Evaluate $\int_{0}^{\pi/4}x\ln^{2}(\sin(x))dx$ Here is a some what challenging log sine integral.  
$$I=\int_{0}^{\pi/4}x\ln^{2}(\sin(x))dx$$
The upper limit of integration is $\frac{\pi}{4}$ instead of the usual $\frac{\pi}{2}$. 
I tried the famous identity $\displaystyle\ln(\sin(x))=-\sum_{k=1}^{\infty}\frac{\cos(2kx)}{k}-\ln(2)$
But, the log is squared and I do not think Parseval is applicable here. But, I may be wrong.  By Parseval, I mean $\displaystyle\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{\cos(kx)\cos(nx)}{kn}$.  
I have tried other means.  I made the sub $t=\sin(x)$. This led to $\displaystyle\int_{0}^{\frac{1}{\sqrt{2}}}\frac{\sin^{-1}(t)\ln^{2}(t)}{\sqrt{1-t^{2}}}dt$
Now, I thought perhaps the famous sum $\displaystyle\frac{\sin^{-1}(t)}{\sqrt{1-t^{2}}}=\sum_{k=1}^{\infty}\frac{(2t)^{2k-1}}{k\binom{2k}{k}}$ could be used.
So, I integrated and got the form
$$I=\frac{\ln^{2}(2)}{16}\sum_{k=1}^{\infty}\frac{(k!)^{2}2^{k}}{k^{2}(2k)!}+\frac{\ln(2)}{8}\sum_{k=1}^{\infty}\frac{(k!)^{2}2^{k}}{k^{3}(2k)!}+\frac{1}{8}\sum_{k=1}^{\infty}\frac{(k!)^{2}2^{k}}{k^{4}(2k)!}$$
These sums actually evaluate to the correct integral result, so it would appear they are equivalent to said integral. But evaluating them may be even more daunting. Though the first one from the left evaluates to $\frac{{\pi}^{2}}{8}$ and the center one evaluates to 
$$\frac{\ln(2)}{8}\left(\pi G-\frac{35}{16}\zeta(3)+\frac{{\pi}^{2}}{8}\ln(2)\right)\;.$$ 
This center one looks to be related to the integral:
$\displaystyle8\ln(2)\int_{0}^{\frac{\pi}{4}}\ln(\sin(x))dx=\frac{35}{16}\ln(2)\zeta(3)-\pi\ln(2)G-\frac{{\pi}^{2}\ln^{2}(2)}{4}$, which I got by expanding the square of the log sine and integrating. 
This just leaves the sum to the right.  It is tougher and I can find no closed form.
Alas, If I could evaluate it then I think the integral would be solved.
I also thought...how about a closed form of some sort for 
$$\int_{0}^{\frac{\pi}{4}}x\sin^{a}(x)dx$$
that could be differentiated w.r.t 'a' and applied?. I do not know of a closed form for 
this, though.  
I got an encouraging looking solution by expanding, making the sub $t=x/2$, then solving for the integral on the far right (which is the integral in question):
$\displaystyle  \int_{0}^{\frac{\pi}{2}}x\ln^{2}(2\sin(x/2))dx$
$=\displaystyle \ln^{2}(2)\int_{0}^{\frac{\pi}{2}}xdx+8\ln(2)\int_{0}^{\frac{\pi}{4}}x\ln(\sin(x))dx+\color{red}4\int_{0}^{\frac{\pi}{4}}x\ln^{2}(\sin(x))dx$ 
But, it did not check numerically. (Error has now been fixed. - Editor) Numerically, the integral in question should evaluate to $I = 0.263605....$
Then again, I may just have a computation error in all of that. 
So, does anyone know of a method for evaluating this log-sine integral?. 
EDIT:  Here is what Mathematica gave for the solution.  It may be too involved to be worth the effort. 
$\displaystyle I=\frac{37{\pi}^{4}}{92160}-\frac{\pi}{4}\sum_{k=1}^{\infty}\frac{\sin(\pi k/4)}{2^{k/2}k^{3}}$ $\displaystyle +\frac{\pi}{8}\ln(2)G+\frac{25}{768}{\pi}^{2}\ln^{2}(2)+\frac{\ln^{4}(2)}{384}+\sum_{k=1}^{\infty}\frac{\cos(\pi k/4)}{2^{k/2}k^{4}}-\frac{35}{128}\ln(2)\zeta(3)$.  
Sometimes with these math engines the solution can be simplified down. But, maybe not in this case. Those complex polylogs may pose a problem. 
 A: Previously I mentioned the three series I arrived at while trying to evaluate the log-sin integral in question.  I have their closed forms via Mathematica. They add up to the required result. That is the best I could do.
The first one can be done rather easily using the classic arc sin sum identity. 
