,
$$\mathrm{P.V.}\int_{-\infty}^\infty f(x)dx=\int_{-\infty}^\infty f(x)dx=2\int_0^\infty f(x)dx.$$

Let us consider an even function $f(x)$ of the form $f(x)=\frac{p(x)}{q(x)}$, where $p(x)$, $q(x)$ are polynomials with real coefficients no factors in common. Furthermore, we assume that $q(z)$ has no real zeros but has at least one zero above the real axis. Let us consider a positively oriented upper semicircle $C_R$ whose radius $R$ is large enough to contain all the zeros above the real axis as shown in the figure below.

$C_R$ together with the interval $[-R,R]$ form a positively oriented simple closed contour. Then by Cauchy’s Residue Theorem we have
$$\int_{-R}^R f(x)dx+\int_{C_R} f(z)dz=2\pi i\sum_{k=1}^n\mathrm{Res}_{z=z_k}f(z),$$
i.e.
$$\int_{-R}^R f(x)dx=2\pi i\sum_{k=1}^n\mathrm{Res}_{z=z_k}f(z)-\int_{C_R} f(z)dz.$$
If $\lim_{R\to\infty}\int_{C_R} f(z)dz=0$ then
$$\mathrm{P.V.}\int_{-\infty}^\infty f(x)dx=2\pi i\sum_{k=1}^n\mathrm{Res}_{z=z_k}f(z).$$
If in addition $f(x)$ is even,
$$\int_{-\infty}^\infty f(x)dx=2\pi i\sum_{k=1}^n\mathrm{Res}_{z=z_k}f(z)$$
or
$$\int_0^\infty f(x)dx=\pi i\sum_{k=1}^n\mathrm{Res}_{z=z_k}f(z).$$

Example. Let us consider the improper integral
$$\int_0^\infty\frac{x^2}{x^6+1}dx.$$
Let $f(z)=\frac{z^2}{z^6+1}$ has isolated singularities at the zeros of $z^6+1$, and is analytic everywhere else. $z^6=-1$ has solutions (the sixth roots of $-1$)
$$c_k=\exp\left[i\left(\frac{\pi}{6}+\frac{2k\pi}{6}\right)\right],\ k=0,1,\cdots,5.$$
The first three roots
$$c_0=e^{i\frac{\pi}{6}},\ c_1=i,\ c_2=e^{i\frac{5\pi}{6}}$$
lie in the upper half plane. Let us consider a positively oriented upper semicircle $C_R$ whose radius $R$ is greater than $1$.

Then
$$\int_{-R}^Rf(x)dx=2\pi(B_0+B_1+B_2)-\int_{C_R}f(z)dz,$$
where $B_k$ is the residue of $f(z)$ at $c_k$, $k=0,1,2$. $B_k$ can be found as we studied here
$$B_k=\mathrm{Res}_{z=c_k}\frac{z^2}{z^6+1}=\frac{c_k^2}{6c_k^5}=\frac{1}{6c_k^3},\ k=0,1,2.$$
Thus, we obtain
$$2\pi(B_0+B_1+B_2)=2\pi\left(\frac{1}{6i}-\frac{1}{6i}+\frac{1}{6i}\right)=\frac{\pi}{3}$$
and hence,
$$\int_{-R}^R f(x)dx=\frac{\pi}{3}-\int_{C_R}f(z)dz.$$
On $C_R$, $|z|=R$ so
$$|z^6+1|\geq ||z|^6-1|=|R^6-1|=R^6-1$$
and thereby we obtain
$$|f(z)|\leq\frac{R^2}{R^6-1}.$$
Since the length of $C_R$ is $\pi R$,
$$\left|\int_{C_R} f(z)dz\right|\leq\frac{R^2}{R^6-1}\cdot\pi R\to 0$$
as $R\to\infty$. Hence,
$$\mathrm{P.V.}\int_{-\infty}^\infty\frac{x^2}{x^6+1}dx=\lim_{R\to\infty}\frac{x^2}{x^6+1}dx=\frac{\pi}{3}.$$
Since the integrand is even,
$$\int_{-\infty}^\infty\frac{x^2}{x^6+1}dx=\frac{\pi}{3}$$
and
$$\int_0^\infty\frac{x^2}{x^6+1}dx=\frac{\pi}{6}.$$

Zeros and poles are closely related and their relationship may be used to calculates residues. First we introduce two theorems without proof. (Their proofs can be found, for instance, in [1].)

Theorem 1. A function $f$ is that is analytic at a point $z_0$ has a zero of order $m$ there if and only if there is a function $g$, which is analytic and nonzero at $z_0$, such that
$$f(z)=(z-z_0)^mg(z).$$

Theorem 2. Suppose that:

(i) two functions $p$ and $q$ are analytic at a point $z_0$;

(ii) $p(z_0)\ne 0$ and $q$ has a zero of order $m$ at $z_0$.

