3.9 Applications of Residues to Improper Integration

One application of the theory of residues is to improper integrals. Before going into the general theory, let’s do an example.

Proposition 3.42

We have $\displaystyle \int _0^{\infty } \frac{x^2}{x^6 + 1}\, dx = \frac{\pi }{6}$.

Proof ▼

First, let’s recall what the improper integral is. By definition, we have

\[ \int _0^{\infty } \frac{x^2}{x^6 + 1}\, dx = \lim _{R\to +\infty } \int _0^{R} \frac{x^2}{x^6 + 1}\, dx. \]

Now, by symmetry, we have

\[ \int _0^{R} \frac{x^2}{x^6 + 1}\, dx = \frac{1}{2}\int _{-R}^{R} \frac{x^2}{x^6 + 1}\, dx. \]

So, as it will turn out to be easier to deal with, let’s work with the integral on the right-hand-side instead of the one on the left.

For each $R{\gt}0$, let’s write $C_R$ for the contour that goes from $-R$ to $R$ only the real axis, and the around the top of the circle of radius $R$ centered about $0$ from $R$ back to $-R$ again. See the picture in Figure 3.1.

$\begin{tikzpicture} \node (A) at (-1, 0) {$\bullet$}; \node (AL) at (-1, -.3) {$-R$}; \node (B) at (1,0) {$\bullet$}; \node (BL) at (1, -.3) {$R$}; \node (C) at (0,1) {$\bullet$}; \node (CL) at (.2, 1.1) [right] {$Ri$}; \draw[-latex] (0,0) -- (1,0) arc[start angle = 0, end angle = 180, radius = 1cm] -- (0,0); \end{tikzpicture}$

Figure 3.1 The contour $C_R$

Now, let’s write $C_R^1$ for the portion of $C_R$ on the bottom, i.e., the part that lies on the real line and goes straight from $-R$ to $R$, and $C_R^2$ for the portion that goes from $R$ back to $-R$ along the top half of the circle. We then have

\begin{align} \int _{C_R} \frac{z^2}{z^6 + 1}\, dz & = \int _{C_R^1} \frac{z^2}{z^6 + 1}\, dz + \int _{C_R^2} \frac{z^2}{z^6 + 1}\, dz\label{CR}\\ \int _{C_R^1} \frac{z^2}{z^6 + 1}\, dz & = \int _{-R}^{R} \frac{z^2}{z^6 + 1}\, dz\label{CR1}. \end{align}

The idea is to carry out the following steps:

Show that $\displaystyle \lim _{R\to \infty } \int _{C_R^2} \frac{z^2}{z^6 + 1}\, dz = 0$.
Use the Residue Theorem to compute $\displaystyle \int _{C_R} \frac{z^2}{z^6 + 1}\, dz$.
Conclude that $\displaystyle \int _0^{\infty } \frac{z^2}{z^6 + 1}\, dz = \frac{1}{2}\lim _{R\to \infty } \int _{C_R} \frac{z^2}{z^6 + 1}\, dz$.

For (a), we use the fact that, for $R {\gt} 1$, the norm of the integrand is at most $\displaystyle R^2/(R^6 - 1)$. It follows that

\begin{equation} |\int _{C_R^2} \frac{z^2}{z^6 + 1}\, dz| \leq \frac{R^2}{R^6 -1} \pi R= \frac{\pi R^{3}}{R^6 -1}. \end{equation}

3.45

So, by the squeeze theorem, $\displaystyle \lim _{R\to \infty } \int _{C_R^2} \frac{z^2}{z^6 + 1}\, dz = \lim _{R\to \infty } \frac{\pi R^{3}}{R^6 -1}= 0$.

For (b), set $f(z) = z^2/(z^6 + 1)$. If $R {\gt} 1$, then $f$ has three poles contained in $C_R$ at $z_1 = e^{\pi i/6}$, at $z_2 = i$ and at $z_3 = e^{5\pi i/6}$. Setting $p(z) = z^2$ and $q(z) = z^6 + 1$, we see that

\[ \operatorname*{\mathrm{Res}}_{z=z_k} \frac{p}{q} = \frac{p(z_k)}{q'(z_k)} = \frac{z_k^2}{6z_k^5} = z_k^{-3}/6. \]

\begin{align*} \int _{C_R} \frac{z^2}{z^6 + 1}\, dz & = 2\pi i\sum _{k=1}^3 \operatorname*{\mathrm{Res}}_{z=z_k} \frac{z^2}{z^6 + 1} \\ & = \frac{2\pi i}{6}\left(i - i + i\right)\\ & = \frac{\pi }{3}. \end{align*}

