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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 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

    \begin{equation} I(a) = I(-a) \end{equation}
  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

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

    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
    \begin{equation}
    I(a)= \frac{1}{2}I(a^2)
    \end{equation}
    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.