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

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 :)

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.

• Why is $-J/(2\log 2)$ being the sum you proposed? It looks suspiciously like you subtracted an $\int \log 2\,\mathrm{d}x=\infty$ instead... – user10354138 May 15 at 3:50
• because I didn't want to keep writing $$J=-2\ln 2\sum ...$$ (I'm lazy :)) – clathratus May 15 at 3:50
• It seems that you have used $\ln(2(1-\cos(x^2))) = \ln(2) \ln(1-\cos(x^2))$ instead of $\ln(2(1-\cos(x^2))) = \ln(2) + \ln(1-\cos(x^2))$. This is why some of the resulting integrals, i.e. $\int_0^\infty \cos^{2k}(x^2) \, \mathrm{d} x$, are divergent. – ComplexYetTrivial May 15 at 10:42

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.
$$2\int_0^\infty \log(2 - 2 \cos(x^2))=2\int_0^\infty \log(2 - e^{iy}-e^{-iy})dy^{1/2}$$ $$=\lim_{a \to 0} 2\int_0^\infty \Re(\log( 1-e^{-( a-i)y}))y^{-1/2}dy$$
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 $$2\int_0^\infty \log(2 - 2 \cos(x^2))= \Re(-2 e^{-i\pi / 4} \Gamma(1/2) \zeta(3/2)) = - \sqrt{2\pi} \zeta(3/2)$$