Proving $\int_0^\infty \log\left (1-2\frac{\cos 2\theta}{x^2}+\frac{1}{x^4} \right)dx =2\pi \sin \theta$ 
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.
 A: For $\theta \in [0;\pi]$,
\begin{align}
J(\theta)&=\int_0^\infty \ln\left(1-\frac{2\cos(2\theta)}{x^2}+\frac{1}{x^4}\right) \,dx
\end{align}
Perform the change of variable $y=\dfrac{1}{x}$,
\begin{align}
J(\theta)&=\int_0^\infty \frac{\ln\left(1-2\cos(2\theta)x^2+x^4\right)}{x^2} \,dx
\end{align}
For $a\geq -1$, define the function $F$ by,
\begin{align}F(a)&=\int_0^\infty \frac{\ln\left(1+2ax^2+x^4\right)}{x^2} \,dx\\
&=\left[-\frac{\ln\left(1+2ax^2+x^4\right)}{x}\right]_0^\infty+\int_0^\infty \frac{4\left(x^2+a\right)}{1+2ax^2+x^4}\,dx\\
&=\int_0^\infty \frac{4\left(x^2+a\right)}{1+2ax^2+x^4}\,dx\\
\end{align} 
Perform the change of variable $y=\dfrac{1}{x}$, 
\begin{align}F(a)&=\int_0^\infty \frac{4\left( \frac{1}{x^2}+a\right) }{x^2\left(1+\frac{2a}{x^2}+\frac{1}{x^4}\right) } \,dx\\
&=\int_0^\infty \frac{4\left( 1+ax^2\right) }{x^4+2ax^2+1 } \,dx\\
\end{align}
Therefore,
\begin{align}F(a)&=\int_0^\infty \frac{2(a+1)\left( 1+x^2\right) }{x^4+2ax^2+1 } \,dx\\
&=2(a+1)\int_0^\infty \frac{\left(1+\frac{1}{x^2}\right)}{x^2+\frac{1}{x^2}+2a } \,dx\\
&=2(a+1)\int_0^\infty \frac{\left(1+\frac{1}{x^2}\right)}{\left(x-\frac{1}{x}\right)^2+2(a+1) } \,dx\\
\end{align}
Perform the change of variable $y=x-\dfrac{1}{x}$,
\begin{align}F(a)&= 2(a+1)\int_{-\infty}^{+\infty}\frac{1}{x^2+2(a+1)}\,dx\\
&=4(a+1)\int_{0}^{+\infty}\frac{1}{x^2+2(a+1)}\,dx\\
&=\left[2\sqrt{2(a+1)}\arctan\left( \frac{x}{\sqrt{2(a+1)}} \right)\right]_0^\infty\\
&=\boxed{\pi\sqrt{2(1+a)}}
\end{align}
Observe that, $J(\theta)=F\big(-\cos(2\theta)\big)$.
\begin{align} 2(1-\cos(2\theta))&=2(1-\cos^2(\theta)+\sin^2 (\theta))\\
&=2\times 2\sin^2 (\theta)\\
&=4\times \sin^2 (\theta)\\
\end{align} 
Since, for $\theta \in [0;\pi],\sin(\theta)\geq 0$ then $\sqrt{2(1-\cos(2\theta))}=2\sin(\theta)$
Therefore,
\begin{align}\boxed{J(\theta)=2\pi \sin(\theta)}\end{align}
A: To complement Zacky's answer I'll add a proof that $I(0)=0$. Applying $x\mapsto\frac{1}{x}$ for $x\ge 1$ gives $$I(x)=2\int_0^\infty\ln|1-x^{-2}|dx=2\int_0^1\left[(1+\frac{1}{x^2})\ln(1-x^2)-2\ln x\right]\\=-2\int_0^1\left[(1+x^2)\sum_{n\ge 0}\frac{x^{2n}}{n+1}+2\ln x\right]dx=-2\int_0^1\left[\sum_{n\ge 0}\left(\frac{x^{2n}}{n+1}+\frac{x^{2n+2}}{n+1}\right)+2\ln x\right]dx\\=-2\left[\sum_{n\ge 0}\left(\frac{x^{2n+1}}{(n+1)(2n+1)}+\frac{x^{2n+3}}{(n+1)(2n+3)}\right)+2x\ln x-2x\right]_0^1\\=-2\left[\sum_{n\ge 0}\left(\frac{4}{(2n+1)(2n+3)}\right)-2\right],$$which vanishes by partial fractions.
A: Assume that $\theta \in (0, \pi).$
Then
$$\begin{align}\int_{0}^{\infty} \log \left(1 -\frac{2 \cos (2\theta)}{x^{2}} +\frac{1}{x^{4}} \right) \, \mathrm dx &= \int_{0}^{\infty} \frac{ \log \left(1- 2 \cos(2 \theta) u^{2} +u^{4}\right)}{u^{2}} \, \mathrm du \\ &= 2\, \Re \int_{0}^{\infty} \frac{ \log \left( 1-u^{2}e^{2 i \theta} \right)}{u^{2}} \, \mathrm du \\ &= -4 \, \Re \int_{0}^{\infty} \frac{e^{2 i \theta}}{1-u^{2}e^{2i \theta}} \, \mathrm du \tag{1}\\ &= -4 \, \Re \left(  e^{i \theta} \operatorname{artanh} \left(ue^{i \theta} \right) \Big|_{0}^{\infty} \right) \\ &=-4 \, \Re \left(e^{i \theta} \, \frac{i \pi}{2} \right) \tag{2} \\ &= 2 \pi \sin(\theta). \end{align}$$

$(1)$ Integrate by parts.
$(2)$ If $ \theta \in (0, \pi )$, then $$\lim_{u \to +\infty} \operatorname{artanh}(ue^{i \theta})=\lim_{u \to +\infty}\frac{1}{2} \, \log \left( \frac{1 +ue^{i \theta}}{1-ue^{i \theta}} \right)   =  \lim_{u \to +\infty}\frac{1}{2} \, \log \left( \frac{\frac{1}{u} +e^{i \theta}}{\frac{1}{u}-e^{i \theta}} \right)  = \frac{i\pi}{2}$$ because the expression inside the $\log$ is approaching $-1$ from above the branch cut.
A: 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:$$I(0)=0+C\Rightarrow C=0 \Rightarrow \boxed{I(\theta)=2\pi\sin(\theta)}$$
