twist on classic log of sine and cosine integral

I ran across this integral and have not been able to evaluate it.

$\displaystyle \int_{0}^{\frac{\pi}{2}}x\ln(\sin(x))\ln(\cos(x))dx=\frac{{\pi}^{2}\ln^{2}(2)}{8}-\frac{{\pi}^{4}}{192}$

I had some ideas. Perhaps some how arrive at $\displaystyle\frac{1}{2}\sum_{k=0}^{\infty}\frac{1}{(2k+1)^{4}}=\frac{{\pi}^{4}}{192}$.

and $\displaystyle \ln(2)\int_{0}^{\frac{\pi}{2}}x\ln(2)dx=\frac{{\pi}^{2}\ln^{2}(2)}{8}$

by using the identity $\displaystyle\sum_{k=1}^{\infty}\frac{x\cos(2kx)}{k}=-x\ln(\sin(x))-x\ln(2)$

and/or $\displaystyle \ln(\cos(x))=-\ln(2)-\sum_{k=1}^{\infty}\frac{(-1)^{k}\cos(2kx)}{k}$

I have used the first one to evaluate $\displaystyle\int_{0}^{\frac{\pi}{2}}x\ln(\sin(x))dx$, so I thought perhaps it could be used in some manner here.

I see some familiar things in the solution, but how to get there?.

Does anyone have any clever ideas?.

Thanks.

$$\zeta(4):=\sum_{n=1}^\infty\frac{1}{n^4}=\frac{\pi^4}{90}\Longrightarrow \zeta_2(4):=\sum_{n=1}^\infty\frac{1}{(2n)^4}=\frac{1}{16}\zeta(4)=\frac{\pi^4}{16\cdot 90}\Longrightarrow$$

$$\Longrightarrow\sum_{n=0}^\infty\frac{1}{(2n+1)^4}=\zeta(4)-\zeta_2(4)=\frac{15}{16}\frac{\pi^4}{90}=\frac{\pi^4}{96}$$

And you have your first question answered.

I thought I would come back and show what I done. I am rather uneasy about this solution and you'll see why.

Using the identities mentioned previously:

$\displaystyle -\ln(\sin(x))=\sum_{k=1}^{\infty}\frac{\cos(2kx)}{k}+\ln(2)$

and $\displaystyle -\ln(\cos(x))=\sum_{k=1}^{\infty}\frac{(-1)^{k}\cos(2kx)}{k}+\ln(2)$

I subbed them in and arrived at:

$\displaystyle\int_{0}^{\frac{\pi}{2}}\left(\sum_{k=1}^{\infty}\frac{x\cos(2kx)}{k}+x\ln(2)\right)\left(\sum_{k=1}^{\infty}\frac{(-1)^{k}\cos(2kx)}{k}+\ln(2)\right)dx$

$=\displaystyle \int_{0}^{\frac{\pi}{2}}(\sum_{k=1}^{\infty}\frac{(-1)^{k}\cos(2kx)}{k}\sum_{k=1}^{\infty}\frac{x\cos(2kx)}{k} $ +$\displaystyle \ln(2)\sum_{k=1}^{\infty}\frac{x\cos(2kx)}{k}+\ln(2)\sum_{k=1}^{\infty}\frac{(-1)^{k}x\cos(2kx)}{k}+x\ln^{2}(2))dx$

Now, here I made an otherwise 'illegal' move. I took the product of the sums under one summation.

$\displaystyle\int_{0}^{\frac{\pi}{2}}(\sum_{k=1}^{\infty}\frac{(-1)^{k}x\cos^{2}(2kx)}{k^{2}}$ $+\displaystyle\ln(2)\sum_{k=1}^{\infty}\frac{x\cos(2kx)}{k}$ $+\displaystyle\ln(2)\sum_{k=1}^{\infty}\frac{(-1)^{k}x\cos(2kx)}{k}$ $+x\ln^{2}(2))dx$

Switch the sum and integral:

$\displaystyle\sum_{k=1}^{\infty}(\underbrace{\int_{0}^{\frac{\pi}{2}}\frac{(-1)^{k}x\cos^{2}(2kx)}{k^{2}}dx}_{\text{[1]}} $ $+\displaystyle\underbrace{\ln(2)\int_{0}^{\frac{\pi}{2}}\frac{x\cos(2kx)}{k}dx}_{\text{[2]}}$ $+\displaystyle\underbrace{\ln(2)\int_{0}^{\frac{\pi}{2}}\frac{(-1)^{k}x\cos(2kx)}{k}dx}_{\text{[3]}}$ $+\underbrace{\ln^{2}(2)\int_{0}^{\frac{\pi}{2}}xdx}_{\text{[4]}})dx$

$[1]:\displaystyle \frac{{\pi}^{2}}{16}\sum_{k=1}^{\infty}\frac{(-1)^{k}}{k^{2}}=\frac{-{\pi}^{4}}{192}$

