How to evaluate this integration $\int_0^{\frac{\pi}{2}} \frac{x^2\ln(2\cos x)}{(x^2+\ln^2(2\cos x))^2}dx$?

Question

Here is my question: $$\int_0^{\frac{\pi}{2}} \frac{x^2\ln(2\cos x)}{(x^2+\ln^2(2\cos x))^2}dx=\frac{7\pi}{192}-\frac{\gamma\pi}{96}+\frac{\pi}{8}\ln A$$(The A is Glaisher-Kinkelin constant) $$$$My Attempt: $$\ln(2\cos x)=\ln(1+e^{2ix})-ix$$ we have $$\frac{x^2\ln(2\cos x)}{(x^2+\ln^2(2\cos x))^2}=\Im\left(\frac{1}{\ln^2(1+e^{2ix})}-e^{-2ix}-e^{-4ix}\right)$$ Then I don't know how to proceed, I tried to use complex integrals but failed, could anybody give me some hints? Thanks!

Answer 1

Here's an approach by Amdeberhan and Moll found here. For $0<a< \ln(2)$ we have $$ \frac{4}{\pi}\int_{0}^{\frac{\pi}{2}}\frac{x^2}{x^2 + \ln^2\left(2e^{-a}\cos(x)\right)}\mathrm{d}x = \frac{\gamma}{a} + \frac{a+\ln(1-e^{-a})-\gamma -\ln(a)}{1-e^{-a}} + \frac{a}{1-e^{-a}}\int_{0}^{1}e^{-at}\ln\left(\Gamma(t)\right)\,\mathrm{d}t $$ Differentiating with respect to $a$ gives $$ \frac{8}{\pi}\int_{0}^{\frac{\pi}{2}}\frac{x^2\ln\left(2e^{-a}\cos(x)\right)}{\left(x^2 + \ln^2\left(2e^{-a}\cos(x)\right)\right)^2}\mathrm{d}x =- \frac{\gamma }{a^2}+ \frac{e^a (1 + (a-1) e^a)}{a (e^a-1)^2} + \frac{e^a \left(-a + \gamma + \ln(a) - \ln\left(1 - e^{-a}\right)\right)}{(e^a-1)^2} - \frac{a}{1-e^{-a}}\int_{0}^{1}e^{-at}t\ln\left(\Gamma(t)\right)\,\mathrm{d}t + \frac{e^a (-1 - a + e^a)}{(e^a-1)^2} \int_{0}^{1}e^{-at}\ln\left(\Gamma(t)\right)\,\mathrm{d}t $$ Taking $a \to 0^+$ thus results in $$ \int_{0}^{\frac{\pi}{2}}\frac{x^2\ln\left(2\cos(x)\right)}{\left(x^2 + \ln^2\left(2\cos(x)\right)\right)^2}\mathrm{d}x =\frac{7\pi}{192} - \frac{\pi \gamma}{96} -\frac{\pi}{8}\int_{0}^{1}t\ln\left(\Gamma(t)\right)\,\mathrm{d}t +\frac{\pi}{16} \int_{0}^{1}\ln\left(\Gamma(t)\right)\,\mathrm{d}t $$ Recalling that $\int_{0}^{1}\ln\Gamma(t)\mathrm{d}t =\frac{\ln(2\pi)}{2}$ and $\int_{0}^{1}t\ln\Gamma(t)\mathrm{d}t =\frac{\ln(2\pi)}{4}-\ln(A)$ gives the desired result.