The closed form of ∫π/40log(1−x)tan2(x)1−xtan2(x) dx\int_0^{\pi/4}\frac{\log(1-x) \tan^2(x)}{1-x\tan^2(x)} \ dx

What tools or ways would you propose for getting the closed form of this integral?

π/40log(1x)tan2(x)1xtan2(x) dx

EDIT: It took a while since I made this post. I’ll give a little bounty for the solver of the problem, 500 points bounty.

Supplementary question:

Calculate

π/40log(1x)log(x)log(1+x)tan2(x)1xtan2(x) dx

Answer

Just a few notes for a series development, because a “closed” formula is very unlikely.

π/40ln(1x)tan2x1xtan2xdx=n=0nk=01nk+1π/40xn+1(tanx)2k+2dx

Or using π/40ln(1x)x(1xtan2x)dx it becomes a little more handsomely:

π/40ln(1x)tan2x1xtan2xdx=Li2(π4)n=0nk=01nk+1π/40xn(tanx)2kdx

Then we have: π/40xn(tanx)2kdx=(1)k(πn+1(n+1)4n+1+An,kBn,kCn,k) for:

A_{n,k} := \left(\sin \frac{\pi n}{2}\right)\frac{n!}{2^{n+1}} \sum\limits_{~j=0 \\ j~\text{odd}}^{2k-1} c_{2k-1,j}~\eta(n-j+1)\\
B_{n,k} := \frac{1}{2^{2n+2}}\sum\limits_{~v=0 \\ v~\text{odd}}^n \left(\sin \frac{\pi v}{2}\right)\frac{n!\pi^{n-v}}{(n-v)!} \sum\limits_{~j=0 \\ j~\text{odd}}^{2k-1} 2^j c_{2k-1,j}~\eta(v-j+1)\\
C_{n,k} := \frac{1}{2^{2n+1}}\sum\limits_{~v=0 \\ v~\text{even}}^n \left(\cos \frac{\pi v}{2}\right)\frac{n!\pi^{n-v}2^v}{(n-v)!} \sum\limits_{~j=0 \\ j~\text{odd}}^{2k-1} c_{2k-1,j}~\beta(v-j+1)\\
c_{n,j} := \frac{1}{n!} \sum\limits_{v=0}^{n+1} {\binom {n+1} v} \sum\limits_{l=j}^n \begin{bmatrix}{n+1}\\{l+1}\end{bmatrix}{\binom l j}(-v)^{l-j}

and the Stirling numbers of the first kind \begin{bmatrix}n\\k\end{bmatrix} defined by: \sum\limits_{k=0}^n\begin{bmatrix}n\\k\end{bmatrix}x^k:=\prod\limits_{k=0}^{n-1}(x+k),

so that c_{ \text{odd},\text{even} }=0 and c_{ \text{even},\text{odd} }=0. Some values \,c_{n,j}\, can be seen here .

The needed analytical continuation for Dirichlet eta function \eta(s) and Dirichlet beta function \beta(s) can be seen in my answer of the question here . With the additional information

\sum\limits_{n=0}^\infty\frac{\pi^{n+1}}{(n+1) 4^{n+1}}\sum\limits_{k=0}^n \frac{(-1)^k}{n-k+1} = \\ \text{Li}_2\left(\frac{\pi}{4}\right) + \text{Li}_2 \left(\frac{1}{2} -\frac{\pi}{8} \right) + \left(\ln\left(\frac{1}{2} +\frac{\pi}{8} \right)\right)\ln\left(1-\frac{\pi}{4}\right)

we get:

\int\limits_0^{\pi/4} \frac{\ln(1-x) \tan^2 x}{1-x\tan^2 x} dx = \\
-\text{Li}_2 \left(\frac{1}{2} -\frac{\pi}{8} \right) – \left(\ln\left(\frac{1}{2} +\frac{\pi}{8} \right)\right)\ln\left(1-\frac{\pi}{4}\right) – \sum\limits_{n=0}^\infty \sum\limits_{k=0}^n \frac{(-1)^k}{n-k+1}(A_{n,k} – B_{n,k} – C_{n,k} )

Attribution
Source : Link , Question Author : user 1591719 , Answer Author : user90369

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