For (c), we then get

\begin{align*} \int _0^{\infty } \frac{x^2}{x^6 + 1}\, dx & = \lim _{R\to +\infty } \int _0^{R} \frac{x^2}{x^6 + 1}\, dx \\ & = \frac{1}{2} \lim _{R\to +\infty } \int _{-R}^{R} \frac{x^2}{x^6 + 1}\, dx \\ & = \frac{1}{2} \lim _{R\to \infty }\int _{C_R} \frac{z^2}{z^6 + 1}\, dz - \frac{1}{2}\lim _{R\to \infty } \int _{C_R^2} \frac{z^2}{z^6 + 1}\, dz \\ & = \left(\frac{1}{2}\right)\left(\frac{\pi }{3}\right) - 0 = \frac{\pi }{6}. \end{align*}

Here’s another, example.

Exercise 3.46

Compute $\displaystyle \int _{-\infty }^{\infty } \frac{\sin ^2 x}{x^2}\, dx$.

Solution ▼

Here, we have

\begin{align*} \sin ^2 z & = \left(\frac{e^{iz} - e^{-iz}}{2i}\right)^2 = \frac{e^{2iz} -2 + e^{-2iz}}{-4} \\ & = \frac{1 - e^{2iz}}{4} + \frac{1 - e^{-2iz}}{4}. \end{align*}

The idea we want to use here is that, if $z = x + iy$ is in the upper-half-plane, then $|e^{2iz}| = e^{-y}$ is small, while, if $z$ is in the lower-half-plane, then $|e^{-2iz}| = e^{y}$ is small.

We also want to use the idea that, since $\displaystyle g(z) := \frac{\sin ^2}{z^2}$ has a removable singularity at $z = 0$, for any $r{\gt}0$, $\displaystyle \int _{-r}^r \frac{\sin ^2}{z^2}\, dz = \int _{L_r} \frac{\sin ^2 z}{z^2}\, dz$ where $L_r$ is the contour going from $-r$ to $r$ around the lower half of the circle of radius $r$ centered at $0$.

From this, we see that

\begin{equation} \label{Lr} \int _{-r}^r \frac{\sin ^2}{z^2}\, dz = \int _{L_r} \frac{1 - e^{2iz}}{4z^2}\, dz + \int _{L_r} \frac{1 - e^{-2iz}}{4z^2}\, dz. \end{equation}

3.47

Now, for $z\in L_r$, we have $\displaystyle |\frac{1-e^{-2iz}}{4z^2}| \leq 2/4r^2$. So

\begin{equation} \label{lbound} |\int _{L_r} \frac{1 - e^{-2iz}\, dz}{4z^2}| \leq \frac{\pi }{4r}. \end{equation}

3.48

Therefore,

\begin{equation} \label{lowz} \lim _{r\to \infty }\int _{L_r} \frac{1 - e^{-2iz}\, dz}{4z^2} = 0. \end{equation}

3.49

On the other hand, let’s write $U_r$ for the contour going from $r$ to $-r$ along the upper half of the circle of radius $r$ centered at $0$, and $C_r$ for the contour $L_r$ followed by $U_r$ or (in other words) the circle of radius $r$ centered at $0$. Then, for similar reasons as for 3.48, we see that

\begin{equation} \label{ubound} |\int _{U_r} \frac{1 - e^{2iz}\, dz}{4z^2}| \leq \frac{\pi }{4r}. \end{equation}

3.50

So,

\begin{equation} \label{highz} \lim _{r\to \infty }\int _{U_r} \frac{1 - e^{2iz}\, dz}{4z^2} = 0. \end{equation}

3.51

Moreover, by the Residue Theorem,

\begin{align*} \int _{C_r} \frac{1 - e^{2iz}}{4z^2}\, dz & = 2\pi i\operatorname*{\mathrm{Res}}_{z=0} \frac{1 - e^{2iz}}{4z^2} \\ & = \frac{2\pi i (-2i)}{4} = \pi , \end{align*}

It follows that

\begin{align*} \lim _{r\to \infty } \int _{L_r} \frac{1 - e^{2iz}}{4z^2}\, dz & = \pi - \lim _{r\to \infty } \int _{U_r} \frac{1 - e^{2iz}}{4z^2}\, dz \\ & = \pi . \end{align*}

But then, putting everything together,

\begin{align*} \int _{-\infty }^{\infty } \frac{\sin ^2 x}{x^2}\, dx & = \lim _{r\to \infty } \int _{L_r} \frac{1 - e^{2iz}}{4z^2}\, dz + \lim _{r\to \infty } \int _{L_r} \frac{1 - e^{-2iz}}{4z^2}\, dz \\ & = \pi + 0 = \boxed {\pi .} \end{align*}