How to evaluate $\int _0^{\frac{\pi }{2}}x\ln \left(\sin \left(x\right)\right)\:dx$ How can i evaluate $$\int _0^{\frac{\pi }{2}}x\ln \left(\sin \left(x\right)\right)\:dx$$
I started like this
$$\int _0^{\frac{\pi }{2}}x\ln \left(\sin \left(x\right)\right)\:dx=\frac{x^2\ln \left(\sin \left(x\right)\right)}{2}|^{\frac{\pi }{2}}_0-\frac{1}{2}\int _0^{\frac{\pi }{2}}x^2\cot \left(x\right)\:dx$$
but this way doesnt turn things any simpler, i also tried using the substitution $t=\tan \left(\frac{x}{2}\right)$ and got this,
$$4\int _0^{1}\arctan \left(t\right)\ln \left(\frac{2t}{1+t^2}\right)\:\frac{1}{1+t^2}\:dt$$
$$=4\ln \left(2\right)\int _0^{1}\frac{\arctan \left(t\right)}{1+t^2}\:dt+4\int _0^{1}\frac{\arctan \left(t\right)\ln \left(t\right)}{1+t^2}\:dt-4\int _0^{1}\frac{\arctan \left(t\right)\ln \left(1+t^2\right)}{1+t^2}\:dt$$
That first integral is very simple but the rest look very difficult, could you help me evaluate this one?
 A: $$\int_0^{\pi/2}x\ln(\sin x)dx=\int_0^{\pi/2}x\left(-\ln2-\sum_{n=1}^\infty\frac{\cos(2nx)}{n}\right)dx$$
$$=-\frac{\pi^2}{8}\ln2-\sum_{n=1}^\infty\frac{1}{n}\int_0^{\pi/2}x\cos(2nx)dx$$
$$=-\frac{\pi^2}{8}\ln2-\sum_{n=1}^\infty\frac{1}{n}\left(\frac{\cos(n\pi)}{4n^2}+\frac{\pi\sin(n\pi)}{4n}-\frac{1}{4n^2}\right)$$
$$=-\frac{\pi^2}{8}\ln2-\sum_{n=1}^\infty\frac{1}{n}\left(\frac{(-1)^n}{4n^2}+\frac{0}{4n}-\frac{1}{4n^2}\right)$$
$$=-\frac{\pi^2}{8}\ln2-\frac14\text{Li}_3(-1)+\frac14\zeta(3)$$
$$=-\frac{\pi^2}{8}\ln2+\frac{7}{16}\zeta(3)$$

Bonus: With subbing $x\to \pi/2-x$ we have
$$\int_0^{\pi/2}x\ln(\cos x)dx=\int_0^{\pi/2}(\pi/2-x)\ln(\sin x)dx$$
$$=\frac{\pi}{2}\int_0^{\pi/2}\ln(\sin x)dx-\int_0^{\pi/2}x\ln(\sin x)dx$$
$$=\frac{\pi}{2}\left(-\frac{\pi}{2}\ln2\right)-\left(-\frac{\pi^2}{8}\ln2+\frac{7}{16}\zeta(3)\right)$$
$$=-\frac{\pi^2}{8}\ln(2)-\frac7{16}\zeta(3)$$
Or we can use the Fourier series of $\ \ln(\cos x)=-\ln2-\sum_{n=1}^\infty\frac{(-1)^n\cos(2nx)}{n}$.
Also by subtracting the two integrals gives
$$\int_0^{\pi/2}x\ln(\tan x)dx=\frac78\zeta(3)$$
Or we can use the Fourier series of $\ \ln(\tan x)=-2\sum_{n=1}^\infty\frac{\cos((4n-2)x)}{2n-1}.$
A: An incomplete solution requiring some more interesting work:
$$I=\int_{0}^{\pi/2} x \ln (\sin x) dx= \int_{0}^{1} \ln t ~\frac{\sin^{-1} t}{\sqrt{1-t^2}} dt.$$
Using the MacLaurin series for $$\frac{\sin^{-1} t}{\sqrt{1-t^2}}= \sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!} t^{2n+1}$$
See:
Deriving Maclaurin series for $\frac{\arcsin x}{\sqrt{1-x^2}}$.
Then
$$I=\sum_{n=0}^{\infty} \int_{0}^{1}\ln t ~\frac{(2n)!!}{(2n+1)!!} t^{2n+1}=-\sum_{n=0}^{\infty}  \frac{(2n)!!}{(2n+1)!!}~ \int_{0}^{\infty}u~e^{-(2n+2)u}~du~~( t=e^{-u})$$
$$\implies I=-\sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!} \frac{1}{(2n+2)^2}$$
I have numerically confirmed using Mathematica that $I$ is nothing but
$$\frac{1}{16}[-\pi^2 \ln 4+7 \zeta(3)]$$
The same as obrained @Ali Shather in his very nice solution above.
Can some one fill the gap here! I may come back.
A: Assuming that you could enjoy polylogarithms, the antiderivative does exist (have a look here)
$$I=\int x\log \left(\sin \left(x\right)\right)\:dx$$ Using the bounds, the results are

