evaluate $\int_0^{\pi/2} x^2\log(\sin x)\,dx$

Question

I am a high school student , I know how to evaluate $\int_0^{\pi/2} x\log(\sin x)\,dx$. It would be great if someone can help me evaluating $\int_0^{\pi/2} x^2\log(\sin x)\,dx$ and tell me if this integral is elementary or non elementary . I tried using the "a-x" property but it resulted in $0=0$

Answer 1

Using this, we have $$ \log(\sin(x))=-\log 2-\sum_{k=1}^{+\infty}\frac{\cos(2kx)}{k} $$ Therefore, $$ \int_0^{\pi/2}x^2\log(\sin x)dx=-\frac{\pi^3}{24}\log 2-\sum_{k=1}^{+\infty}\frac{1}{k}\int_0^{\pi/2}x^2\cos(2kx)dx $$ Using integration by parts, we have $$ \int_0^{\pi/2}x^2\cos(2kx)dx = \frac{(-1)^k\pi}{4k^2} $$ Thus, $$ \sum_{k=1}^{+\infty}\frac{1}{k}\int_0^{\pi/2}x^2\cos(2kx)dx=\frac{\pi}{4}\sum_{k=1}^{+\infty}\frac{(-1)^k}{k^3}=-\frac{\pi}{4}\eta(3)=-\frac{3\pi}{16}\zeta(3) $$ where $\eta$ is the Dirichlet eta function. Finally, $$ \int_0^{\pi/2}x^2\log(\sin x)dx=\frac{3\pi}{16}\zeta(3)-\frac{\pi^3}{24}\log 2 $$

Answer 2

Let

$$S=\int_0^{\pi/2} x^2\log(2\sin x)\,dx,\>\>\>\>\> C=\int_0^{\pi/2} x^2\log(2\cos x)\,dx$$ Since you already knew $\int_0^{\pi/2} x\log(2\cos x)\,dx=-\frac7{16}\zeta(3)$, apply the variable change $x\to\frac\pi2-x$ to the $S$-integral to get $$S - C =-\pi\int_0^{\pi/2} x\log(2\cos x)\,dx=\frac{7\pi}{16}\zeta(3)\tag1 $$ Also \begin{align} S+C & = \int_0^{\pi/2} x^2\log(2\sin 2x)\,dx \overset{2x\to x}=\frac18\int_0^{\pi} x^2\log(2\sin x)\,dx\\ &=\frac18S + \frac18\int_{\pi/2}^{\pi} x^2\log(2\sin x)\overset{x\to\frac\pi2+x}{ dx}\\ &= \frac18(S+C) + \frac\pi8 \int_0^{\pi/2} x\log(2\cos x)\,dx = -\frac\pi{16}\zeta(3)\tag2 \end{align} Combine (1) and (2) to obtain $S= \frac{3\pi}{16}\zeta(3)$, or

$$\int_0^{\pi/2} x^2\log(\sin x)\,dx= \frac{3\pi}{16}\zeta(3)-\frac{\pi^3}{24}\ln2$$

Answer 3

Another possibility is to write $$\log[\sin(x)]=\sum_{n=1}^\infty (-1)^n \frac{2^{2 n-3}\,\, (E_{2 n-1}(1)-E_{2 n-1}(0))}{n \,(2 n-1)!}\left(x-\frac{\pi }{2}\right)^{2 n}$$ where appear the Euler polynomial and use $$\int_0^{\frac \pi 2}x^2\left(x-\frac{\pi }{2}\right)^{2 n}\,dx=\frac{\left(\frac{\pi }2\right)^{2 n+3}}{(n+1) (2 n+1) (2 n+3)}$$ This gives

$$I_2=\frac 1{16}\sum_{n=1}^\infty (-1)^n\frac{\pi ^{2 n+3} (E_{2 n-1}(1)-E_{2 n-1}(0))}{ (2 n+3)!}$$

which, after a series of tedious manipulations, leads to the result already given by @Tuvasbien.

Numerically, this summation does not converge quite fast. Considering the partial (from $n=1$ to $n=p$) $$\left( \begin{array}{cc} p & \Sigma_p \\ 1 & -0.159385 \\ 2 & -0.178112 \\ 3 & -0.183246 \\ 4 & -0.185204 \\ 5 & -0.186108 \\ 6 & -0.186581 \\ 7 & -0.186852 \\ 8 & -0.187019 \\ 9 & -0.187127 \\ 10 & -0.187200 \\ \cdots & \cdots \\ \infty & -0.187426 \end{array} \right)$$

You could easily generalized the result for $$I_p=\int_0^{\frac \pi 2}x^p\log[\sin(x)]\,dx$$ since $$\int_0^{\frac \pi 2}x^p\left(x-\frac{\pi }{2}\right)^{2 n}\,dx=\left(\frac{\pi }{2}\right)^{2 n+p+1}\frac{(2 n)!\,\,p!}{(2 n+p+1)!}$$

Edit

Numerically, if you look at this question of mine, we could have very good approximations writing $$I_2\sim\frac 1{16}\sum_{n=1}^p (-1)^n\frac{\pi ^{2 n+3} (E_{2 n-1}(1)-E_{2 n-1}(0))}{ (2 n+3)!}-\frac {\pi^3}8\sum_{n=p+1}^\infty \frac{1}{n (n+1) (2 n+1) (2 n+3)}$$ and $$\sum_{n=p+1}^\infty \frac{1}{n (n+1) (2 n+1) (2 n+3)}=\frac{1}{3} \left(H_{p+\frac{3}{2}}-H_p\right)-\left(H_{p+1}-H_{p+\frac{1}{2}}\right)$$

Using $p=5$ gives an absolute error equal to $9.54\times 10^{-10}$.