Integral $\int_{-\infty}^{\infty}\ln(2-2\cos(x^2))dx=-\sqrt{2\pi}\zeta(3/2)$

Question

Prove that
$$\int_{-\infty}^{\infty}\ln(2-2\cos(x^2))dx=-\sqrt{2\pi}\zeta(3/2)$$

I was given this integral in my post Request for crazy integrals. I have never seen an integral like this before and I need help evaluating it.

Here's what I've tried.

Setting $$J=\int_{-\infty}^{\infty}\ln(2-2\cos(x^2))dx$$
We have that
$$I=-\frac{J}{2\ln2}=\sum_{n\geq1}\frac1{n}\int_0^\infty \cos^n(x^2)dx$$
from the series
$$-\ln(1-x)=\sum_{n\geq1}\frac{x^n}{n}$$
Then set $$p_n=\int_0^{\infty}\cos^n(x^2)dx$$
so that
$$I=p_1+\sum_{k\geq1}\frac{p_{2k+1}}{2k+1}+\frac12\sum_{k\geq1}\frac{p_{2k}}{k}$$
We have from Wikipedia that if $n$ is odd then
$$\cos^n x=2^{1-n}\sum_{k=0}^{(n-1)/2}{n\choose k}\cos[(n-2k)x]$$
And for even $n$,
$$\cos^n x=\frac1{2^n}{n\choose n/2}+2^{1-n}\sum_{k=0}^{n/2-1}{n\choose k}\cos[(n-2k)x]$$
So
$$p_{2k+1}=\frac1{4^k}\sum_{\ell=0}^{k} {2k+1\choose \ell}\int_0^\infty \cos[(2k-2\ell+1)x^2]dx$$
Then wolfram provides
$$\int_0^\infty \cos(ax^2)dx=\frac{\sqrt{\pi}}{2\sqrt{2|a|}}$$
Which I know how to prove. Anyway,
$$p_{2k+1}=\frac{\sqrt\pi}{2^{2k+3/2}}\sum_{\ell=0}^{k}\frac{{2k+1\choose \ell}}{\sqrt{2k-2\ell+1}}$$
Thus $$I=\frac12\sqrt{\frac\pi2}\left[1+\sum_{k\geq1}\frac1{4^k(2k+1)}\sum_{\ell=0}^{k}\frac{{2k+1\choose \ell}}{\sqrt{2k-2\ell+1}}\right]+\frac12\sum_{k\geq1}\frac{p_{2k}}{k}$$
The evaluation of $p_{2k}$ may be significantly more difficult from potential convergence issues. In any case, this doesn't seem to getting me anywhere close to $\zeta(3/2)$, so I would like to see how it's done. Thanks 🙂

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Edit:

Okay, from @ComplexYetTrivial's comment, we have that essentially everything I have done so far (apart from exploiting symmetry) is wrong. So yeah, I'm happy someone caught that.

Answer 1

$2\int_0^\infty \log(2 - 2 \cos(x^2))=2\int_0^\infty \log(2 - e^{iy}-e^{-iy})dy^{1/2}$ $=\lim_{a \to 0} 2\int_0^\infty \Re(\log( 1-e^{-( a-i)y}))y^{-1/2}dy$

Then use the Taylor expansion of $\log(1-z)$ and that $\lim_{a \to 0}\int_0^\infty e^{-(a-i)n y} y^{-1/2}dy =n^{-1/2} e^{-i\pi / 4} \Gamma(1/2)$ obtaining $2\int_0^\infty \log(2 - 2 \cos(x^2))= \Re(-2 e^{-i\pi / 4} \Gamma(1/2) \zeta(3/2)) = - \sqrt{2\pi} \zeta(3/2)$