Articles
[ Place and/or view comments (1) ] | [ Back ]
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.
- Making the substitution $x \mapsto \pi - x$ we note that
\begin{equation} I(a) = I(-a) \end{equation} - 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.
[ Place and/or view comments (1) ] | [ Back ]