Integral of the day XII

Fun integral posted by @integralsbot

Today we evaluate this fun integral posted by @integralsbot \[\int_{0}^{\pi} \frac{\ln (1-2a\cos x + a^2)}{1-2a\cos x +a^2} dx = \frac{2\pi \ln(1-a^2)}{1-a^2} \quad |a|<1 \] To proof this result we will use Fourier series and contour integration.

Proof

Firt, we need the following Fourier series: \[ \sum_{k=1}^{\infty} \frac{a^k\cos(kx)}{k} = -\frac{1}{2} \ln (1-2a\cos x + a^2) \quad x\in(0,2\pi), \quad a^2\leq 1 \tag{1} \] The proof is almost straight: take principal branch of the $\ln(z)$ function, then: \begin{align*} \sum_{k=1}^{\infty} \frac{a^k\cos(kx)}{k} =& \frac{1}{2}\sum_{k=1}^{\infty} \frac{a^k}{k}(e^{ikx} + e^{-ikx})\\ =& \frac{1}{2}\sum_{k=1}^{\infty}\frac{(ae^{ix})^k}{k} + \frac{1}{2}\sum_{k=1}^{\infty}\frac{(ae^{-ix})^k}{k} \\ =& -\frac{1}{2}\ln\left(1-ae^{ix}\right) -\frac{1}{2}\ln\left(1-ae^{-ix}\right)\\ =& -\frac{1}{2}\ln\left((1-ae^{ix})(1-ae^{-ix})\right)\\ =& -\frac{1}{2}\ln\left(1-a(e^{-ix}+e^{ix})+a^2\right)\\ =& -\frac{1}{2}\ln(1-2a\cos(x) +a^2) \end{align*} Back to the integral: \begin{align*} \int_{0}^{\pi} \frac{\ln (1-2a\cos x + a^2)}{1-2a\cos x +a^2} dx = -2\int_{0}^{\pi} \sum_{k=1}^{\infty}\frac{a^k\cos(kx)}{k(1-2a\cos x +a^2)} dx\\ = -2 \sum_{k=1}^{\infty}\frac{a^k}{k}\int_{0}^{\pi} \frac{\cos(kx)}{1-2a\cos x +a^2}dx \tag{2} \end{align*} Now note that \[f(x) = \frac{\cos(kx)}{1-2a\cos x +a^2}\] is an even function.

Hence \[ \int_{0}^{\pi} \frac{\cos(kx)}{1-2a\cos x +a^2}dx = \frac{1}{2}\int_{-\pi}^{\pi} \frac{\cos(kx)}{1-2a\cos x +a^2}dx = \frac{1}{2}\Re{\left(\int_{-\pi}^{\pi} \frac{e^{ixk}}{1-2a\cos x +a^2}dx\right)} \tag{3}\] If we make the substitution \[ \cos x = \frac{z+z^{-1}}{2}\] \[ e^{ix} = z\] \[ dx = \frac{dz}{zi}\] Our integral is transformed in a contour integral round the unit complex circle: \[\int_{-\pi}^{\pi} \frac{e^{ixk}}{1-2a\cos x +a^2}dx = i\oint_{|z|=1} \frac{z^{k}}{(z-a)(az-1)}dz \tag{4}\]

The function $\displaystyle f(z) = \frac{z^{k}}{(z-a)(az-1)}$ has two poles at $z=a$ and $\displaystyle z=\frac{1}{a}$.

The conditions in (1) assure that $z=a$ with $|a|<1$ is the only pole inside the unit circle.

Then, by the Cauhcy integral formula \[ i\oint_{|z|=1} \frac{z^{k}}{(z-a)(az-1)}dz = 2\pi \frac{ a^k}{1-a^2} \tag{5}\] Therefore \begin{align*} \int_{0}^{\pi} \frac{\ln (1-2a\cos x + a^2)}{1-2a\cos x +a^2} dx =& -2\sum_{k=1}^{\infty}\frac{a^k}{k}\int_{0}^{\pi} \frac{\cos(kx)}{1-2a\cos x +a^2}dx \quad \textrm{from (2) }\\ =& -\sum_{k=1}^{\infty}\frac{a^k}{k}\int_{-\pi}^{\pi} \frac{\cos(kx)}{1-2a\cos x +a^2}dx \\ =& -\sum_{k=1}^{\infty}\frac{a^k}{k}\Re{\left(\int_{-\pi}^{\pi} \frac{e^{ixk}}{1-2a\cos x +a^2}dx\right)} \quad \textrm{from (3) }\\ =& -\sum_{k=1}^{\infty}\frac{a^k}{k}\Re{\left(i\oint_{|z|=1} \frac{z^{k}}{(z-a)(az-1)}dz\right)} \quad \textrm{from (4)}\\ =& -\sum_{k=1}^{\infty}\frac{a^k}{k}\Re{\left(2\pi \frac{ a^k}{1-a^2}\right)} \quad \textrm{from (5)}\\ =& \frac{2\pi}{1-a^2}\left(-\sum_{k=1}^{\infty}\frac{a^{2k}}{k}\right)\\ =& \frac{2\pi \ln(1-a^2)}{1-a^2} \end{align*} Hence, we can conclude \[\boxed{ \int_{0}^{\pi} \frac{\ln (1-2a\cos x + a^2)}{1-2a\cos x +a^2} dx = \frac{2\pi \ln(1-a^2)}{1-a^2} \quad |a|<1 }\]