The Log-Sine Affair

Since $B(\alpha,\beta) \; = \; 2\int_0^{\frac12\pi} \sin^{2\alpha-1}\theta\,\cos^{2\beta-1}\theta\,d\theta$ we deduce that $\begin{aligned} \int_0^{\frac12\pi} (\ln\,\sin \theta)^2\,d\theta & = \; \tfrac18\frac{\partial^2 B}{\partial \alpha^2}(\alpha,\beta) \Big|_{\alpha = \beta = \tfrac12} \\ & = \; \tfrac18B(\alpha,\beta)\Big[(\psi'(\alpha) - \psi'(\alpha+\beta)) + (\psi(\alpha) - \psi(\alpha+\beta))^2\Big] \Big|_{\alpha = \beta = \tfrac12} \\ & = \; \tfrac18B(\tfrac12,\tfrac12)\big[(\psi'(\tfrac12) - \psi'(1)) + (\psi(\tfrac12) - \psi'(1))^2\big] \; = \; \tfrac18\Gamma(\tfrac12)^2\big[ \tfrac12\pi^2 - \tfrac16\pi^2 + (-\gamma - 2\ln2 + \gamma)^2\big] \\ & = \; \tfrac18\pi\big[\tfrac13\pi^2 + 4\ln^22\big] \; = \; \tfrac{1}{24}\pi^3 + \tfrac12\pi \ln^22 \; = \; \tfrac{1}{24}\pi^3 + \pi \ln^2\big(\sqrt{2}^{\sqrt{2}}\big) \end{aligned}$ making the answer $\sqrt{\frac{24 \times 3}{2}} = \boxed{6}$ .

We consider the beta integral of the form $B(x,y)=\int^{\frac{\pi}{2}}_02\sin^{2x-1}(t)\cos^{2y-1}(t)\ dt.$ Differentiating two times with respect to $x,$ we have $\begin{aligned} \frac{\partial^2}{\partial x^2}B(x,y)&=&\int^{\frac{\pi}{2}}_08(\ln\sin x)^2\sin^{2x-1}(t)\cos^{2y-1}(t)\ dt\\ &=&B(x,y)\left(\left(\psi(x)-\psi(x+y)\right)^2+\psi_1(x)-\psi_1(x+y)\right)\\ \end{aligned}$ where $\psi(x)$ is digamma function and $\psi_1(x):=\frac{d}{dx}\psi(x)$ is trigamma function. By substituting $x=y=\frac{1}{2},$ and applying the facts that $\psi(\frac{1}{2})=-\gamma-\ln4, \psi(1)=\gamma, \psi_1\left(\frac{1}{2}\right)=\frac{\pi^2}{2}$ and $\psi_1(1)=\frac{\pi^2}{6},$ the result follows, i.e. $\int^{\frac{\pi}{2}}_0(\ln\sin x)^2\ dx=\pi\ln^2\left(\sqrt{2}^{\sqrt{2}}\right)+\frac{\pi^3}{24}.$