Then $\frac{p(z)}{q(z)}$ has a pole of order $m$ at $z_0$.

Now we discuss our main theorem in this lecture.

Theorem. Let two functions $p$ and $q$ be analytic at a point $z_0$. If
$$p(z_0)\ne 0,\ q(z_0)=0,\ \mbox{and}\ q'(z_0)\ne 0,$$
then  $z_0$ is a simple pole of $\frac{p(z)}{q(z)}$ and
$$\mathrm{Res}_{z=z_0}\frac{p(z)}{q(z)}=\frac{p(z_0)}{q'(z_0)}.$$

Proof. From the conditions, we see that $q(z)$ has a zero of order $1$, so by theorem 1 it can be written as
$$q(z)=(z-z_0)g(z)$$
where $g(z)$ is analytic at $z=z_0$ and $g(z_0)\ne 0$. This can be in fact readily seen without quoting theorem 1. Since $q(z)$ is analytic at $z_0$, it can be written as a Taylor series expansion
\begin{align*}
q(z)&=q(z_0)+\frac{q'(z_0)}{1!}(z-z_0)+\frac{q^{\prime\prime}(z_0)}{2!}(z-z_0)^2+\cdots\\
&=\frac{q'(z_0)}{1!}(z-z_0)+\frac{q^{\prime\prime}(z_0)}{2!}(z-z_0)^2+\cdots\\
&=(z-z_0)\left\{\frac{q'(z_0)}{1!}+\frac{q^{\prime\prime}(z_0)}{2!}(z-z_0)+\cdots\right\}.
\end{align*}
Set $g(z)=\frac{q'(z_0)}{1!}+\frac{q^{\prime\prime}(z_0)}{2!}(z-z_0)+\cdots$. Then $g(z)$ is analytic at $z=z_0$ and that $g(z_0)=q'(z_0)\ne 0$.

Now, theorem 2 implies that $\frac{p(z)}{q(z)}$ has a simple pole at $z=z_0$, but without quoting theorem 2, $\frac{p(z)}{q(z)}=\frac{\frac{p(z)}{g(z)}}{z-z_0}$, $\frac{p(z)}{g(z)}$ is analytic and nonzero at $z=z_0$. So, $\frac{p(z)}{q(z)}$ has a simple pole at $z=z_0$ as we studied here. Thus,
$$\mathrm{Res}_{z=z_0}\frac{p(z)}{q(z)}=\frac{p(z_0)}{g(z_0)}=\frac{p(z_0)}{q'(z_0)}.$$
This completes the proof.

Example. Find the residue of the functions
$$f(z)=\frac{z}{z^4+4}$$
at the isolated singularity $z_0=\sqrt{2}e^{\frac{i\pi}{4}}=1+i$.

Solution. Let $p(z)=z$ and $q(z)=z^4+4$. Then $p(z_0)=p(1+i)=1+i\ne 0$, $q(z_0)=0$, and $q'(z_0)=4z_0^3=4(1+i)^3\ne 0$. Hence, $f(z)$ has a simple pole at $z_0$. The residue $B_)$ is found by
$$B_0=\mathrm{Res}_{z=z_0}f(z)=\frac{p(z_0)}{q'(z_0)}=\frac{z_0}{3z_0^3}=\frac{1}{4z_0^2}=-\frac{i}{8}.$$

Example. Evaluate the contour integral
$$\int_C\frac{e^{zt}}{\sinh z}dz,$$
where $C$ is the positively oriented circle $|z|=8$.

Solution. $\sinh z=0$ if and only if $e^{2z}=1$. Let $z=x+iy$. Then it follows from $e^{2z}=1$ that $$e^{2x}\cos 2y=1,\ e^{2x}\sin 2y=0.$$
The solutions are $x=0$ and $y=n\pi$, $n=0,\pm 1,\pm 2,\cdots$. Thus, $\sinh z=0$ if and only if $z=n\pi i$, $ n=0,\pm 1,\pm 2$. Among these, only $z_0=0$, $z_1=\pi i$, $z_2=-\pi i$, $z_3=2\pi i$, and $z_4=-2\pi i$ lie within the interior of the circle $|z|=8$. Let $p(z)=e^{zt}$  and $q(z)=\sinh z$. Then for each $i=0,1,2,3,4$, $p(z_i)\ne 0$, $q(z_i)=0$ and $q(z_i)\ne 0$. So each $z_i$ is a simple pole of $f(z)=\frac{e^{zt}}{\sinh z}$. If we denote each residue $\mathrm{Res}_{z=z_i}f(z)$ by $B_i$, then we have
\begin{align*}B_0&=\frac{p(0)}{q'(0)}=1,\\
B_1&=\frac{p(\pi i)}{q'(\pi i)}=-e^{\pi it},\ B_2=\frac{p(-\pi i)}{q'(-\pi i)}=-e^{-\pi it},\\
B_3&=\frac{p(2\pi i)}{q'(2\pi i)}=e^{2\pi it},\ B_4=\frac{p(-2\pi i)}{q'(-2\pi i)}=e^{-2\pi it}.\end{align*}
By Cauchy’s Residue Theorem, we obtain
\begin{align*}
\int_C\frac{e^{zt}}{\sinh z}dz&=2\pi i\sum_{i=0}^4B_i\\
&=2\pi i(1-2\cos\pi t+2\cos 2\pi t).
\end{align*}