$\displaystyle \frac{\ln^{2}(2)}{16}\sum_{k=1}^{\infty}\frac{(k!)^{2}2^{k}}{k^{2}(2k)!}=\frac{\ln^{2}(2)}{16}\cdot \frac{{\pi}^{2}}{8}$
The second one can be derived from the arcsin identity as well, though it is a wee bit tougher.
$\displaystyle\frac{\ln(2)}{8}\sum_{k=1}^{\infty}\frac{(k!)^{2}2^{k}}{k^{3}(2k)!}=\frac{\pi\ln(2)G}{8}-\frac{35}{128}\zeta(3)\ln(2)+\frac{{\pi}^{2}}{64}\ln(2)$
The third one is nastier.
$\displaystyle\frac{1}{8}\sum_{k=1}^{\infty}\frac{(k!)^{2}2^{k}}{k^{4}(2k)!}=\frac{7}{768}{\pi}^{2}\ln^{2}(2)+\frac{\ln^{4}(2)}{384}+\frac{37{\pi}^{4}}{92160}-\frac{\pi}{4}\sum_{k=1}^{\infty}\frac{\sin(\pi k/4)}{2^{k/2}k^{3}}+\sum_{k=1}^{\infty}\frac{\cos(k\pi/4)}{2^{k/2}k^{4}}$.
One then sums them up. 
A: Thanks for the input, Fred. I managed to evaluate these types of integrals by using
$\displaystyle \ln(1-e^{ix})=\ln(2\sin(x/2))-i/2(\pi-x)$.  Mathematica often gives 
hypergeometric solutions,  which I have no interest in. These log-sin integrals can be done, 
but they are extremely onerous. I managed to work out the above integral except the upper limit was $\frac{\pi}{2}$.  
It was extremely tedious, but does have a closed form in terms of 
polylogs, zeta, and so on. As does the $\frac{\pi}{4}$ case. 
For instance, evaluating $\displaystyle \int_{0}^{\frac{\pi}{2}}x\ln^{2}(2\sin(x/2))dx$ gives 
$\displaystyle \frac{5}{96}\ln^{4}(2)-\frac{{\pi}^{2}\ln^{2}(2)}{48}+\frac{5}{4}Li_{4}(1/2)+\frac{19{\pi}^{4}}{1152}+\pi\mathfrak{I} \left[Li_{3}\left(1/2+i/2\right)\right]+\frac{35}{32}\zeta(3)\ln(2)-\frac{\pi\ln(2)}{2}G$.
This required a lot of patience and work on my part to arrive at this. The $\frac{\pi}{4}$ case is even worse to evaluate. 
But, regarding the math engine result: take the indefinite integral 
$\displaystyle \int\frac{1-x}{(1+x)\sqrt{x^{3}+x^{2}+x}}dx$.  
This has a rather simple closed form of 
$\displaystyle-\sin^{-1}\left(\frac{x^{2}+1}{(x+1)^{2}}\right)-\frac{\pi}{2}$,
but all I could get Mathematica and Maple to return was a long, complicated EllipticF mess. 
A: It's been a while since I posted this problem. Well, I didn't give up. I doubt if anyone cares at this point, but I found a way to evaluate integrals of this form. It is tedious to say the least.  
$\frac{\pi}{3}$ being the upper limit of integration is about the easiest, but it is even kind of difficult to do. Not to mention, the $\pi/2$ and notoriously difficult $\pi/4$ case.
It uses $\ln(1-e^{ix})=\ln(2\sin(x/2))-\frac{i}{2}(\pi-x)$ and various polylog identities.
I managed to evaluate $\displaystyle \int_{0}^{\frac{\pi}{3}}x\left(\ln(2\sin(x/2))-\frac{i}{2}(\pi-x)\right)^{2}dx=\frac{-7{\pi}^{4}}{4860}-\frac{8{\pi} i\zeta(3)}{9}+2L_{4}(e^{\frac{\pi i}{3}})+\frac{2{\pi}^{2}i}{9}\sum_{n=1}^{\infty}\frac{\sin(\frac{k\pi}{3})}{k^{2}}$. 
If need be, the Clausen sum can be written in a closed form involving derivatives of digamma and what not.
This checks numerically with Mathematica which I was happy to see finally worked out. 
This is the EASY one. Other upper limits, such as $\frac{\pi}{2}, \;\ \frac{\pi}{4}$ are monstrosities. 
It is related to $\displaystyle \int_{0}^{x}\frac{\ln^{a}(x)\ln^{b}(1-x)}{x}dx=i^{a}\int_{0}^{x}t^{a}\left(\ln(2\sin(t/2))-\frac{i}{2}(\pi-t)\right)^{b}dt$.
