Evaluating $\int_0^{\frac{\pi}{2}}\ln\left(\frac{\ln^2\sin\theta}{\pi^2+\ln^2\sin\theta}\right)\,\frac{\ln\cos\theta}{\tan\theta}\,d\theta$ Prove
$$\int_0^{\frac{\pi}{2}}\ln\left(\frac{\ln^2\sin\theta}{\pi^2+\ln^2\sin\theta}\right)\,\frac{\ln\cos\theta}{\tan\theta}\,d\theta = \frac{\pi^2}{4}$$
 A: I am also a proponent of the opinion that the proposed rules are against the way how we communicate ideas in this community. At the same time, however, the mathematical part of OP is something worth it to be dealt with. So here is a solution:

1. Preliminary
Before the calculation we make some preliminary results:

Lemma 1. For any $u > 0$ and $n > 0$, we have
  $$\frac{1}{n^{2}} \log \left(1 + \frac{4\pi^{2}n^{2}}{u^{2}} \right) = \pi^{2} \int_{u/2}^{\infty} \frac{2}{s^{2} + n^{2}\pi^{2}} \, \frac{ds}{s}.$$

Proof. Differentiating both sides with respect to $u$, we check that they must equal up to a constant. Taking $u \to \infty$, we find that this constant should equal zero. ////

Lemma 2. For any real $x$, we have
  $$ \sum_{n=1}^{\infty} \frac{2}{s^{2} + n^{2}\pi^{2}} = \frac{s \coth s - 1}{s^{2}}. $$

Although non-trivial, this is a standard result in complex analysis. So we omit the proof.

Lemma 3. Let $f(s) = (1 - e^{-2s})(s\coth s - 1)$. Then
  
  
*
  
*$f(s) = (s-1) + (s+1)e^{-2s}$ and hence $f''(s) = 4s e^{-2s}$. 
  
*$f(s)/s^{2}$ and $f'(s)/s$ converges to $0$ as $s \to 0$ and $s \to +\infty$.
  

Proof. The first assertion is just a simple calculation. To prove the second assertion, it suffices to look into the McLaurin series expansion $f(s) = \frac{2}{3}s^{3} - \frac{2}{3}s^{4} + \cdots$. ////
2. Calculation
Now we are ready to calculate the integral. Let $I$ denote the integral. Then with the substitution $\sin^{2}\theta = e^{-t}$ (so that $d\theta/\tan\theta = -dt/2t$), we have
\begin{align*}
I
&= \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \log \left( \frac{\log^{2} \sin^{2}\theta}{4\pi^{2} + \log^{2}\log^{2}\theta} \right) \frac{\log^{2}\cos^{2}\theta}{\tan\theta} \, d\theta \\
&= \frac{1}{4} \int_{0}^{\infty} \log(1 - e^{-t}) \log\left( \frac{t^{2}}{4\pi^{2} + t^{2}} \right) \, dt\\
&= \frac{1}{4} \int_{0}^{\infty} \sum_{n=1}^{\infty} \frac{e^{-nt}}{n} \log \left(1 + \frac{4\pi^{2}}{t^{2}} \right) \, dt.
\end{align*}
Now we utilize the Tonelli's theorem to interchange the summation and integral. Then
\begin{align*}
I
&= \frac{1}{4} \sum_{n=1}^{\infty} \int_{0}^{\infty} \frac{e^{-nt}}{n} \log \left(1 + \frac{4\pi^{2}}{t^{2}} \right) \, dt \\
&= \frac{1}{4} \sum_{n=1}^{\infty} \int_{0}^{\infty} \frac{e^{-u}}{n^{2}} \log \left(1 + \frac{4\pi^{2}n^{2}}{u^{2}} \right) \, du, \quad (u = nt) \\
&= \frac{\pi^{2}}{4} \sum_{n=1}^{\infty} \int_{0}^{\infty} e^{-u} \left( \int_{u/2}^{\infty} \frac{2}{s^{2} + n^{2}\pi^{2}} \, \frac{ds}{s} \right) \, du,
\end{align*}
where the last equality follows from Lemma 1. Applying the Tonelli's theorem again, Lemma 2 shows that
\begin{align*}
I
&= \frac{\pi^{2}}{4} \int_{0}^{\infty} e^{-u} \left( \int_{u/2}^{\infty} \frac{s \coth s - 1}{s^{2}} \, \frac{ds}{s} \right) \, du \\
&= \frac{\pi^{2}}{4} \int_{0}^{\infty} \left( \int_{0}^{2s} e^{-u} \, du \right) \frac{s \coth s - 1}{s^{3}} \, ds \\
&= \frac{\pi^{2}}{4} \int_{0}^{\infty} \frac{f(s)}{s^{3}} \, ds,
\end{align*}
where we applied Tonelli's theorem again in the second line, and $f(s)$ denotes the function in Lemma 3. So it suffices to prove that the last integral, without the constant $\pi^{2}/4$, equals 1. Indeed, Lemma 3 shows that
\begin{align*}
\int_{0}^{\infty} \frac{f(s)}{s^{3}} \, ds
&= \left[ -\frac{f(s)}{2s^{2}} \right]_{0}^{\infty} + \frac{1}{2} \int_{0}^{\infty} \frac{f'(s)}{s^{2}} \, ds \\
&= \left[ -\frac{\smash{f'}(s)}{2s} \right]_{0}^{\infty} + \frac{1}{2} \int_{0}^{\infty} \frac{f''(s)}{s} \, ds \\
&= \int_{0}^{\infty} 2e^{-2s} \, ds
 = 1
\end{align*}
and therefore we get $I = \pi^{2}/4$ as desired.
A: Note: I will be making this post Community Wiki as it bears resemblance to Chris's sis's answer.

