Htykuut

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

$$int_0^{pi/4}frac{log(1-x) tan^2(x)}{1-xtan^2(x)} dx$$

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

$$int_0^{pi/4}frac{log(1-x)log(x)log(1+x) tan^2(x)}{1-xtan^2(x)} dx$$

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

$$intlimits_0^{pi/4} frac{ln(1-x) tan^2 x}{1-xtan^2 x} dx = -sumlimits_{n=0}^infty sumlimits_{k=0}^n frac{1}{n-k+1} intlimits_0^{pi/4} x^{n+1} (tan x)^{2k+2} dx$$

Or using $intlimits_0^{pi/4} frac{ln(1-x)}{x (1-xtan^2 x)} dx$ it becomes a little more handsomely:

$$ intlimits_0^{pi/4} frac{ln(1-x) tan^2 x}{1-xtan^2 x} dx = text{Li}_2left(frac{pi}{4}right) -sumlimits_{n=0}^infty sumlimits_{k=0}^n frac{1}{n-k+1} intlimits_0^{pi/4} x^n (tan x)^{2k} dx$$

Then we have: $$ intlimits_0^{pi/4} x^n (tan x)^{2k} dx = (-1)^k left(frac{pi^{n+1}}{(n+1)4^{n+1}} + A_{n,k} - B_{n,k} - C_{n,k} right) enspace$$ for:

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

and the Stirling numbers of the first kind $begin{bmatrix}n\kend{bmatrix}$ defined by: $$sumlimits_{k=0}^nbegin{bmatrix}n\kend{bmatrix}x^k:=prodlimits_{k=0}^{n-1}(x+k),$$

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

$$ sumlimits_{n=0}^inftyfrac{pi^{n+1}}{(n+1) 4^{n+1}}sumlimits_{k=0}^n frac{(-1)^k}{n-k+1} = \ text{Li}_2left(frac{pi}{4}right) + text{Li}_2 left(frac{1}{2} -frac{pi}{8} right) + left(lnleft(frac{1}{2} +frac{pi}{8} right)right)lnleft(1-frac{pi}{4}right)$$

we get:

$$ intlimits_0^{pi/4} frac{ln(1-x) tan^2 x}{1-xtan^2 x} dx = \
-text{Li}_2 left(frac{1}{2} -frac{pi}{8} right) - left(lnleft(frac{1}{2} +frac{pi}{8} right)right)lnleft(1-frac{pi}{4}right) - sumlimits_{n=0}^infty sumlimits_{k=0}^n frac{(-1)^k}{n-k+1}(A_{n,k} - B_{n,k} - C_{n,k} )$$