*

*at $\frac \pi 2$, $\frac{1}{48} \left(9 \zeta (3)+i \pi ^3-6 \pi ^2 \log (2)\right)$

*at $0$, $\frac{1}{48} \left(i \pi ^3-12 \zeta (3) \right)$
and, then, the result.
A: This solution is based on Cauchy's integral theorem.
Integrate
\begin{equation*}
 f(z) = \log(z)\dfrac{\log(1-z^2)}{z}
\end{equation*}
where
\begin{equation*}
 \log(z)=\ln|z|+i\arg(z), \qquad -\pi<\arg(z)<\pi,
\end{equation*}
over the boundary $\gamma$ of the unit circle in the first quadrant. Let $\gamma = \gamma_1+\gamma_2+\gamma_3$. Here
\begin{alignat*}{1}
\gamma_1(x)&=x,\, 0\le x\le 1\\
\gamma_2(t)&=e^{it}, \, 0 \le t \le {\pi}/2\\
\gamma_3(y)&=iy,\, y \mbox{ from } 1 \mbox{ to } 0.
\end{alignat*}
From Cauchy's integral theorem we get
\begin{gather*}
 0 =  \int_{\gamma_1}f(z)\,\mathrm{d}z +  \int_{\gamma_2}f(z)\,\mathrm{d}z + \int_{\gamma_3}f(z)\,\mathrm{d}z=\\[2ex]
 \int_{0}^{1}\ln(x)\dfrac{\log(1-x^2)}{x} \,\mathrm{d}x+
 \int_{0}^{\pi/2}\log(e^{it})\dfrac{\log(1-e^{i2t})}{e^{it}}ie^{it} \,\mathrm{d}t-
 \int_{0}^{1}\log(iy)\dfrac{\log(1+y^2)}{iy}i\, \mathrm{d}y=\\[2ex]
  \int_{0}^{1}\ln(x)\dfrac{\log(1-x^2)}{x} \,\mathrm{d}x+
  \int_{0}^{\pi/2}i^2t\left(\ln(2\sin(t))+i\arg\left(1-e^{i2t}\right)\right)\,\mathrm{d}t-\\[2ex]
   \int_{0}^{1}\left(\ln(y)+i\dfrac{\pi}{2}\right)\dfrac{\log(1+y^2)}{y}\, \mathrm{d}y.
\end{gather*}
We extract the real part of every integral.
\begin{gather*}
0 = -\int_{0}^{1}\left(\sum_{k=1}^{\infty}\dfrac{x^{2k-1}\ln(x)}{k}\right)\, \mathrm{d}x -
\int_{0}^{\pi/2}t\ln(2)\, \mathrm{d}t -\int_{0}^{\pi/2}t\ln(\sin(t))\, \mathrm{d}t-\\[2ex]
-\int_{0}^{1}\left(\sum_{k=1}^{\infty}(-1)^{k-1}\dfrac{y^{2k-1}\ln(y)}{k}\right)\, \mathrm{d}y =\\[2ex]
\sum_{k=1}^{\infty}\dfrac{1}{4k^3}-\dfrac{\pi^2}{8}\ln(2) -\int_{0}^{\pi/2}t\ln(\sin(t))\, \mathrm{d}t +\sum_{k=1}^{\infty}\dfrac{(-1)^{k-1}}{4k^3}=\\[2ex]
\dfrac{1}{4}\zeta(3) -\dfrac{\pi^2}{8}\ln(2) -\int_{0}^{\pi/2}t\ln(\sin(t))\, \mathrm{d}t + \dfrac{3}{16}\zeta(3).
\end{gather*}
Consequently
\begin{equation*}
 \int_{0}^{\pi/2}t\ln(\sin(t))\, \mathrm{d}t = \dfrac{7}{16}\zeta(3)-\dfrac{\pi^2}{8}\ln(2).
\end{equation*}
A: Using the trapezioidal rule like numerical integration:
$$\displaystyle{\int \limits _{1}^{\frac \pi2}x\ln(\sin (x))\,dx\approx \frac{1}{2}h\left(f(1)+f\left(\frac\pi2\right)\right)=-0.03672410\\h=\frac \pi2-1}$$
I think that this is a short way to find the value of the $ \int \limits _{1}^{\frac \pi2}x\ln(\sin (x))\,dx$.
