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### A logarithmic Poisson integral

Posted: Sat Oct 12, 2019 12:26 pm
A logarithmic Poisson integral
by Tolaso J Kos

Let $a \geq 0$. We will prove that $$I(a)=\int_{0}^{\pi}\ln\left(1-2a\cos x+a^2\right) \, \mathrm{d}x = \left\{\begin{matrix} 0 & , & \left | a \right | \leq 1 \\ 2 \pi \ln \left | a \right | &, & \text{otherwise} \end{matrix}\right.$$ Background: This particular integral first appeared in a Mathematical...

### Re: A logarithmic Poisson integral

Posted: Sat Oct 12, 2019 12:46 pm
It is closely related to another famous integral; namely

$$\int_{0}^{\pi}\frac{\mathrm{d} \theta}{1-2a\cos \theta+a^2}\quad , \quad |a|<1$$

Evaluation of the integral:

For $|a|<1$ we have successively:

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

due to the Poisson Kernel.

Evaluation of the initial integral.

We could use that to evaluate the integral at hand when $|a|\leq 1$. Consider the function

$$I(a)=\int_{0}^{\pi}\ln\left(1-2a\cos x+a^2\right) \, \mathrm{d}x \quad, \quad |a| \leq 1$$

Due to equations $(1)$ , $(3)$ it is $I(-1)=I(1)=0$. So , we are only interested in the case $|a|<1$. Differentiating under the integral sign we get

\begin{align*}
I'(a) &= \int_{0}^{\pi} \frac{2a-2\cos x}{1-2a\cos x+a^2} \, \mathrm{d}x\\
&= \frac{1}{a} \int_{0}^{\pi} \frac{2a^2-2a\cos x}{1-2a\cos x + a^2} \, \mathrm{d}x\\
&=\frac{1}{a} \int_{0}^{\pi} \frac{\left ( a^2+1 \right ) + a^2-1-2a\cos x}{1-2a\cos +a^2} \\
&=\frac{\pi}{a} + \frac{1}{a} \int_{0}^{\pi} \frac{a^2-1}{1-2a\cos x +1} \, \mathrm{d}x \\
&\!\!\!\overset{|a|<1}{=\! =\! =\! =\!} \; \frac{\pi}{a} - \frac{\pi}{a} \\
&=0
\end{align*}

meaning that $I$ is constant. Since $I(0)=0$ we conclude that $I(a)=0$ for all $|a| \leq 1$.