Computing $\int^{\pi}_{0}x^2\ln(\sin x)\mathrm dx$

Question

Find $\displaystyle \int^{\pi}_{0}x^2\ln(\sin x)\mathrm dx$.

My attempt:

I try

Assume $\displaystyle I =\int^{\pi}_{0}x^2\ln(\sin x)\mathrm dx=\int^{\pi}_{0}x^2\ln\left(2\sin\frac{x}2\cos \frac{x}2\right)\mathrm dx$

$\implies\displaystyle I=\int^{\pi}_{0}x^2\ln(\sin \frac{x}{2})\mathrm dx+\int^{\pi}_{0}x^2\ln(\cos \frac{x}{2})\mathrm dx+\int^{\pi}_{0}x^2\ln(2)\mathrm dx$

$\implies\displaystyle I=8\int^{\frac{\pi}{2}}_{0}t^2\ln\sin(t)\mathrm dt+8\int^{\frac{\pi}{2}}_{0}t^2\ln\cos(t)\mathrm dt+2\int^{\frac{\pi}{2}}_{0}\ln(2)\mathrm dt$

I do not understand how can I solve it.

Answer 1

Hint Begin by $$ \displaystyle \int^{v}_{u}x^2\ln(\sin x)dx=\displaystyle \int^{v}_{u}\Big(\frac{x^3}{3}\Big)'\ln(\sin x)dx $$ and use integration by parts.

Of course, you will have to prove the convergence when $u\to 0$ and $v\to \pi$.

You can also use the Fourier series for $0<x<\pi$ $$ -\log(\sin(x))=\sum_{k=1}^\infty\frac{\cos(2kx)}{k}+\log(2) $$ If needed, look at the very elegant derivation of it here by user17762.
And then use, as was indicated, the technique of Jacky Chong

For your information, computer algebra gives (same as Dr. Sonnhard Graubner) $$ -1/6\pi^3\log(4) - 1/2\pi\zeta(3)\ . $$

Answer 2

Note that
$$I=\int^{\pi}_{0}x^2\ln(2\sin x)dx = \int_0^{\pi/2} (\pi^2-2\pi x +2x^2)\ln(2\sin x)dx$$ where $\int^{\pi}_{0}\ln(2\sin x)dx=0$ $$\int^{\pi/2}_{0}x\ln(2\sin x)dx=\frac7{16}\zeta(3) $$ $$\int^{\pi/2}_{0}x^2\ln(2\sin x)dx=\frac{3\pi}{16}\zeta(3) $$ which lead to $I=-\frac\pi2 \zeta(3)$ and $$\int^{\pi}_{0}x^2\ln(\sin x)dx= I-\int_0^\pi x^2\ln2\ dx = -\frac\pi2 \zeta(3)-\frac{\pi^3}3\ln2$$