Dtjytyk

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

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

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

Any hints? Thanks in advance.

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

For $theta in [0;pi]$,
begin{align}
J(theta)&=int_0^infty lnleft(1-frac{2cos(2theta)}{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{lnleft(1-2cos(2theta)x^2+x^4right)}{x^2} ,dx
end{align}

For $ageq -1$, define the function $F$ by,
begin{align}F(a)&=int_0^infty frac{lnleft(1+2ax^2+x^4right)}{x^2} ,dx\
&=left[-frac{lnleft(1+2ax^2+x^4right)}{x}right]_0^infty+int_0^infty frac{4left(x^2+aright)}{1+2ax^2+x^4},dx\
&=int_0^infty frac{4left(x^2+aright)}{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{4left( frac{1}{x^2}+aright) }{x^2left(1+frac{2a}{x^2}+frac{1}{x^4}right) } ,dx\
&=int_0^infty frac{4left( 1+ax^2right) }{x^4+2ax^2+1 } ,dx\
end{align}
Therefore,
begin{align}F(a)&=int_0^infty frac{2(a+1)left( 1+x^2right) }{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[2sqrt{2(a+1)}arctanleft( frac{x}{sqrt{2(a+1)}} right)right]_0^infty\
&=boxed{pisqrt{2(1+a)}}
end{align}

Observe that, $J(theta)=Fbig(-cos(2theta)big)$.

begin{align} 2(1-cos(2theta))&=2(1-cos^2(theta)+sin^2 (theta))\
&=2times 2sin^2 (theta)\
&=4times sin^2 (theta)\
end{align}
Since, for $theta in [0;pi],sin(theta)geq 0$ then $sqrt{2(1-cos(2theta))}=2sin(theta)$

Therefore,
begin{align}boxed{J(theta)=2pi sin(theta)}end{align}

To complement Zacky's answer I'll add a proof that $I(0)=0$. Applying $xmapstofrac{1}{x}$ for $xge 1$ gives $$I(x)=2int_0^inftyln|1-x^{-2}|dx=2int_0^1left[(1+frac{1}{x^2})ln(1-x^2)-2ln xright]\=-2int_0^1left[(1+x^2)sum_{nge 0}frac{x^{2n}}{n+1}+2ln xright]dx=-2int_0^1left[sum_{nge 0}left(frac{x^{2n}}{n+1}+frac{x^{2n+2}}{n+1}right)+2ln xright]dx\=-2left[sum_{nge 0}left(frac{x^{2n+1}}{(n+1)(2n+1)}+frac{x^{2n+3}}{(n+1)(2n+3)}right)+2xln x-2xright]_0^1\=-2left[sum_{nge 0}left(frac{4}{(2n+1)(2n+3)}right)-2right],$$which vanishes by partial fractions.