TITLE: Find : $\int_0^{\pi/4}x\ln(\sin x)\mathrm dx$
QUESTION [7 upvotes]: I'm try to find this integral 
$$\int_0^{\pi/4}x\ln(\sin x)\mathrm dx$$
My try use : 
$\ln(\sin x)=-\ln2-\sum\limits_{n=1}^{\infty}\frac{\cos (2nx)}{n}$
But I don't know how to complete summation ...  
I will happy if someone help me
Thanks!

REPLY [6 votes]: A self-contained answer:
\begin{align} I&=\int_0^{\frac{\pi}{4}}x\ln\left( \sin x\right)\,dx\\
J&=\int_0^{\frac{\pi}{4}}x\ln\left( \cos x\right)\,dx\\
I+J&=\int_0^{\frac{\pi}{4}}x\ln\left( \sin x\cos x\right)\,dx\\
&=\int_0^{\frac{\pi}{4}}x\ln\left( \frac{\sin(2x)}{2}\right)\,dx\\
&=\int_0^{\frac{\pi}{4}}x\ln\left( \sin\left (2x\right)\right)\,dx-\frac{\pi^2}{32}\ln 2\\
\end{align}
Perform the change of variable $\displaystyle y=2x$,
\begin{align}I+J&=\frac{1}{4}\int_0^{\frac{\pi}{2}}x\ln\left( \sin x\right)\,dx-\frac{\pi^2}{32}\ln 2\\
&=\frac{1}{4}\int_0^{\frac{\pi}{4}}x\ln\left( \sin x\right)\,dx+\frac{1}{4}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}x\ln\left( \sin x\right)\,dx-\frac{\pi^2}{32}\ln 2\\
\end{align}
In the second integral perform the change of variable $y=\dfrac{\pi}{2}-x$,
\begin{align}I+J&=\frac{1}{4}\int_0^{\frac{\pi}{4}}x\ln\left( \sin x\right)\,dx+\frac{1}{4}\int_0^{\frac{\pi}{4}}\left(\frac{\pi}{2}-x  \right)\ln\left( \cos x\right)\,dx-\frac{\pi^2}{32}\ln 2\\
&=\frac{1}{4}I-\frac{1}{4}J+\frac{\pi }{8}\int_0^{\frac{\pi}{4}}\ln\left( \cos x\right)\,dx-\frac{\pi^2}{32}\ln 2
\end{align}
Therefore,
\begin{align}3I+5J&=\frac{\pi }{2}\int_0^{\frac{\pi}{4}}\ln\left( \cos x\right)\,dx-\frac{\pi^2}{8}\ln 2\end{align}
\begin{align}A&=\int_0^{\frac{\pi}{4}}\ln\left( \sin x\right)\,dx\\
B&=\int_0^{\frac{\pi}{4}}\ln\left( \cos x\right)\,dx\\
A+B&=\int_0^{\frac{\pi}{4}}\ln\left( \frac{\sin(2x)}{2}\right)\,dx\\
&=\int_0^{\frac{\pi}{4}}\ln\left( \sin(2x)\right)\,dx-\frac{\pi}{4}\ln 2
\end{align}
Perform the change of variable $\displaystyle y=2x$,
\begin{align}A+B&=\frac{1}{2}\int_0^{\frac{\pi}{2}}\ln\left( \sin x\right)\,dx-\frac{\pi}{4}\ln 2\\
&=\frac{1}{2}\int_0^{\frac{\pi}{4}}\ln\left( \sin x\right)\,dx+\frac{1}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\ln\left( \sin x\right)\,dx-\frac{\pi}{4}\ln 2\\
\end{align}
In the second integral perform the change of variable $y=\dfrac{\pi}{2}-x$,
\begin{align}A+B&=\frac{1}{2}\int_0^{\frac{\pi}{4}}\ln\left( \sin x\right)\,dx+\frac{1}{2}\int_0^{\frac{\pi}{4}}\ln\left( \cos x\right)\,dx-\frac{\pi}{4}\ln 2\\
\end{align}
Therefore,
\begin{align}A+B&=-\frac{\pi}{2}\ln 2\end{align}
\begin{align}A-B&=\int_0^{\frac{\pi}{4}}\ln\left( \tan x\right)\,dx\end{align}
Perform the change of variable $\displaystyle y=\tan x$,
\begin{align}A-B&=\int_0^1\frac{\ln x}{1+x^2}\,dx\\
&=-\text{G}\end{align}
$\text{G}$ is the Catalan constant.Therefore,
\begin{align}A&=-\frac{\pi}{4}\ln 2-\frac{1}{2}\text{G} \\
B&=\frac{1}{2}\text{G}-\frac{\pi}{4}\ln 2\\
3I+5J&=\frac{1}{4}\pi\text{G}-\frac{1}{4}\pi^2\ln 2
\end{align}
\begin{align}I-J&=\int_0^{\frac{\pi}{4}}x\ln\left(\tan x\right)\,dx\end{align}
Perform the change of variable $\displaystyle y=\tan x$,
\begin{align}I-J&=\int_0^1\frac{\arctan x\ln x}{1+x^2}\,dx\end{align}
Define the function $R$ on $[0;1]$ by:
\begin{align}R(x)&=\int_0^x \frac{\ln t}{1+t^2}\,dt\\
&=\int_0^1 \frac{x\ln(tx)}{1+t^2x^2}\,dt\\\end{align}
Observe that, $R(0)=0,R(1)=-\text{G}$.
