Evaluate integral $\int_0^\frac{\pi}{2}x\ln(\sin x)~dx$

Question

I've corrected typing error in the integral. I apologize for my mistake.

Reedited question:

Can anybody solve integral: $$\int_0^\frac{\pi}{2}x\ln(\sin x)~dx$$ I'm just trying to guess some simple formula for $\zeta(3)$. My "strategy" is simple: Find some conjectures and check them numerically.

Value of similar integral is well known: $$\int_0^\pi x\ln(\sin x)~dx=-\dfrac{\pi^2}{2}\ln(2) $$ Is there any idea for antiderivative of $x\ln(\sin x)$? It's too difficult to solve it for me. Any ideas?

Answer 1

The integral from $0$ to $\pi/2$ of $\ln(\sin x)$ is one of Euler's integrals - look here for some information. This paper explains why $$\int_{0}^{\pi/2}x\ln(\sin x)\ dx=\frac7{16}\zeta(3)-\frac{\pi^2}8\ln2. $$

Answer 2

$$I=\int^{\frac{\pi}{2}}_0 x \log|\sin(x )| \, dx =\int^{\frac{\pi}{2}}_0 x \log|2\sin(x )| -\frac{\pi^2}{8}\log(2)$$

We relate the integral to the Clausen function \begin{align} \int^{\frac{\pi}{2}}_0 x \log|2\sin(x )| dx &=\frac{1}{2} \int^{\frac{\pi}{2}}_0 \mathrm{Cl}_2(2\theta)\, d\theta\\ &=\frac{1}{2} \int^{\frac{\pi}{2}}_0 \sum_{n=1}^{\infty}\frac{\sin(2n\theta)}{n^2}\, d\theta \\ &=-\frac{1}{4}\sum_{n=1}\frac{(-1)^n}{n^3}+\frac{1}{4}\sum_{n=1}\frac{1}{n^3}\\ &=\frac{7}{16}\zeta(3) \end{align}

Collecting that together we have

$$I=\frac{7}{16}\zeta(3)-\frac{\pi^2}{8}\log(2)$$

I considered a more general case in this thread

Answer 3

Note

\begin{align}\int_0^\frac{\pi}{2}x\ln(2\sin x)~dx =\frac12 \int_0^\frac{\pi}{2}x\ln(2\sin 2x)~dx + \frac12 \int_0^\frac{\pi}{2}x\ln\frac{\sin x}{\cos x}~dx\tag1\\ \end{align} where \begin{align} &\int_0^\frac{\pi}{2} \overset{2x\to x} {x\ln(2\sin 2x)}~dx =\frac14 \int_0^{\pi} \overset{x\to\pi-x} {x\ln(2\sin x)}~{dx} =\frac\pi8 \int_0^{\pi}\ln(2\sin x)~dx =0 \end{align} and \begin{align} &\int_0^\frac{\pi}{2}x\ln\frac{\sin x}{\cos x} ~dx \overset{t=\tan x}= \int_0^\infty \frac{\ln t\tan^{-1}t}{1+t^2} dt \\ =& \int_0^\infty \frac{\ln t}{1+t^2} \int_0^1 \frac t{1+ t^2 y^2}dy ~dt \overset{t^2\to t}=\frac14\int_0^1 \int_0^\infty \frac{\ln t}{(1+t)(1+ t y^2)} \overset{t\to 1/(y^2t)}{dt ~dy}\\ =& \frac14\int_0^1 \int_0^\infty \frac{-\ln t -\ln y^2}{(1+t)(1+ t y^2)}dt ~dy = -\frac14\int_0^1 \int_0^\infty \frac{\ln y}{(1+t)(1+ t y^2)}dt ~dy\\ =& \frac12\int_0^1 \frac{\ln^2 y}{1-y^2}dy = \frac12\int_0^1 \frac{\ln^2 y}{1-y}dy - \frac12\int_0^1 \frac{y \ln^2 y}{1-y^2}\overset{y^2\to y}{dy}\\ =& \frac7{16}\int_0^1 \frac{\ln^2 y}{1-y}dy = \frac7{16}\cdot 2\zeta(3)=\frac 78\zeta(3) \end{align} Substitute above results into (1) to get $\int_0^\frac{\pi}{2}x\ln(2\sin x)~dx= \frac7{16}\zeta(3)$, which leads to $$ \int_0^\frac{\pi}{2}x\ln(\sin x)~dx= \frac7{16}\zeta(3)-\frac{\pi^2}8\ln2$$