Proving $\int_0^\infty \log\left (1-2\frac{\cos 2\theta}{x^2}+\frac{1}{x^4} \right)dx =2\pi \sin \theta$

Question

Prove $$\int_0^\infty \log \left(1-2\frac{\cos 2\theta}{x^2}+\frac{1}{x^4} \right)dx =2\pi \sin \theta$$where $\theta\in[0,\pi]$.

I've met another similar problem,
$$ \int_0^{2\pi} \log(1-2r\cos \theta +r^2) d\theta=2\pi \log^+(r^2) $$
I am curious whether there is any relationship between them.

And I got stuck on the proposition in the title. I found that
$$1-2\frac{\cos 2\theta}{x^2}+\frac{1}{x^4} =\left(\frac{1}{x}-e^{i\theta}\right)\left(\frac{1}{x}+e^{i\theta}\right)\left(\frac{1}{x}-e^{-i\theta}\right)\left(\frac{1}{x}+e^{-i\theta}\right)$$
But I couldn't move on.

Any hints? Thanks in advance.

Answer 1

We start off by some $x\rightarrow \frac{1}{x}$ substitutions while derivating under the integral sign: $$I(\theta)=\int_0^\infty \log \left(1-2\frac{\cos 2\theta}{x^2}+\frac{1}{x^4} \right)dx\overset{x\rightarrow \frac{1}{x}}=\int_0^\infty \frac{\ln(1- 2\cos(2\theta) x^2 +x^4)}{x^2}dx$$ $$I'(\theta)=4\int_0^\infty \frac{\sin(2\theta)}{x^4-2\cos(2\theta)x^2+1}dx\overset{x\rightarrow \frac{1}{x}}=4\int_0^\infty \frac{\sin(2\theta)x^2}{x^4-2\cos(2\theta)x^2+1}dx$$ Now summing up the two integrals from above gives us: $$\Rightarrow 2I'(\theta)=4\int_0^\infty \frac{\sin(2\theta)(1+x^2)}{x^4-2\cos(2\theta)x^2+1}dx=4\int_0^\infty \frac{\sin(2\theta)\left(\frac{1}{x^2}+1\right)}{x^2+\frac{1}{x^2}-2\cos(2\theta)}dx$$ $$\Rightarrow I'(\theta)=2\int_0^\infty \frac{\sin(2\theta)\left(x-\frac{1}{x}\right)'}{\left(x-\frac{1}{x}\right)^2 +2(1-\cos(2\theta))}dx\overset{\large x- \frac{1}{x}=t}=2\int_{-\infty}^\infty \frac{\sin(2\theta)}{t^2 +4\sin^2 (\theta)}dt$$ $$=2 \frac{\sin(2\theta)}{2\sin(\theta)}\arctan\left(\frac{t}{2\sin(\theta)}\right)\bigg|_{-\infty}^\infty=2\cos(\theta) \cdot \pi$$ $$\Rightarrow I(\theta) = 2\pi \int \cos(\theta) d\theta =2\pi \sin \theta +C$$ But $I(0)=0$ (see J.G. answer), thus:$$I(0)=0+C\Rightarrow C=0 \Rightarrow \boxed{I(\theta)=2\pi\sin(\theta)}$$