Perform integration by parts,
\begin{align}I-J&=\Big[R(x)\arctan x\Big]_0^1-\int_0^1 \frac{R(x)}{1+x^2}\,dx\\
&=-\frac{1}{4}\pi\text{G}-\int_0^1\int_0^1 \frac{x\ln(tx)}{(1+x^2)(1+t^2x^2)}\,dt\,dx\\
&=-\frac{1}{4}\pi\text{G}-\int_0^1 \int_0^1\frac{x\ln t}{(1+x^2)(1+t^2x^2)}\,dt\,dx-\int_0^1\int_0^1 \frac{x\ln x}{(1+x^2)(1+t^2x^2)}\,dt\,dx\\
&=-\frac{1}{4}\pi\text{G}-\int_0^1 \left[\frac{\ln t}{2(1-t^2)}\ln\left(\frac{1+x^2}{1+t^2x^2}\right)\right]_{x=0}^{x=1}\,dt-\int_0^1 \Big[\frac{\ln x\arctan(tx)}{1+x^2}\Big]_{t=0}^{t=1}\,dx\\
&=-\frac{1}{4}\pi\text{G}-\int_0^1 \frac{\ln t}{2(1-t^2)}\ln\left(\frac{2}{1+t^2}\right)\,dt-\int_0^1 \frac{\ln x\arctan x}{1+x^2}\,dx\\
&=-\frac{1}{4}\pi\text{G}-\frac{1}{2}\ln 2\int_0^1 \frac{\ln t}{1-t^2}\,dt+\frac{1}{2}\int_0^1 \frac{\ln t\ln(1+t^2)}{1-t^2}\,dt-\left(I-J\right)\end{align}
Therefore,
\begin{align}I-J&=-\frac{1}{8}\pi\text{G}-\frac{1}{4}\ln 2\int_0^1 \frac{\ln t}{1-t^2}\,dt+\frac{1}{4}\int_0^1 \frac{\ln t\ln(1+t^2)}{1-t^2}\,dt\\
C&=\int_0^1 \frac{\ln x\ln(1+x^2)}{1-x^2}\,dx
\end{align}
Define the function $S$ on $[0;1]$ by:
\begin{align}S(x)&=\int_0^x \frac{\ln t}{1-t^2}\,dt\\
&=\int_0^1 \frac{x\ln(tx)}{1-t^2x^2}\,dt\end{align}
Observe that $\displaystyle S(0)=0$.