Of course, to evaluate $\displaystyle\int_{0}^{\frac{\pi}{3}}x\ln^{2}(2\sin(x/2))dx$, one could expand the square and solve for the intended integral.  But, this can be rather laborious because of $\displaystyle\pi \int_{0}^{\frac{\pi}{3}}x\ln(2\sin(x/2))dx=\frac{2\pi\zeta(3)}{3}-\frac{{\pi}^{2}}{3}\sum_{k=1}^{\infty}\frac{\sin(\frac{k\pi}{3})}{k^{2}}$ and
$\displaystyle \int_{0}^{\frac{\pi}{3}}x^{2}\ln(2\sin(x/2))dx=\mathcal{R}\left[-\frac{2\pi\zeta(3)}{9}-\frac{{\pi}^{2}}{9}\sum_{k=1}^{\infty}\frac{\sin(\frac{\pi k}{3})}{k^{2}}-2iL_{4}(e^{\frac{\pi i}{3}})\right]$
Note:  I do not know any nice closed forms for $L_{4}$ polylogs.  
i.e $\displaystyle L_{3}(e^{\frac{\pi i}{3}})=\frac{\zeta(3)}{3}+\frac{5{\pi}^{3}i}{162}$ and
$\displaystyle L_{2}(e^{\frac{\pi i}{3}})=\frac{{\pi}^{2}}{36}+i\sum_{k=1}^{\infty}\frac{\sin(\frac{\pi k}{3})}{k^{2}}$
I am sorry.  I do not know how this post got displayed before the one several weeks ago.
A: Here is what I got from Mathematica:
$$\frac{1}{8} \, _5F_4\left(1,1,1,1,1;\frac{3}{2},2,2,2;\frac{1}{2}\right)+\frac{1}{8} \log (2) \, _4F_3\left(1,1,1,1;\frac{3}{2},2,2;\frac{1}{2}\right)+\frac{1}{128} \pi ^2 \log ^2(2) = 0.263605$$
A: It is
$$
\log(\cos x)=-\log 2+ix-\sum^{\infty}_{k=1}\frac{(-1)^k}{k}e^{-2ikx}
$$
Squaring we get
$$
\log^2(\cos x)=\log^2 2-x^2-2x\sum^{\infty}_{k=1}\frac{(-1)^k}{k}\sin(2kx)+2\log 2 \sum^{\infty}_{k=1}\frac{(-1)^k}{k}\cos(2kx)+
$$
$$
+2\sum^{\infty}_{k=1}\frac{(-1)^k}{k}\cos(2kx)H_{k-1},
$$
since
$$
\sum_{1\leq k,l\leq n,k+l=n}\frac{1}{kl}=\frac{2}{n}H_{n-1},
$$
where $H_n=1^{-1}+2^{-1}+\ldots+n^{-1}$. Hence
$$
\int^{\pi/4}_{0}x\log^2(\cos x)dx=\frac{\pi^4}{1024}-\frac{\pi}{4}G\log 2+\frac{\pi^2}{32}\log^22+\frac{21}{64}\zeta(3)\log 2+S_1,\tag 1
$$
where
$$
S_1=2\sum^{\infty}_{k=1}(-1)^k\frac{-2+2\cos(k\pi/2)+k\pi\sin(k\pi/2)}{8k^3}H_{k-1}.
$$
Hence
$$
S_1:=\frac{1}{2}\sum^{\infty}_{k=0}\frac{H_{2k}}{(2k+1)^3}-\frac{\pi}{4}\sum^{\infty}_{k=0}\frac{(-1)^kH_{2k}}{(2k+1)^2}-\frac{1}{8}\sum^{\infty}_{k=0}\frac{H_{4k+1}}{(2k+1)^3}.\tag 2
$$
In the same way
$$
\log^2(\sin x)=-\frac{\pi^2}{4}+\pi x-x^2+\log^2 2+(\pi-2x)\sum^{\infty}_{k=1}\frac{\sin(2kx)}{k}+
$$
$$
+2\log 2\sum^{\infty}_{k=1}\frac{\cos(2kx)}{k}+2\sum^{\infty}_{k=1}\frac{\cos(2kx)}{k}H_{k-1}
$$
Hence as with cosine
$$
\int^{\pi/4}_{0}x\log^2(\sin x)dx=\frac{11\pi^4}{3072}+\frac{\pi}{4}G\log 2+\frac{\pi^2}{32}\log^2 2-\frac{35}{64}\zeta(3)\log 2+S_2,\tag 3
$$
where
$$
S_2=-\frac{1}{2}\sum^{\infty}_{k=0}\frac{H_{2k}}{(2k+1)^3}+\frac{\pi}{4}\sum^{\infty}_{k=0}\frac{(-1)^kH_{2k}}{(2k+1)^2}-\frac{1}{8}\sum^{\infty}_{k=0}\frac{H_{4k+1}}{(2k+1)^3}.\tag 4
$$
