TITLE: Bronstein Integral 21.42
QUESTION [8 upvotes]: Good morning.
I came across the following integral in some field theory calculation:
$\int_0^\pi dx\,\log\left(a^2+b^2-2ab\cos x\right)=2\pi\log\left(\max\lbrace a,b\rbrace\right)$
for $0<a,b\in\mathbb{R}$. The notation has been adapated to make contact with Bronstein's integral 21.42.
Despite 6 pp.of calculations, I was not able to figure out a way to prove the identity (integration by parts, substitution, derivative trick, contour integration...). Does anyone have a proof?
Best regards,
David

REPLY [2 votes]: Suppose that $b>a$. Then
$$
\int_{0}^{\pi}\log(a^{2}+b^{2}-2ab\cos x)\, dx = 2\pi\log b + \int_{0}^{\pi}\log(c^{2}+ 1-2c\cos x)\, dx
$$
where $c = \dfrac{a}{b}$ and $0 \le c < 1$.
We will prove that
$$
I = \int_{0}^{\pi}\log(c^{2}+ 1-2c\cos x)\, dx = 0.
$$
If $\log$ is the principal branch then $f(z) = \dfrac{\log(1-z)}{z}$ is an analytic function in |z| <1.
Consequently
$$
 0 = \oint_{|z|=c}f(z)\, dz = \int_{-\pi}^{\pi}\dfrac{\log(1-ce^{it})}{ce^{it}}ice^{it}\, dt = i\int_{-\pi}^{\pi}\log(1-ce^{it})\, dt
$$
The imaginary part of the right hand side is
$$
0 = \int_{-\pi}^{\pi}\log|1-ce^{it}|\, dt = \int_{-\pi}^{\pi}\log(\sqrt{c^{2}+1-2c\cos t})\, dt = \int_{0}^{\pi}\log(c^{2}+1-2c\cos t)\, dt = I.
$$