I will use the following result:
$$\lim_{N\to\infty}\left[\sum^\infty_{k=1}\frac{(-1)^{k+1}}{2k(2k)!}N^{2k}-\ln{N}\right]=\gamma$$
It's derivation can be found here.

Letting $\ln(\sin{\theta})=-x$ and using $\mathcal{I}$ to denote the integral in question,
\begin{align}
\mathcal{I}
=&\frac{1}{2}\int^\infty_0\ln(1-e^{-2x})\ln\left(\frac{x^2}{\pi^2+x^2}\right)\ {\rm d}x\\
=&-\frac{\partial}{\partial a}\Bigg{|}_{a=0}\sum^\infty_{n=1}\frac{1}{n}\int^\infty_0x^ae^{-2nx}\ {\rm d}x+\frac{1}{2}\sum^\infty_{n=1}\frac{1}{n}\int^\infty_0e^{-2nx}\ln(\pi^2+x^2)\ {\rm d}x\\
=&-\frac{\partial}{\partial a}\Bigg{|}_{a=0}\frac{\Gamma(a+1)\zeta(a+2)}{2^{a+1}}+\frac{1}{2}\sum^\infty_{n=1}\frac{\ln{\pi}}{n^2}+\frac12\sum^\infty_{n=1}\frac{1}{n^2}\int^\infty_0\frac{xe^{-2nx}}{\pi^2+x^2}{\rm d}x\\
=&-\frac{1}{2}\Gamma'(1)\zeta(2)-\frac{1}{2}\Gamma(1)\zeta'(2)+\frac{1}{2}\Gamma(1)\zeta(2)\ln{2}+\frac{1}{2}\zeta(2)\ln{\pi}\\
&+\frac{1}{2}\sum^\infty_{n=1}\frac{1}{n}\int^\infty_0\int^\infty_0e^{-2nx}e^{-xy}\cos{\pi y}\ {\rm d}x\ {\rm d}y\\
=&\frac{\pi^2}{12}\left(\gamma+\ln{2\pi}\right)-\frac{1}{2}\zeta'(2)+\frac12\sum^\infty_{n=1}\frac{1}{n^2}\int^\infty_0\frac{\cos{\pi y}}{y+2n}{\rm d}y\\
=&\frac{\pi^2}{12}\left(\gamma+\ln{2\pi}\right)-\frac{1}{2}\zeta'(2)+\frac12\sum^\infty_{n=1}\frac{1}{n^2}\int^\infty_0\frac{\cos{(y+2n\pi)}}{y+2n\pi}{\rm d}y\\
=&\frac{\pi^2}{12}\left(\gamma+\ln{2\pi}\right)-\frac{1}{2}\zeta'(2)+\frac12\sum^\infty_{n=1}\frac{1}{n^2}\int^\infty_{2n\pi}\frac{\cos{y}}{y}{\rm d}y\\
=&\frac{\pi^2}{12}\left(\gamma+\ln{2\pi}\right)-\frac{1}{2}\zeta'(2)+\frac12\sum^\infty_{n=1}\frac{1}{n^2}\left[\left(\int^\infty_0-\int^{2n\pi}_0\right)\frac{\cos{y}-1}{y}{\rm d}y+\ln{y}\Bigg{|}^\infty_{2n\pi}\right]\\
=&\color{grey}{\frac{\pi^2}{12}\left(\gamma+\ln{2\pi}\right)-\frac{1}{2}\zeta'(2)}+\frac12\sum^\infty_{n=1}\frac{1}{n^2}\int^{2n\pi}_0\frac{1-\cos{y}}{y}{\rm d}y\color{grey}{-\frac{1}{2}\sum^\infty_{n=1}\frac{\ln(2n\pi)}{n^2}}\\
&+\color{grey}{\frac{1}{2}\lim_{N\to\infty}\sum^\infty_{n=1}\frac{1}{n^2}\left[\ln{N}-\sum^\infty_{k=1}\frac{(-1)^{k+1}}{2k(2k)!}N^{2k}\right]}\\
=&\frac{1}{2}\sum^\infty_{n=1}\frac{1}{n^2}\int^1_0\frac{1-\cos{2n\pi y}}{y}{\rm d}y\\
=&\frac{1}{2}\int^1_0\left(\frac{\pi^2}{6y}-\pi^2y+\pi^2-\frac{\pi^2}{6y}\right)\ {\rm d}y=\frac{\pi^2}{2}\int^1_0(1-y)\ {\rm d}y=\Large{\frac{\pi^2}{4}}
\end{align}
as was to be shown.
A: Here's a solution using complex analysis. As other people have shown, the integral is equivalent to:
$$I=\frac{1}{2}\int_0^\infty\log\bigg(\frac{x^2}{x^2+\pi^2}\bigg)\log(1-e^{-2x})dx$$
I'll integrate the function:
$$f(z)=\log z \log(1-e^{-2z})$$ 
On this contour:

I'll spare you the gory details of showing that for $\epsilon \rightarrow 0$ and $R \rightarrow \infty$ the integrals on $\gamma_2$, $\gamma_4$ and $\gamma_6$ vanish. We are left with:
$$\int_0^\infty \log x\log(1-e^{-2x})dx +\int_\infty^0\log(x+i\pi)\log(1-e^{-2x})dx+i\int_\pi^0\log (iy) \log(1-e^{-2iy})dy=0$$
I'll call the sum of the first two integrals $I_1$ and the last one $I_2$. For $I_1$ we have:
$$I_1=\int_0^\infty \log\bigg(\frac{x}{\sqrt{x^2+\pi^2}}\bigg)\log(1-e^{-2x})dx-i\int_0^\infty \arg(x+i\pi)\log(1-e^{-2x})dx$$
Of which, we can see, the real part is very interesting. Now, for $I_2$ we have
$$I_2=-i\int_0^\pi\log(iy)\log(1-e^{-2iy})dy=-i\int_0^\pi \bigg[\log y+i\frac{\pi}{2}\bigg]\bigg[\log(2-2\cos 2y)+i \arg (1-e^{-2iy})\bigg]dy$$
Of which we are interested in the real part, which is:
$$\Re I_2=\frac{\pi}{2}\int_0^\pi \log(2-2\cos 2y)dy + \int_0^\pi \log y \arg (1-e^{-2iy})dy=J_1+J_2$$ 
Let's evaluate the first one. 
$$J_1=\frac{\pi}{2}\int_0^\pi \log(2-2\cos 2y)dy=\frac{\pi}{2}\int_0^\pi \log(4\sin^2 y)dy=0$$
The last line follows from the well known integral of $\log(\sin x)$. Let's evaluate $J_2$. 
$$1-e^{-2iy}=1-\cos (2y) + i \sin (2y)= 2 \sin^2 y + 2i \sin y \cos y=2 \sin y(\sin y +i \cos y)=2i \sin y e^{-iy}$$
We can now see that, up to unimportant costants, the multiplication by $2i\sin y$ is a rotation by $\pi/2$, so in terms of argument it's equivalent to a multiplication by $e^{i\pi/2}$. So:
$$\arg(1-e^{-2iy})=\arg (e^{i(\pi/2-y)})=\frac{\pi}{2}-y$$
So $J_2$ becomes:
$$J_2=\int_0^\pi \log y\frac{\pi}{2}dy-\int_0^{\pi}y\log y dy=-\frac{\pi^2}{4}$$
The last line follows from elementary integration techniques. Putting everything together we have:
$$\Re I_1=-\Re I_2$$
$$\int_0^\infty \log\bigg(\frac{x}{\sqrt{x^2+\pi^2}}\bigg)\log(1-e^{-2x})dx = \frac{\pi^2}{4} $$
So, multiplying by $2$ and remembering the original $1/2$ we have: 
$$I=\frac{\pi^2}{4}$$
As desired.
A: First, let $-\log(\sin(\theta))=y$ that yields 
$$\underbrace{\int_0^{\infty} \log(y) \log(1-e^{-2 y}) \ dy}_{\displaystyle \pi^2 \log(A)-\frac{1}{12}\pi^2 \log(\pi)}-\frac{1}{2}\underbrace{\int_0^{\infty} \log(\pi^2+y^2) \log(1-e^{-2 y}) \ dy}_{\displaystyle\sum_{k=1}^{\infty} \frac{\operatorname{Ci}(2k \pi)}{k^2} -\sum_{k=1}^{\infty}\frac{\log(\pi)}{k^2}}$$
where the series result is got by combining the series of $\log(1-e^{-2 y})$ and the exponential integral.
Making use of the $1.22 a$ from http://arxiv.org/pdf/1008.0040.pdf, we conclude that 
$$\begin{equation}\large{\int_0^{\Large\frac{\pi}{2}}}\ln\left(\frac{\ln^2\sin\theta}{\pi^2+\ln^2\sin\theta}\right)\,\frac{\ln\cos\theta}{\tan\theta}\,d\theta=\frac{\pi^2}{4}\end{equation}$$
Q.E.D. (note I only used well-known old results - isn't it too easy for a contest?)