$[2]: \ln(2)\left(\frac{-1}{4}\sum_{k=1}^{\infty}\frac{1}{k^{3}}+\frac{1}{4}\sum_{k=1}^{\infty}\frac{(-1)^{k}}{k^{3}}\right)$

$[3]: \displaystyle \ln(2)\left(\frac{-1}{4}\sum_{k=1}^{\infty}\frac{(-1)^{k}}{k^{3}}+\frac{1}{4}\sum_{k=1}^{\infty}\frac{1}{k^{3}}\right)$

$[4]: \displaystyle \ln^{2}(2)\int_{0}^{\frac{\pi}{2}}xdx=\frac{{\pi}^{2}\ln^{2}(2)}{8}$

[2] and [3] cancel one another out and I arrive at:

$\displaystyle\frac{{\pi}^{2}\ln^{2}(2)}{8}-\frac{{\pi}^{4}}{192}$

This worked out beautifully. Is it a fluke or can one manipulate sums, like I done above, under certain conditions?. Or did I actually manage to come up with a clever solution?.

Also sorry for the undersized parentheses. Every time I tried enlarging them, the Latex would not display. I have been wrestling with this for sometime trying to get it all to display. Thanks All.

Here is an approach that avoids using the Fourier series expansions for $\ln (\cos x)$ and $\ln (\sin x)$.

Consider $$I = \int_0^{\pi/2} x^2 \ln (\sin x) \ln (\cos x) \, dx.$$ Enforcing a substitution of $x \mapsto \pi/2 - x$ gives \begin{align*} I &= \int_0^{\pi/2} \left (\frac{\pi}{2} - x \right )^2 \ln \left [\sin \left (\frac{\pi}{2} - x \right ) \right ] \ln \left [\cos \left (\frac{\pi}{2} - x \right ) \right ] \, dx\\ &= \int_0^{\pi/2} \left (\frac{\pi}{2} - x \right )^2 \ln (\cos x) \ln (\sin x) \, dx\\ &= \frac{\pi^2}{4} \int_0^{\pi/2} \ln (\sin x) \ln (\cos x) \, dx - \pi \int_0^{\pi/2} x \ln (\sin x) \ln (\cos x) \, dx\\ & \qquad + \int_0^{\pi/2} x^2 \ln (\sin x) \ln (\cos x) \, dx, \end{align*} or $$\int_0^{\pi/2} x \ln (\sin x) \ln (\cos x) \, dx = \frac{\pi}{4} \int_0^{\pi/2} \ln (\sin x) \ln (\cos x) \, dx.$$

The integral appearing on the right can be found by differentiating the beta function. As $$\text{B}(x,y) = 2 \int_0^{\pi/2} \cos^{2x - 1} \theta \sin^{2y - 1} \theta \, d\theta,$$ we see that $$\int_0^{\pi/2} \ln (\sin \theta) \ln (\cos \theta) \, d\theta = \frac{1}{8} \partial_x \partial_y \text{B}(x,y) \Big{|}_{x=y=1/2},$$ giving $$\int_0^{\pi/2} x\ln (\sin x) \ln (\cos x) \, dx = \frac{\pi}{32} \partial_x \partial_y \text{B}(x,y) \Big{|}_{x=y=1/2}.$$

Since $$\partial_x \text{B}(x,y) = \text{B}(x,y) [\psi(x) - \psi (x + y)] \quad \text{and} \quad \partial_y \text{B}(x,y) = \text{B}(x,y) [\psi(y) - \psi (x + y)],$$ where $\psi (z)$ is the digamma function, we have $$\partial_x \partial_y \text{B}(x,y) = \text{B}(x,y) \left [ \left \{\psi(x) - \psi (x + y) \right \} \left \{\psi(y) - \psi (x + y) \right \} - \psi^{(1)}(x + y) \right ].$$ Thus \begin{align*} \int_0^{\pi/2} x\ln (\sin x) \ln (\cos x) \, dx &= \frac{\pi}{32} \text{B} \left (\frac{1}{2}, \frac{1}{2} \right ) \left [ \left \{ \psi \left (\frac{1}{2} \right ) - \psi (1) \right \}^2 - \psi^{(1)}(1) \right ]. \end{align*}

Since $$\text{B} \left (\frac{1}{2}, \frac{1}{2} \right ) = \frac{\Gamma (1/2) \Gamma (1/2)}{\Gamma (1)} = \pi,$$ and $$\psi^{(1)}(1) = \zeta (2) = \frac{\pi^2}{6} \quad \text{and} \quad \psi \left (\frac{1}{2} \right ) - \psi (1) = - 2\ln (2),$$ we have $$\int_0^{\pi/2} x\ln (\sin x) \ln (\cos x) \, dx = \frac{\pi^2}{8} \ln^2 (2) - \frac{\pi^4}{192},$$ as expected.