References:

[1] James Brown and Ruel Churchill, Complex Variables and Applications, 8th Edition, McGraw-Hill, 2008

When $f(z)$ has a pole of order $m$, we may be able to find the residue of $f(z)$ at $z_0$ without expanding $f(z)$ into a Laurent series at $z=z_0$. This gives a great computational advantage.

Suppose that $z_0$ is a pole of order $m$ of $f(z)$. Then $f(z)$ has a Laurent series expansion
$$f(z)=\sum_{n=0}^\infty a_n(z-z_0)^n+\frac{b_1}{z-z_0}+\frac{b_2}{(z-z_0)^2}+\cdots+\frac{b_m}{(z-z_0)^m}\ (b_m\ne 0)$$
valid in a punctured disk $0<|z-z_0|<R$.
Define
$$\phi(z)=\left\{\begin{array}{ccc}
(z-z_0)^mf(z) & \mbox{if} & z\ne z_0,\\
b_m & \mbox{if} & z=z_0.
\end{array}\right.$$
Then $\phi(z)$ has a power series representation
$$\phi(z)=b_m+b_{m-1}(z-z_0)+\cdots+b_2(z-z_0)^{m-2}+b_1(z-z_0)^{m-1}+\sum_{n=0}^\infty(z-z_0)^{m+n},$$
for $|z-z_0|<R$. That is, $\phi(z)$ is analytic at $z=z_0$ and in the disk $|z-z_0|<R$. We find that
$$
b_1=\left\{\begin{array}{ccc}\frac{\phi^{(n-1)}(z_0)}{(m-1)!} & \mbox{if} & m\geq 2,\\
\phi(z_0) & \mbox{if} & m=1.
\end{array}\right.$$

Conversely, suppose that $f(z)$ can be written in the form
$$f(z)=\frac{\phi(z)}{(z-z_0)^m},$$
where $\phi(z)$ is analytic and nonzero at $z=z_0$. $\phi(z)$ has a Taylor series expansion at $z_0$
\begin{align*}
\phi(z)=&\phi(z_0)+\frac{\phi'(z_0)}{1!}(z-z_0)+\frac{\phi^{\prime\prime}(z_0)}{2!}(z-z_0)^2\\
&+\cdots+\frac{\phi^{(m-1)}(z_0)}{(m-1)!}(z-z_0)^{m-1}+\sum_{n=m}^\infty\frac{\phi^{(n)}(z_0)}{n!}(z-z_0)^n
\end{align*}
in some neighbourhood $|z-z_0|<\epsilon$. Since $\phi(z_0)\ne 0$, $z_0$ is a pole of order $m$ of $f(z)$. Clearly, we have
$$
\mathrm{Res}_{z=z_0}f(z)=\left\{\begin{array}{ccc}
\phi(z_0) & \mbox{if} & m=1,\\
\frac{\phi^{(n-1)}(z_0)}{(m-1)!} & \mbox{if} & m\geq 2.
\end{array}\right.$$
Therefore, we have the following theorem holds.

Theorem. An isolated singularity $z_0$ of a function $f$ is a pole of order $m$ if and only if $f(z)$ can be written as
$$f(z)=\frac{\phi(z)}{(z-z_0)^m},$$
where $\phi(z)$ is analytic and nonzero at $z=z_0$. Moreover,
$$
\mathrm{Res}_{z=z_0}f(z)=\left\{\begin{array}{ccc}
\phi(z_0) & \mbox{if} & m=1,\\
\frac{\phi^{(n-1)}(z_0)}{(m-1)!} & \mbox{if} & m\geq 2.
\end{array}\right.$$

Example. $f(z)=\frac{z+1}{z^2+9}$ has an isolated singularity at $z=3i$. $f(z)$ can be written as
$$f(z)=\frac{\frac{z+1}{z+3i}}{z-3i}.$$
Then $\phi(z)=\frac{z+1}{z+3i}$ is analytic at $z=3i$ and $\phi(3i)=\frac{3-i}{6}\ne 0$. Hence, $z=3i$ is a simple pole of $f(z)$ and
$$\mathrm{Res}_{z=3i}f(z)=\phi(3i)=\frac{3-i}{6}.$$
$z=-3i$ is also a simple pole of $f(z)$ and
$$\mathrm{Res}_{z=-3i}f(z)=\frac{3+i}{6}.$$

