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

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

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 $2\int_0^\infty \log(2 - 2 \cos(x^2))= \Re(-2 e^{-i\pi / 4} \Gamma(1/2) \zeta(3/2)) = - \sqrt{2\pi} \zeta(3/2)$

We can use the half-angle formula to write $$ -I = \int \limits_{-\infty}^{\infty} -\ln[2(1-\cos(x^2))] \, \mathrm{d} x = 2 \int \limits_0^\infty -\ln\left[4\sin^2\left(\frac{x^2}{2}\right)\right] \mathrm{d}x \stackrel{x^2 = t}{=} 2 \int \limits_0^\infty \frac{-\ln\left(2 \left\lvert\sin \left(\frac{t}{2}\right)\right\rvert\right)}{\sqrt{t}} \, \mathrm{d} t \, .$$ Then we integrate by parts using the Clausen function and plug in its series expansion: \begin{align} -I &= \int \limits_0^\infty \frac{\operatorname{Cl}_2(t)}{t^{3/2}} \, \mathrm{d} t = \sum \limits_{n=1}^\infty \frac{1}{n^2} \int \limits_0^\infty \frac{\sin(nt)}{t^{3/2}} \, \mathrm{d} t \stackrel{nt = s}{=} \sum \limits_{n=1}^\infty \frac{1}{n^{3/2}} \int \limits_0^\infty \frac{\sin(s)}{s^{3/2}} \, \mathrm{d} s \\ &= \operatorname{\zeta} \left(\frac{3}{2}\right) \operatorname{\mathcal{M}[\sin]}\left(-\frac{1}{2}\right) = \operatorname{\zeta} \left(\frac{3}{2}\right) \sin \left(-\frac{\pi}{4}\right) \operatorname{\Gamma} \left(-\frac{1}{2}\right) = \sqrt{2 \pi} \operatorname{\zeta} \left(\frac{3}{2}\right) \, . \end{align} The Mellin transform of the sine is discussed in this question.