$\int_{0}^{\pi/2}\ln^3(\sin x)\,\mathrm dx$

Question

There are closed forms for $\int_{0}^{\pi/2}\ln(\sin x)\,\mathrm dx\,$ and $\,\int_{0}^{\pi/2}\ln^2(\sin x)\,\mathrm dx\,$ but I can’t seem to find a closed form for $$\int_{0}^{\pi/2}\ln^3(\sin x)\,\mathrm dx\;.$$

How would I calculate it?

Answer 1

Noting that $$ \int_0^{\frac{\pi}{2}} \ln ^3(\sin x) d x=\left.\frac{\partial^3}{\partial a^3} I(a)\right|_{a=0} $$ where $$ I(a)=\int_0^{\frac{\pi}{2}} \sin ^a x d x=\frac{1}{2}B\left (\frac{a+1}{2},\frac{1}{2} \right)$$

\begin{aligned} &\quad \frac{\partial^3}{\partial x^3}(\mathrm{~B}(x, y)) \\&=\left[\left(\psi^{(0)}(x)-\psi^{(0)}(x+y)\right)^3+3\left(\psi^{(1)}(x)-\psi^{(1)}(x+y)\right) \left(\psi^{(0)}(x)-\psi^{(0)}(x+y)\right)\\ +\psi^{(2)}(x)-\psi^{(2)}(x+y)\right] \mathrm{B}(x, y) \end{aligned}

Putting $x=\frac{a+1}{2} $ and $y=\frac{1}{2} $ at $a=0$ in the derivative yields

\begin{aligned} & \quad \frac{\partial^3}{\partial a^3} B\left(\frac{a+1}{2}, \frac{1}{2}\right)\\&=\frac{\pi}{8}[(\left.\psi\left(\frac{1}{2}\right)-\psi(1)\right)^3+3\left(\psi^{\prime}\left(\frac{1}{2}\right)-\psi^{\prime}(1)\right)\left(\psi\left(\frac{1}{2}\right)-\psi(1)\right) \\ &\left. \quad +\psi^{(2)}\left(\frac{1}{2}\right)-\psi^2(1)\right] \\ &=\frac{\pi}{8}\left[(-\gamma-\ln 4+\gamma)^3+3\left(\frac{\pi^2}{2}-\frac{\pi^2}{6}\right)(-\gamma-\ln \psi+\gamma)\right. +(-14 \zeta(3)+2 \zeta(3))] \\ &= \frac{\pi}{8}\left(-\ln ^3 4-\pi^2 \ln 4-12 \zeta(3)\right) \end{aligned}

Hence $$\boxed{I=-\frac{\pi}{16}\left(\ln ^3 4+\pi^2 \ln 4+12 \zeta(3) \right)}$$

Answer 2

Mathematica says: $$ -\frac{1}{16}\pi\left(12\zeta(3)+\log^3(4)+\pi^2\log(4)\right)$$