Definite Integral from 0 to Pi of a Squared minus 2 a b Cosine x plus b Squared

Theorem

 * $\displaystyle \int_0^\pi \map \ln {a^2 - 2 a b \cos x + b^2} \rd x = \begin{cases}2 \pi \ln a & a \ge b > 0 \\ 2 \pi \ln b & b \ge a > 0\end{cases}$

Proof
Note that:


 * $\paren {a - b}^2 \ge 0$

so by Square of Sum:


 * $a^2 - 2 a b + b^2 \ge 0$

So:


 * $a^2 + b^2 \ge 2 a b = \size {-2 a b}$

so we may apply Definite Integral from $0$ to $\pi$ of $\map \ln {a + b \cos x}$.

We then have:

Note that if $a \ge b > 0$ we have:

Note that if $b \ge a > 0$ we have: