Evaluating Definite Integrals of Type 2

# Evaluating Definite Integrals of Type 2

We will continue to apply some of results we have recently looked at regarding residues of functions at points to solve definite integrals of real-valued functions that would otherwise be difficult to compute. There are three main types of definite integrals we can solve and the pages involving the results are listed below:

• Evaluating Definite Integrals of Type 2

We will be specifically look at evaluating improper integrals below. We must make an important distinction though. Recall that the improper Riemann integral of $f$ on $(-\infty, \infty)$ is defined as:

(1)
\begin{align} \quad \int_{-\infty}^{\infty} f(x) \: dx = \int_{-\infty}^{0} f(x) \: dx + \int_0^{\infty} f(x) \: dx = \lim_{b \to \infty} \int_{-b}^{0} f(x) \: dx + \lim_{c \to \infty} \int_0^c f(x) \: dx \end{align}

The principle value of the improper Riemann integral of $f$ on $(-\infty, \infty)$ is instead defined as:

(2)
\begin{align} \quad PV \int_{-\infty}^{\infty} f(x) \: dx = \lim_{b \to \infty} \int_{-b}^{b} f(x) \: dx \end{align}

Both are types of improper integrals but they need not be equal.

## Evaluating Definite Integrals of Type 2a

 Theorem 1 (Evaluating Definite Integrals of Type 2a): Let $H$ denote the upper half complex plane. If $f$ is a rational function, $\displaystyle{f(x) = \frac{p(x)}{q(x)}}$ where $\mathrm{deg} (q) > \mathrm{deg} (p) + 2$ and $q$ has no real roots, then $\displaystyle{\int_{-\infty}^{\infty} f(x) \: dx = 2\pi i \sum \mathrm{The \: residues \: of \: f \: in \: H}}$.

For example, suppose that we want to compute the following integral:

(3)
\begin{align} \quad \int_{-\infty}^{\infty} \frac{x + 4}{(x^2 + 2x + 2)(x^2 + 9)} \: dx \quad (*) \end{align}

Let $f(x) = \displaystyle{ \frac{x + 4}{(x^2 + 2x + 2)(x^2 + 9)}}$. Then:

(4)
\begin{align} \quad p(x) = x + 4 \quad , \quad q(x) = (x^2 + 2x + 2)(x^2 + 9) \end{align}

We see that $\mathrm{deg} (p) = 1$ and $\mathrm{deg} (q) = 4$. So clearly $4 = \mathrm{deg} (q) > \mathrm{deg} (p) + 2 = 3$. Furthermore, $q$ has no real roots. So, by Theorem 1, the integral $(*)$ is of type 2a and:

(5)
\begin{align} \quad \int_{-\infty}^{\infty} \frac{x + 4}{(x^2 + 2x + 2)(x^2 + 9)} = 2\pi i \sum \mathrm{The \: residues \: of \: f \: in \: H} \end{align}

The isolated singularities of $f$ are $3i$, $-3i$, $-1 + i$ and $-1 - i$ which are graphed below (the red singularities being the ones lying in the upper half complex plane):

Therefore:

(6)
\begin{align} \quad \int_{-\infty}^{\infty} \frac{x + 4}{(x^2 + 2x + 2)(x^2 + 9)} = 2\pi i \left [ \mathrm{Res} (f, 3i) + \mathrm{Res} (f, -1 + i) \right ] \end{align}

Both $3i$ and $-1 + i$ are pole singularities of order $1$. So by the theorem on The Residue of an Analytic Function at a Pole Singularity page we have that:

(7)
\begin{align} \quad \mathrm{Res} (f, 3i) = \lim_{z \to 3i} f(z)(z - 3i) = \lim_{z \to 3i} \frac{x + 4}{(z - (-1 + i))(z - (-1 - i))(z - 3i)(z + 3i)} (z - 3i) = \lim_{z \to 3i} \frac{z+4}{(z+1-i)(z+1+i)(z+3i)} = ... \end{align}
(8)
\begin{align} \quad \mathrm{Res} (f, -1 + i) = \lim_{z \to -i + 1} f(z)(z - 3i) = ... \end{align}

After the computations are completed we have that:

(9)
\begin{align} \quad \int_{-\infty}^{\infty} \frac{x + 4}{(x^2 + 2x + 2)(x^2 + 9)} \: dx = \frac{13\pi}{51} \end{align}

## Evaluating Definite Integrals of Type 2b

 Theorem 2 (Evaluating Definite Integrals of Type 2b): Let $H$ denote the upper half complex plane. If $f$ is a rational function, $\displaystyle{f(x) = \frac{p(x)}{q(x)}}$ where $\mathrm{deg} (q) > \mathrm{deg} (p) + 2$ and $q$ has finitely many simple poles on the $x$-axis, then $\displaystyle{PV \int_{-\infty}^{\infty} f(x) \: dx = 2\pi i \sum \mathrm{The \: residues \: of \: f \: in \: H} + \pi i \sum \mathrm{The \: residues \: of \: f \: on \: the \:} x-\mathrm{axis}}$.

For example, suppose that we want to give an expression for evaluating the following integral:

(10)
\begin{align} \quad PV \int_{-\infty}^{\infty} \frac{1}{(x^2 + 1)(x^2 - 9)} \: dx \end{align}

The isolated singularities of $\displaystyle{f(z) = \frac{1}{(z^2 + 1)(z^2 - 9)}}$ are $i$, $-i$, $3$, and $-3$. Note that $\displaystyle{f(x) = \frac{p(x)}{q(x)}}$ where $p(x) = 1$, $q(x) = (x^2 + 1)(x^2 - 9)$, and $4 = \mathrm{deg} (q) > \mathrm{deg} (p) + 2 = 2$. So by Theorem 2 we have that:

(11)
\begin{align} \quad PV \int_{-\infty}^{\infty} \frac{1}{(x^2 + 1)(x^2 - 9)} \: dx = 2\pi i \mathrm{Res} (f, i) + \pi i [\mathrm{Res} (f, 3) + \mathrm{Res} (f, -3)] \end{align}