Evaluate the integral:

$$\mathcal{J} = \int_0^\pi \frac{{\rm d}x}{1-2a \cos x + a^2} \quad , \quad \left| a \right| <1$$

We are recalling the Poisson kernel. Thus:

\int_{0}^{\pi} \frac{{\rm d}x}{1-2a \cos x+a^2} &= \frac{1}{2} \int_{-\pi}^{\pi} \frac{{\rm d}x}{1-2a \cos x + a^2} \\

&= \frac{1}{2 \left ( 1-a^2 \right )} \int_{-\pi}^{\pi} \sum_{n=-\infty}^{\infty} a^ \left|n \right|e^{in x} \, {\rm d}x\\

&= \frac{2\pi}{2 \left ( 1-a^2 \right )}\\

&= \frac{\pi}{1-a^2}

\end{align*}

$$\mathcal{J} = \int_0^\pi \frac{{\rm d}x}{1-2a \cos x + a^2} \quad , \quad \left| a \right| <1$$

**Solution**We are recalling the Poisson kernel. Thus:

**\begin{align*}**\int_{0}^{\pi} \frac{{\rm d}x}{1-2a \cos x+a^2} &= \frac{1}{2} \int_{-\pi}^{\pi} \frac{{\rm d}x}{1-2a \cos x + a^2} \\

&= \frac{1}{2 \left ( 1-a^2 \right )} \int_{-\pi}^{\pi} \sum_{n=-\infty}^{\infty} a^ \left|n \right|e^{in x} \, {\rm d}x\\

&= \frac{2\pi}{2 \left ( 1-a^2 \right )}\\

&= \frac{\pi}{1-a^2}

\end{align*}

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