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..

$$I=\frac{1}{2}\int_{0}^{\frac{\pi }{2}}\frac{\ln(\sin y)\ln(\cos y)}{\sin y\cos y}dy$$

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.

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 $$S = 2 \zeta(3)$$

I will assume that your $dx$ is indeed a type of $dy$, otherwise the answer is clear.

Let $t = \sin^2 y$. Then $dt = 2 \sin y \cos y \, dy$ and we have

$$I = \frac{1}{16} \int_{0}^{1} \frac{\log t \log(1-t)}{t(1-t)} \, dt. $$

Then thanks to the Taylor expansion

$$ \frac{\log(1-t)}{t} = -\sum_{n=1}^{\infty} \frac{t^{n-1}}{n}, $$

we have

\begin{align*} I &= -\frac{1}{16} \int_{0}^{1} \frac{\log t}{1-t} \sum_{n=1}^{\infty} \frac{t^{n-1}}{n} \, dt = -\frac{1}{16} \sum_{n=1}^{\infty} \frac{1}{n}\int_{0}^{1} \frac{t^{n-1} \log t}{1-t} \, dt \\ &= -\frac{1}{16} \sum_{n=1}^{\infty} \frac{1}{n} \sum_{m=0}^{\infty} \int_{0}^{1} t^{n+m-1} \log t \, dt \\ &= \frac{1}{16} \sum_{n=1}^{\infty} \sum_{m=0}^{\infty} \frac{1}{n} \frac{1}{(m+n)^2} \\ &= \frac{1}{16}\zeta(3) + \frac{1}{16}\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{n} \frac{1}{(m+n)^2}. \end{align*}

Now we focus on the double summation in the last line. By interchanging the role of $m$ and $n$,

\begin{align*} \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{n} \frac{1}{(m+n)^2} &= \frac{1}{2}\sum_{m,n=1}^{\infty} \left\{ \frac{1}{n} \frac{1}{(m+n)^2} + \frac{1}{m} \frac{1}{(m+n)^2} \right\} \\ &= \frac{1}{2}\sum_{m,n=1}^{\infty} \frac{1}{mn(m+n)} = \frac{1}{2}\sum_{n=1}^{\infty} \frac{1}{n^2}\sum_{m=1}^{\infty} \left\{ \frac{1}{m} - \frac{1}{m+n}\right\} \\ &= \frac{1}{2}\sum_{n=1}^{\infty} \frac{H_{n}}{n^2} \end{align*}

To this end, it is natural to consider the following power series.

$$ \sum_{n=1}^{\infty} \frac{H_n}{n^2} x^n. $$

Associated with this series we easily observe that

$$ \sum_{n=1}^{\infty} H_n x^n = -\frac{\log(1-x)}{1-x}. $$

Thus we have

\begin{align*} \sum_{n=1}^{\infty} \frac{H_n}{n} x^n &= -\int_{0}^{x} \frac{\log(1-t)}{t(1-t)} \, dt = -\int_{0}^{x} \left\{ \frac{\log(1-t)}{t} + \frac{\log(1-t)}{1-t} \right\} \, dt \\ &= \mathrm{Li}_2 (x) + \frac{1}{2}\log^2(1-x), \end{align*}

where $\mathrm{Li}_s (x)$ is the polylogarithm defined by

$$ \mathrm{Li}_s (x) = \sum_{n=1}^{\infty} \frac{x^n}{n^s}. $$

Thus we must have

\begin{align*} \sum_{n=1}^{\infty} \frac{H_n}{n^2} x^n &= \int_{0}^{x} \frac{1}{t}\left\{ \mathrm{Li}_2 (t) + \frac{1}{2}\log^2(1-t) \right\} \, dt \\ &= \mathrm{Li}_3 (x) + \frac{1}{2} \int_{0}^{x} \frac{\log^2(1-t)}{t} \, dt. \end{align*}

Let $I(x)$ denote the latter integral. Then by integration by parts,

\begin{align*} 2I(x) = \int_{0}^{x} \frac{\log^2(1-t)}{t} \, dt = -\mathrm{Li}_2(x) \log(1-x) - \int_{0}^{x} \frac{\mathrm{Li}_2(t)}{1-t} \, dt. \end{align*}

But we know that $\mathrm{Li}_2 (x)$ satisfies the Euler reflection formula

$$ \mathrm{Li}_2 (x) + \mathrm{Li}_2 (1-x) = \zeta(2) - \log x \log (1-x), $$

which is easily proved by differentiation both sides with respect to $x$. Thus we have

\begin{align*} \int_{0}^{x} \frac{\mathrm{Li}_2 (t)}{1-t} \; dt & = \int_{0}^{x} \frac{1}{1-t} \left[ \zeta(2) - \log t \log (1-t) - \mathrm{Li}_2 (1-t) \right] \; dt \\ & = \int_{0}^{x} \frac{\zeta(2) - \mathrm{Li}_2 (1-t)}{1-t} \; dt - \int_{0}^{x} \frac{\log t \log (1-t)}{1-t} \; dt. \end{align*}

But the fist integral is easily calculated as we can find its antiderivative:

\begin{align*} \int_{0}^{x} \frac{\zeta(2) - \mathrm{Li}_2 (1-t)}{1-t} \; dt &= \left[ -\zeta(2)\log(1-t) + \mathrm{Li}_{3} (1-t) \right]_{0}^{x} \\ &= -\zeta(2) \log(1-x) + \mathrm{Li}_{3}(1-x) - \zeta(3). \end{align*}

The second integral reduces to a formula involving $I(x)$:

\begin{align*} -\int_{0}^{x} \frac{\log t \log (1-t)}{1-t} \; dt &= \left[ \frac{1}{2} \log^2 (1-t) \log t \right]_{0}^{x} - I(x) \\ &= \frac{1}{2} \log^2 (1-x) \log x - I(x). \end{align*}

From this calculation, we obtain a closed form for $I(x)$ as follows:

$$ I(x) = \zeta(3) - \mathrm{Li}_3 (1-x) + \{ \zeta(2) - \mathrm{Li}_2 (x) \} \log (1 - x) - \frac{1}{2} \log^2 (1-x) \log x. $$

Plugging $x = 1$, we then have

$$ I(1) = \zeta(3).$$

Therefore we have

$$ \sum_{n=1}^{\infty} \frac{H_n}{n^2} = 2 \zeta(3)$$

and hence

\begin{align*} I = \frac{1}{16}\zeta(3) + \frac{1}{16} \zeta(3) = \frac{1}{8}\zeta(3). \end{align*}

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.