How to calculate $I=\frac{1}{2}\int_{0}^{\frac{\pi }{2}}\frac{\ln(\sin y)\ln(\cos y)}{\sin y\cos y}dy$? How do I integrate this guy? I've been stuck on this for hours..
$$I=\frac{1}{2}\int_{0}^{\frac{\pi }{2}}\frac{\ln(\sin y)\ln(\cos y)}{\sin y\cos y}dy$$
 A: According to Mathematica, $$I=\frac{\zeta(3)}{8}$$where $\zeta$ is the zeta function. (See also this about the integral). This is approximately 0.150257.
A: Let $x = \sin^2y$.
(It seems we've started like @Sasha here and like @sos440.)
Then 
$$\begin{eqnarray*}
I &=& \frac{1}{16} \int_0^1 dx\, \frac{\log(x)\log(1-x)}{x(1-x)} \\
    &=& \frac{1}{16} \int_0^1 dx\, \frac{\log(x)\log(1-x)}{x}
    + \underbrace{\frac{1}{16} \int_0^1 dx\, \frac{\log(x)\log(1-x)}{1-x}}_{x\to 1-x}
    \qquad (\textrm{partial fractions}) \\
    &=& \frac{1}{8} \int_0^1 dx\, \frac{\log(x)\log(1-x)}{x}
    \qquad (\textrm{integral linked above}) \\
    &=& \frac{1}{8} \int_0^1 dx\, \frac{\log(x)}{x} \left(-\sum_{k=1}^\infty \frac{x^k}{k}\right)
    \qquad (\textrm{Taylor expansion for }\log(1-x) ) \\
    &=& -\frac{1}{8} \sum_{k=0}^\infty \frac{1}{k+1} \underbrace{\int_0^1 dx\, x^k \log x}_{-1/(k+1)^2} 
    \qquad (\textrm{standard integral involving log}) \\
    &=& \frac{1}{8} \sum_{k=0}^\infty \frac{1}{(k+1)^3} \\
    &=& \frac{\zeta(3)}{8}. 
\end{eqnarray*}$$
Addendum: 
Note that $\csc(y)\sec(y) = \cot y + \tan y$. 
Then 
$$\begin{eqnarray*}
I &=& \frac{1}{2} \int_{0}^{\pi/2} dy\,
    \frac{\ln(\sin y)\ln(\cos y)}{\sin y\cos y} \\
    &=& \frac{1}{2} \int_{0}^{\pi/2} dy\,
    \frac{\ln(\sin y)\ln(\cos y)}{\tan y}
    + \underbrace{\frac{1}{2} \int_{0}^{\pi/2} dy\,
    \frac{\ln(\sin y)\ln(\cos y)}{\cot y}}_{y\to \pi/2-y} \\
    &=& \int_{0}^{\pi/2} dy\,
    \frac{\ln(\sin y)\ln(\cos y)}{\tan y}.
\end{eqnarray*}$$
This is exactly the integral linked above. 
A: Let $\sin(y) = t$. Recall that $\cos^2(y) = 1-t^2$. Then $\cos(y) dy = dt$. Hence,
$$I = \dfrac14 \int_0^1 \dfrac{\ln(\sin(y)) \ln(\cos^2(y)) \cos(y) dy}{\sin(y) \cos^2(y)}$$
$$4I = \int_0^1 \dfrac{\ln(t) \ln(1-t^2) dt}{t (1-t^2)}$$
$$\dfrac{\ln(t) \ln(1-t^2)}{t (1-t^2)} = -\dfrac{\ln(t)}{t(1-t^2)} \sum_{k=1}^{\infty} \dfrac{t^{2k}}k = - \sum_{k=1}^{\infty}\dfrac{\ln(t)}{(1-t^2)} \dfrac{t^{2k-1}}k$$
$$\dfrac{\ln(t)}{(1-t^2)} \dfrac{t^{2k-1}}k = \dfrac1k \sum_{l=0}^{\infty}t^{2k+2\ell-1} \ln(t)$$
Hence,
$$\dfrac{\ln(t) \ln(1-t^2)}{t (1-t^2)} = -\sum_{k=1}^{\infty} \dfrac1k\sum_{l=0}^{\infty}t^{2k+2\ell-1} \ln(t)$$
Recall that $$\int_0^1 t^m \log(t) dt = -\dfrac1{(m+1)^2}$$
Hence, $$4I = \sum_{k=1}^{\infty} \sum_{\ell=0}^{\infty} \dfrac1k \dfrac1{(2k+2 \ell)^2} = \dfrac14 \sum_{k=1}^{\infty} \sum_{\ell=0}^{\infty} \dfrac1k \dfrac1{(k+\ell)^2} = \dfrac14 2 \zeta(3)$$
Hence, $$I = \dfrac{\zeta(3)}8$$
EDIT
Consider the sum
$$S = \sum_{k=1}^{\infty} \sum_{l=0}^{\infty} \dfrac1k\dfrac1{(k+l)^2}$$
We have that
$$S = \underbrace{\sum_{k=1}^{\infty} \sum_{l=k}^{\infty} \dfrac1k\dfrac1{l^2} = \sum_{l=1}^{\infty} \sum_{k=1}^l \dfrac1k\dfrac1{l^2}}_{\text{Change order of summation}} = \sum_{l=1}^{\infty} \dfrac{H_l}{l^2}$$
Also, note that
$$S = \sum_{k=1}^{\infty} \dfrac1k\dfrac1{k^2} + \sum_{k=1}^{\infty} \sum_{l=1}^{\infty} \dfrac1k\dfrac1{(k+l)^2} = \zeta(3) + \sum_{k=1}^{\infty} \sum_{l=1}^{\infty} \dfrac1k\dfrac1{(k+l)^2}$$
Now let us evaluate the second sum in the above line.(To evaluate this, we follow the technique @sos440 has in his answer.)
\begin{align*}
\sum_{k=1}^{\infty} \sum_{l=1}^{\infty} \frac{1}{k} \frac{1}{(k+l)^2}
&= \frac{1}{2}\sum_{k,l=1}^{\infty} \left\{ \frac{1}{k} \frac{1}{(k+l)^2} + \frac{1}{l} \frac{1}{(k+l)^2} \right\} \\
&= \frac{1}{2}\sum_{k,l=1}^{\infty} \frac{1}{kl(k+l)}
 = \frac{1}{2}\sum_{k=1}^{\infty} \frac{1}{k^2}\sum_{l=1}^{\infty} \left\{ \frac{1}{l} - \frac{1}{k+l}\right\} \\
&= \frac{1}{2}\sum_{k=1}^{\infty} \frac{H_{k}}{k^2}
\end{align*}
Hence, we have $$S = \sum_{n=1}^{\infty} \frac{H_{n}}{n^2} = \zeta(3) + \frac{1}{2}\sum_{k=1}^{\infty} \frac{H_{k}}{k^2}$$
This gives us $$S = 2 \zeta(3)$$
