TITLE: Evaluate $\int_0^{\frac{π}{2}}\tan (x)\ln (\sin x)\ln (\cos x)dx$
QUESTION [6 upvotes]: Evaluate: 
  $$I=\int_0^{\frac{π}{2}}\tan (x)\ln (\sin x)\ln (\cos x)dx$$

My ideas is to use the Fourier series of log sin and log cos: 
$$\ln (2\sin x)=-\sum_{k=1}^{\infty}\frac{\cos (2kx)}{k}$$
$$\ln (2\cos x)=-\sum_{k=1}^{\infty}\frac{(-1)^{k}\cos (2kx)}{k}$$ 
But my problem is that I find difficult integrals like: 
$$\int\tan (x)\cos (2kx)dx$$
My another idea is: 
Use the substation : $y=\tan x$ then $dx=\frac{dy}{1+y^2}$ 
Then where $x=0 \Rightarrow y=0$ and for $x=\frac{π}{2} \Rightarrow y=\infty$ 
So: 
$$I=\frac{1}{2}\int_0^{\infty}\frac{y\ln \left(\frac{y}{\sqrt{1+y^2}}\right)\ln (1+y^2)}{1+y^2}dy$$ 
But now I don't know how to complete.

REPLY [2 votes]: We can actually use the Beta Function with this one. Here we will address your integral:
\begin{equation}
I = \int_0^\frac{\pi}{2} \tan(x) \ln(\sin(x))\ln(\cos(x))\:dx
\end{equation}
We note that:
\begin{equation}
\tan(x) = \sin(x)\cos^{-1}(x),\:\:\lim_{a \rightarrow 0^+} \frac{d}{da} \sin^a(x) = \ln(\sin(x)),\:\:\lim_{b \rightarrow 0^+} \frac{d}{db} \cos^b(x) = \ln(\cos(x))
\end{equation}
Thus $I$ becomes:
\begin{equation}
I = \int_0^\frac{\pi}{2} \sin(x)\cos^{-1}(x) \cdot \lim_{a \rightarrow 0^+} \frac{d}{da} \sin^a(x) \cdot \lim_{b \rightarrow 0^+} \frac{d}{db} \cos^b(x)\:dx
\end{equation}
By the Dominated Convergence Theorem and  Leibniz's Integral Rule:
\begin{align}
I &= \lim_{a \rightarrow 0^+} \frac{d}{da}  \lim_{b \rightarrow 0^+} \frac{d}{db}  \int_0^\frac{\pi}{2} \sin(x)\cos^{-1}(x) \sin^a(x) \cos^b(x)\:dx \nonumber \\ 
&= \lim_{(a,b)\rightarrow (0,0)^+} \frac{\partial^2}{\partial a \partial b} \int_0^\frac{\pi}{2} \sin^{a + 1}(x)\cos^{b - 1}(x)\:dx \nonumber \\
&= \lim_{(a,b)\rightarrow (0,0)^+} \frac{\partial^2}{\partial a \partial b} \left[\frac{1}{2}B\left(\frac{a + 2}{2}, \frac{b}{2} \right) \right]
\end{align}
Where $B(\cdot, \cdot)$ is the Beta Function
From here convert to the Gamma Representation, apply the derivatives and the evaluate at $(a,b) = (0,0)$ and you're done!

EDIT - This doesn't work. The Beta Function is undefined for $b = 0$.