Evaluating Definite Integrals of Type 1

# Evaluating Definite Integrals of Type 1

We will now begin 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 1.
 Theorem 1 (Evaluation of Definite Integrals of Type 1): Let $R(\cos \theta, \sin \theta)$ be a rational function in $\theta$ that contains no poles on the unit circle. Then $\displaystyle{\int_{0}^{2\pi i} R(\cos \theta, \sin \theta) \: d \theta = \int_{\mid z \mid = 1} f(z) \: dz}$ where $\displaystyle{f(z) = \frac{1}{iz} R \left ( \frac{1}{z} \left [ z + \frac{1}{z} \right ] , \frac{1}{2i} \left [ z - \frac{1}{z} \right ] \right ) \: dz}$.

The formula for $f(z)$ can be difficult to remember so we will demonstrate a process to derive it.

Suppose that $x = \cos \theta$ and $y = \sin \theta$. Then $(x, y)$ is a point on the unit circle as $\theta$ ranges from $0$ to $2\pi$. Note that if $z = x + yi$ and $\mid z \mid = 1$ ($z$ is on the unit circle) then we know that $z \cdot \overline{z} = 1$. Therefore $\overline{z} = \frac{1}{z}$. Then:

(1)
\begin{align} \quad x = \frac{1}{2} \left [ z + \overline{z} \right ] = \frac{1}{2} \left ( z + \frac{1}{z} \right ) \end{align}
(2)
\begin{align} \quad y = \frac{1}{2i} \left [ z - \overline{z} \right ] = \frac{1}{2} \left ( z - \frac{1}{z} \right ) \end{align}

And also $\frac{dz}{d\theta} = \frac{d}{d \theta} (\cos \theta + i \sin \theta) = -\sin \theta + i \cos \theta$. Therefore

(3)
\begin{align} \quad dz = (-\sin \theta + i \cos \theta) \: d \theta = i(\cos \theta + i \sin \theta) \: d \theta = iz \: d \theta \end{align}

And:

(4)
\begin{align} \quad d \theta = \frac{dz}{iz} \end{align}

By the change of variables we have that:

(5)
\begin{align} \quad \int_0^{2\pi} R(\cos \theta, \sin \theta) \: d \theta = \int_{\mid z \mid = 1} f(z) \: dz \end{align}

By The Residue Theorem, the rightmost integral is simply equal to $2\pi i$ multiplied by the sum of the residues of the singularities inside the unit circle, i.e., in $D(0, 1)$.

## Example 1

Evaluate the integral $\displaystyle{\int_0^{2\pi} \frac{1}{5 + 3\cos \theta} \: d \theta}$.

We make the following substitutions in the integral above:

(6)
\begin{align} \quad x = \cos \theta = \frac{1}{2} \left ( z + \frac{1}{z} \right ) \quad , \quad y = \sin \theta = \frac{1}{2i} \left ( z - \frac{1}{z} \right ) \quad , \quad d \theta = \frac{dz}{iz} \end{align}

To get:

(7)
\begin{align} \quad \int_0^{2\pi} \frac{1}{5 + 3 \cos \theta} \: d \theta &= \int_{\mid z \mid=1} \frac{1}{5 + 3 \cdot \frac{1}{2} \left ( z + \frac{1}{z} \right )} \: \frac{dz}{iz} \\ &= \int_{\mid z \mid=1} \frac{2z}{3z^2 + 10z + 3} \frac{1}{iz} \: dz \\ &= \int_{\mid z \mid=1} \frac{-2i}{3z^2 + 10z + 3} \: dz \\ &= -\frac{2i}{3} \int_{\mid z \mid=1} \frac{1}{\left ( z + \frac{1}{3} \right ) \left ( z + 3 \right )} \: dz \end{align}

The function inside the integral has singularities at $\displaystyle{z = -\frac{1}{3}}$ and at $z = -3$. The only of such singularity that occurs inside the unit circle is $\displaystyle{z = -\frac{1}{3}}$ (which is a pole of order $1$), and so by applying the residue theorem and the theorem from The Residue of an Analytic Function at a Pole Singularity page:

(8)
\begin{align} \quad \int_0^{2\pi} \frac{1}{5 + 3 \cos \theta} \: d \theta &= -\frac{2i}{3} \cdot 2\pi \mathrm{Res} \left ( \frac{1}{\left ( z + \frac{1}{3} \right ) \left ( z + 3 \right )}, -\frac{1}{3} \right ) \\ &= -\frac{2i}{3} \cdot 2\pi i \cdot \lim_{z \to -\frac{1}{3}} \frac{1}{\left ( z + \frac{1}{3} \right ) \left ( z + 3 \right )} \cdot \left ( (z + \frac{1}{3} \right ) \\ &= - \frac{2i}{3} \cdot 2\pi i \lim_{z \to -\frac{1}{3}} \frac{1}{z + 3} \\ &= \frac{4\pi}{3} \cdot \frac{3}{8} \\ &= \frac{\pi}{2} \end{align}