Example. Let us consider $f(z)=\frac{z^3+2z}{(z-i)^3}$. $\phi(z)=z^3+2z$ is analytic at $z=i$ and $\phi(i)=i\ne 0$. Hence, $z=i$ is a pole of order $3$ and
$$b_1=\mathrm{Res}_{z=i}f(z)=\frac{\phi^{\prime\prime}(i)}{2!}=3i.$$

Recall that if $f(z)$ has an isolated singularity at $z=z_0$, it may be represented by a Laurent series
$$f(z)=\sum_{n=0}^\infty a_n(z-z_0)^n+\frac{b_1}{z-z_0}+\frac{b_2}{(z-z_0)^2}+\cdots+\frac{b_n}{(z-z_0)^n}+\cdots$$
in a puctured disk $0<|z-z_0|<R$. The part of series that contains negative powers of $z-z_0$
$$\frac{b_1}{z-z_0}+\frac{b_2}{(z-z_0)^2}+\cdots+\frac{b_n}{(z-z_0)^n}+\cdots$$
is called the principal part of $f(z)$ at $z_0$.

Suppose that there exists $m\in\mathbb{N}$ such that $b_m\ne 0$ and $b_{m+1}=b_{m+2}=\cdots=0$. Then
$$f(z)=\sum_{n=0}^\infty a_n(z-z_0)^n+\frac{b_1}{z-z_0}+\frac{b_2}{(z-z_0)^2}+\cdots+\frac{b_m}{(z-z_0)^m},$$
where $0<|z-z_0|<R$. In this case, the isolated singularity $z_0$ is called a pole of order $m$. A pole of order 1 is usually called a simple pole.

Example.
\begin{align*}
\frac{z^2-2z+3}{z-2}&=z+\frac{3}{z-2}\\
&=2+(z-2)+\frac{3}{z-2},\ 0<|z-2|<\infty
\end{align*}
has a simple pole at $z_0=2$. The residue at $z_0=2$ is 3.

Example.
\begin{align*}
\frac{\sinh z}{z^4}&=\frac{1}{z^4}\left(z+\frac{z^3}{3!}+\frac{z^5}{5!}+\cdots\right)\\
&=\frac{1}{z^3}+\frac{1}{3!z}+\frac{z}{5!}+\frac{z^3}{7!}+\cdots,\ 0<|z|<\infty
\end{align*}
has a pole of order $m=3$ at $z_0=0$. The residue at $z_0=0$ is $\frac{1}{6}$.

When $b_n=0$ for all $n\geq 1$, so that
$$f(z)=\sum_{n=0}^\infty (z-z_0)^n$$
the point $z_0$ is called a removable singularity.

Example.
\begin{align*}
f(z)&=\frac{1-\cos z}{z^2}\\
&=\frac{1}{z^2}\left[1-\left(1-\frac{z^2}{2!}+\frac{z^4}{4!}-\frac{z^6}{6!}+\cdots\right)\right]\\
&=\frac{1}{2!}-\frac{z^2}{4!}+\frac{z^4}{6!}-\cdots,\ 0<|z|<\infty.
\end{align*}
Thus $z_0=0$ is a removable singularity. Define
$$g(z)=\left\{\begin{array}{ccc}
f(z) & \mbox{if} & z\ne 0,\\
\frac{1}{2!} & \mbox{if} & z=0.
\end{array}\right.$$
Then $g(z)$ is entire.

When an infinite number of the $b_n$ are nonzero, $z_0$ is called an essential singularity.

Example.
\begin{align*}
\exp\left(\frac{1}{z}\right)&=\sum_{n=0}^\infty\frac{1}{n! z^n}\\
&=1+\frac{1}{1!z}+\frac{1}{2!z^2}+\cdots,\ 0<|z|<\infty
\end{align*}
has an essential singularity at $z_0=0$.

Example. $e^z=-1$ when $z=(2n+1)\pi i$ $(n=0,\pm 1, \pm 2,\cdots)$. So $e^{\frac{1}{z}}=-1$ when $\frac{1}{z}=(2n+1)\pi i$ or $z=-\frac{i}{(2n+1)\pi}$ $(n=0,\pm 1,\pm 2,\cdots)$and an infinite number of these points lie in any neighbourhood of the essential singularity $z_0=0$.

Picard’s Theorem. In each neighbourhood of an essential singularity, a function assumes every finite value, with one possible exception, an infinite number of time.

In the above example, since $e^{\frac{1}{z}}\ne 0$ for all $z$, $z=0$ is the exceptional value in Picard’s theorem.