Integral

[Tolaso J Kos]

Let $0 \leq a<1$. Evaluate the integral:

$$\int_0^{\pi} \frac{1 - a^2}{1 - 2a \cos \theta + a^{2}}\, {\rm d} \theta$$

[Hamza Mahmood]

\begin{aligned}
\int \frac{1-a^2}{1-2a \cos x + a^2}\, dx &=(1-a^2)\int \frac{1}{\left ( a^2+1 \right ) \left ( \cos^2 \frac{x}{2}+ \sin^2 \frac{x}{2} \right ) - 2a \left ( \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2} \right )} \, dx \\
&= \left ( 1-a^2 \right ) \int \frac{1}{\cos^2 \frac{x}{2} \left [ \left ( a^2+1 \right )\left ( 1+\tan^2 \frac{x}{2} \right )-2a \left ( 1-\tan^2 \frac{x}{2} \right ) \right ]}\, dx\\
&\overset{t=\tan \frac{x}{2}}{=} (1-a^2)\int \frac{2}{\left ( a^2+1 \right )\left ( 1+t^2 \right )- 2a \left ( 1-t^2 \right )}\, dt \\
&= 2 (1-a^2)\int \frac{1}{t^2 \left ( 1+a \right )^2 + (1-a)^2}\, dt \\
&=\frac{2 (1-a^2)}{\left ( 1+a \right )^2} \int \frac{1}{t^2 + \left ( \frac{1-a}{1+a} \right )^2}\, dt \\
&= \frac{2(1-a)}{1+a}\cdot \frac{1}{\left | \frac{1-a}{1+a} \right |} \tan^{-1}\left ( \frac{1}{\left | \frac{1-a}{1+a} \right |} \right )+c
\end{aligned}

Since $0\leq a<1$ we have that:

$$\int \frac{1-a^2}{1-2a \cos x+ a^2}\, dx = 2 \tan^{-1} \left ( \frac{1+a}{1-a} \tan \frac{x}{2} \right )+c , \; c \in \mathbb{R}$$

Hence

$$\int_{0}^{\pi}\frac{1-a^2}{1-2a \cos x +a^2} \, {\rm d}x = \lim_{x \rightarrow \pi^-} 2\tan ^{-1} \left ( \frac{1+a}{1-a} \tan \frac{x}{2} \right ) - 0 = 2 \cdot \frac{\pi}{2}= \pi$$

[Tolaso J Kos]

Here is another way of doing that by invoking the Poisson kernel. Indeed it is pretty quite known from the theory of Fourier Analysis that:

$$\frac{1-a^2}{1-2a\cos \theta +a^2}= \sum_{n=-\infty}^{\infty}a^{|n|}e^{in \theta}$$

In the unit disk this series converges uniformly and therefore the interchange of integral and summation is allowed. Also, the Poisson kernel (denoted as $P_a(\theta)$) satisfies the following property (among other interesting ones too)

\begin{equation} \frac{1}{2\pi}\int_{-\pi}^{\pi} P_a(\theta)\, {\rm d}\theta=1 \end{equation}

Hence:

\begin{align*}
\int_{0}^{\pi}\frac{1-a^2}{1-2a \cos \theta +a^2}\, {\rm d}\theta &=\frac{1}{2} \int_{-\pi}^{\pi}\frac{1-a^2}{1-2a\cos \theta +a^2}\, {\rm d}\theta \\
&=\frac{1}{2}\int_{-\pi}^{\pi}\sum_{n=-\infty}^{\infty}a^{|n|}e^{in \theta} \, {\rm d}\theta \\
&\overset{(1)}{=} \frac{2\pi}{2}\\
&= \pi
\end{align*}