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# A logarithmic Poisson integral

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 Tripos somewhere in $1930$. Since then it has been studied methodically and analytically.

We shall present a simple yet powerful method to tackle this particular integral. For that reason we break the solution into two main steps.

1. Making the substitution $x \mapsto \pi - x$ we note that

$$I(a) = I(-a)$$
2. We also note that

\begin{align*}
I(a) + I(-a)
&= \int^{\pi}_{0}\log \! \Big ( \left (1 - 2a\cos x + a^2 \right ) \left (1 + 2a\cos x + a^2 \right ) \Big) \, \mathrm{d}x\\
&= \int^{\pi}_{0}\log \! \Big ( \left (1 + a^2 \right )^2 - \left (2a\cos x \right )^2 \Big) \,\mathrm{d}x\\
\end{align*}

Using double angle formulae we get

\begin{align*}
I(a) + I(-a)&= \int^{\pi}_{0}\log \left ( 1 + 2a^2 + a^4 - 2a^2 \left ( 1 + \cos 2x \right ) \right) \, \mathrm{d}x\\
&= \int^{\pi}_{0}\log \left ( 1 - 2a^2\cos 2x + a^4 \right) \, \mathrm{d}x\\
\end{align*}

so by setting $x \mapsto \frac{x}{2}$ we get

$$I(a) + I(-a) = \frac{1}{2}\int^{2\pi}_{0}\log \left ( 1 - 2a^2\cos x + a^4 \right) \, \mathrm{d}x$$

Splitting the integral at $\pi$ and setting $x \mapsto 2\pi -x$ at the second integral we get

\begin{align*}
I(a) + I(-a) &= \frac{1}{2} I\left(a^2\right) + \frac{1}{2}\int^{2\pi}_{\pi}\log \left ( 1 - 2a^2\cos x + a^4 \right) \, \mathrm{d}x\\
&= \frac{1}{2} I\left(a^2\right) + \frac{1}{2}\int^{\pi}_{0}\log \left ( 1 - 2a^2\cos x + a^4 \right) \, \mathrm{d}x\\
&= I(a^2)
\end{align*}

Hence

I(a)= \frac{1}{2}I(a^2)

It follows from $(3)$ that $I(0)=0$ and $I(1)=0$.We now distinguish cases:

(a) $0\leq a <1$ Iterating $n$ times $(3)$ we get:

$$I(a) = \frac{1}{2^n} I \left ( a^{2^{n}} \right )$$

Letting $n \rightarrow +\infty$ we get that $I(a)=0$.

(b) When $a>1$ it follows that $0<\frac{1}{a}<1$ and consequently $I\left(\frac{1}{\alpha}\right)=0$. Thus,

\begin{align*}
I(a) &= \int^\pi_0 \ln \left( a^2 \left (\frac{1}{a^2} + \frac{\cos x}{a} + 1 \right ) \right) \; \mathrm{d} x\\
&= 2\pi\ln a + I\left ( \frac{1}{a} \right )\\
&= 2\pi\ln a
\end{align*}

Extending the result for negative $a$ using $(1)$ we get to our conclusion.