Perform integration by parts,
\begin{align}C&=\Big[S(x)\ln(1+x^2)\Big]_0^1-\int_0^1 \int_0^1 \frac{2x^2\ln(tx)}{(1-t^2x^2)(1+x^2)}\,dt\,dx\\
&=S(1)\ln 2-\int_0^1 \int_0^1 \frac{2x^2\ln t}{(1-t^2x^2)(1+x^2)}\,dt\,dx-\int_0^1 \int_0^1 \frac{2x^2\ln x}{(1-t^2x^2)(1+x^2)}\,dt\,dx\\
&=S(1)\ln 2-\int_0^1 \left[\frac{2\left(\text{arctanh}(tx)-t\arctan(x)\right)\ln t}{t(1+t^2)}\right]_{x=0}^{x=1}\,dx-\\
&\int_0^1 \left[\frac{2x\text{arctanh}(tx)\ln x}{1+x^2}\right]_{t=0}^{t=1}\,dt\\
&=S(1)\ln 2-2\int_0^1 \frac{\text{arctanh}(t)\ln t}{t(1+t^2)}\,dt-\frac{1}{2}\pi\text{G}-2\int_0^1\frac{x\text{arctanh}(x)\ln x}{1+x^2}\,dx\\
\end{align}
Since for $t \neq 0$, $\dfrac{1}{t(1+t^2)}=\dfrac{1}{t}-\dfrac{t}{1+t^2}$ then,
\begin{align}
C&=S(1)\ln 2-2\int_0^1 \frac{\text{arctanh}(t)\ln t}{t}\,dt-\frac{1}{2}\pi\text{G}\\
&=\ln 2\int_0^1 \frac{\ln t}{1-t^2}\,dt+\int_0^1 \frac{\ln\left(\frac{1-t}{1+t}\right)\ln t}{t}\,dt-\frac{1}{2}\pi\text{G}\\
&=\ln 2\int_0^1 \frac{\ln t}{1-t^2}\,dt+\frac{1}{2}\left[\ln^2 t\ln\left(\frac{1-t}{1+t}\right)\right]_0^1+\int_0^1 \frac{\ln^2 t}{1-t^2}\,dt-\frac{1}{2}\pi\text{G}\\
&=\ln 2\int_0^1 \frac{\ln t}{1-t^2}\,dt+\int_0^1 \frac{\ln^2 t}{1-t^2}\,dt-\frac{1}{2}\pi\text{G}\\
&=\ln 2\int_0^1 \frac{\ln t}{1-t^2}\,dt+\int_0^1 \frac{\ln^2 t}{1-t}\,dt-\int_0^1 \frac{t\ln^2 t}{1-t^2}\,dt-
\frac{1}{2}\pi\text{G}\\
\end{align}
In the last integral perform the change of variable $\displaystyle y=t^2$,
\begin{align} C&=\ln 2\int_0^1 \frac{\ln t}{1-t^2}\,dt+\frac{7}{8}\int_0^1 \frac{\ln^2 t}{1-t}\,dt-
\frac{1}{2}\pi\text{G}\\
&=\ln 2\int_0^1 \frac{\ln t}{1-t^2}\,dt+\frac{7}{4}\zeta(3)-
\frac{1}{2}\pi\text{G}\end{align}
Therefore,
\begin{align} I-J&=\frac{7}{16}\zeta(3)-\frac{1}{4}\pi\text{G}\end{align}
Moreover,
\begin{align} 3I+5J&=\frac{1}{4}\pi\text{G}-\frac{1}{4}\pi^2\ln 2
\end{align} 
Therefore,
\begin{align}
\boxed{I=\frac{35}{128}\zeta(3)-\frac{1}{8}\pi\text{G}-\frac{1}{32}\pi^2\ln 2}
\end{align}
and,
\begin{align}J&=-\frac{21}{128}\zeta(3)+\frac{1}{8}\pi\text{G}-\frac{1}{32}\pi^2\ln 2\end{align}
NB:
I assume only,
\begin{align}\int_0^1 \frac{\ln^2 t}{1-t}\,dt=2\zeta(3)\\
\int_0^1 \frac{\ln t}{1+t^2}\,dt=-\text{G}\\
\end{align}
$\text{G}$ is the Catalan constant.
PS:
To answer to Clathratus
One wants to compute:
\begin{align}
J&=\int_0^1 A(x)B(x)\ln x\,dx
\end{align}
Suppose one can compute:
\begin{align}\int_0^1 \left(\int_0^1 xA^\prime(x)B(tx)\ln x\,dt\right)\,dx +\int_0^1 \left(\int_0^1 xA^\prime(x)B(tx)\ln t \,dx\right)\,dt\end{align}
Moreover,
if for $x\in [0;1]$,
\begin{align}R(x)=\int_0^x B(t)\ln t\end{align}
Suppose that one can compute $\displaystyle \lim_{x\rightarrow 0}A(x)R(x)$ and $\displaystyle \lim_{x\rightarrow 1}A(x)R(x)$
Therefore one can compute $J$.
(Use integration by parts)
Where does it come from?
"reverse engineering"
First time, i was using it:
https://math.stackexchange.com/a/1842492/186817
I was wondering how to express $\pi^3$ using integral.
If you omit rational multiplicative factor, 
$\pi^3$ is $\displaystyle \arctan(1)\times \int_0^1 \frac{\ln x}{1+x}\,dx$
That is,
\begin{align}\pi^3&=\text{constant}\times \Big[R(x)\arctan(x)\Big]_{x=0}^{x=1}\end{align}
$R$ is the function defined on $[0;1]$ by,
\begin{align}R(x)=\int_0^x \frac{\ln t}{1+t}\,dt\end{align}
Therefore,
\begin{align}\pi^3=\text{constant}\times \int_0^1 \frac{\partial}{\partial x}\left(R(x)\arctan x\right)\,dx\end{align}