Any hints on this following type of integral: $$I(j) = \int_0^1 \dfrac{\cos{(\frac{\pi}{2} j x)}}{\sin{(\frac{\pi}{4}(1-x))}} \log{\Big( \frac{\sin{(\frac{\pi}{4})}}{\sin{(\frac{\pi x}{4})}} \Big)}dx,$$
where $j \in \mathbb{Z}^+$. For odd $j$'s Mathematica has no trouble finding both analytic and numerical solution. For even ones it complains about singularities.
Example solutions found by Mathematica: $$ I(1) = -\dfrac{2 \sqrt{2} (-2 + \log{(8 (\tan{(\frac{\pi}{8})})^2})}{\pi}$$
$$I(3) = \dfrac{2 (720 - 542 \sqrt{2} + 765 \sqrt{2} \log{2} + 510 \sqrt{2} \log{(\tan{\frac{\pi}{8})}}}{225 \pi}$$
$$ I(5) = - \dfrac{420000 - 448804 \sqrt{2} + 655830 \sqrt{2} \log{2} + 437220 \sqrt{2} \log{(\tan{\frac{\pi}{8}}})}{99225 \pi} $$
I'm interested only in odd $j$ cases.
Obviously there exists a closed form solution to this integral given in terms of a product or a sum that depends on the $j$ variable. Also asymptotic solution for $j \rightarrow \infty$ looks something like $\sim 1/j$ judging from numerics.
My attempts so far:
Integration by parts (singularity problems)
Using identity as in link (results in sums over hypergeometric functions)
Using Feynmanesque trick knowing that $ \partial_{x} a^x = a^{x} \ln{(a)}$ as in link (a lot of seemingly non-trivial trigonometric integrals)
Bottom line question is: Is there a simple closed form solution for this integral for general odd $j$?
Any help would be welcome,
sofista137
