Evaluating Definite Integrals of Type 3

# Evaluating Definite Integrals of Type 3

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 3.

## Evaluating Definite Integrals of Type 3a

 Theorem 1 (Evaluating Definite Integrals of Type 3a): If $\mid f(z) \mid$ converges to $0$ as $z \to \infty$ and if $f$ has no singularities on the real axis then $\displaystyle{\int_{-\infty}^{\infty} e^{ix} f(x) \: dx = 2\pi i \sum \mathrm{The \: residues \: of \:} e^{iz} f(z) \: \mathrm{in \: H}}$.

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

(1)
\begin{align} \quad \int_0^{\infty} \frac{\cos x}{x^2 + 1} \: dx \end{align}

At first glance this integral is clearly not of the appropriate form to be of type 3a. However, note that $\displaystyle{f(x) = \frac{\cos x}{x^2 + 1}}$ is an even function. Therefore:

(2)
\begin{align} \quad \int_0^{\infty} \frac{\cos x}{x^2 + 1} \: dx = \frac{1}{2} \int_{-\infty}^{\infty} \frac{\cos x}{x^2 + 1} \: dx \quad (*) \end{align}

Also note that:

(3)
\begin{align} \quad e^{ix} = \cos x + i \sin x \end{align}

And so:

(4)
\begin{align} \quad \frac{e^{ix}}{x^2 + 1} = \frac{\cos x}{x^2 + 1} + i \frac{\sin x}{x^2 + 1} \end{align}

Hence $\displaystyle{\mathrm{Re} \left ( \frac{e^{ix}}{x^2 + 1} \right ) = \frac{\cos x}{x^2 + 1}}$. So let's find the following integral (which is of the appropriate form):

(5)
\begin{align} \int_{-\infty}^{\infty} \frac{e^{ix}}{x^2+1} \: dx \end{align}

The singularities of integrand are $i$ and $-i$. The only singularity that occurs in the upper half complex plane is $i$. Also $\displaystyle{\lim_{z \to 0} \biggr \lvert \frac{1}{x^2 + 1} \biggr \rvert = 0}$. Therefore:

(6)
\begin{align} \quad \int_{-\infty}^{\infty} \frac{e^{ix}}{x^2+1} \: dx &= 2\pi i \mathrm{Res} \left ( \frac{e^{iz}}{z^2 + 1}, i \right ) \\ &= 2\pi i \lim_{z \to i} \frac{e^{iz}}{z^2 + 1} (z - i) \\ &= 2\pi i \lim_{z \to i} \frac{e^{iz}}{(z + i)} \\ &= 2\pi i \frac{e^{i^2}}{2i} \\ &= \frac{\pi}{e} \end{align}

Therefore:

(7)
\begin{align} \quad \int_{-\infty}^{\infty} \frac{\cos x}{x^2 + 1} \: dx &= \mathrm{Re} \left ( \int_{-\infty}^{\infty} \frac{e^{ix}}{x^2+1} \: dx \right ) \\ &= \mathrm{Re} \left ( \frac{\pi}{e} \right ) \\ &= \frac{\pi}{e} \end{align}

From $(*)$ this means that:

(8)
\begin{align} \quad \int_0^{\infty} \frac{\cos x}{x^2 + 1} \: dx = \frac{\pi}{2e} \end{align}
 Theorem 2 (Evaluating Definite Integrals of Type 3b) If $\mid f(z) \mid$ converges to $0$ as $z \to \infty$ and if $f$ has only finitely many simple pole singularities on the real axis then $\displaystyle{PV \int_{-\infty}^{\infty} e^{ix} f(x) \: dx = 2\pi i \sum \mathrm{The \: residues \: of \:} e^{iz} f(z) \: \mathrm{in \: H} + \pi i \sum \mathrm{The \: residues \: of \:} e^{iz} f(z) \: \mathrm{on \: the} \: x-\mathrm{axis}}$.

For example, suppose we want to evaluate the following integral:

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

The isolated singularities of the integrand are $i$, $-i$, $1$, and $-1$. Furthermore $\displaystyle{\lim_{z \to 0} \biggr \lvert \frac{1}{(x^2 + 1)(x^2 - 1)} \biggr \rvert = 0 }$. There are only finitely many simple pole singularities of $f$ on the real-axis so:

(10)
\begin{align} \quad PV \int_{-\infty}^{\infty} \frac{e^{ix}}{(x^2 + 1)(x^2 - 1)} \: dx = 2\pi i \mathrm{Res} \left ( \frac{e^{ix}}{(x^2 + 1)(x^2 - 1)}, i \right ) + \pi i \left [ \mathrm{Res} \left (\frac{e^{ix}}{(x^2 + 1)(x^2 - 1)}, 1 \right ) + \mathrm{Res} \left (\frac{e^{ix}}{(x^2 + 1)(x^2 - 1)}, -1 \right ) \right ] \end{align}