$$I_n=\int_0^{\pi/2}x\log^n(\tan x)\ dx,\quad n\in\mathbb{Z}$$ I want to evaluate this log-tangent integral using exponential generating functions. My attempt is below.
Define, $$G(t)=\sum_{n=0}^\infty\frac{I_n}{n!}t^n=\int_0^{\pi/2}x\sum_{n=0}^\infty\frac{(t\log(\tan x))^n}{n!}\ dx=\int_0^{\pi/2}x\tan^{t}(x)\ dx$$ rewriting $x$ as $\arcsin(\sin(x))$ and expanding by its Maclaurin series, $$\int_0^{\pi/2}x\tan^{t}(x)\ dx=\int_0^{\pi/2}\arcsin(\sin(x))\frac{\sin^t(x)}{\cos^{t}(x)}\ dx=\int_0^{\pi/2}\sum_{n=0}^\infty\binom{2n}{n}\frac{\sin^{2n+1}(x)}{4^n(2n+1)}\frac{\sin^t(x)}{\cos^{t}(x)}\ dx\\ =\sum_{n=0}^\infty\binom{2n}{n}\frac{4^{-n}}{(2n+1)}\int_0^{\pi/2}\sin^{2n+t+1}(x)\cos^{-t}(x)\ dx=\sum_{n=0}^\infty\binom{2n}{n}\frac{4^{-n}}{2(2n+1)}B\left(\frac{2n+t+2}{2},\frac{1-t}{2}\right)$$ where $B(m,n)$ is the Beta function, $$B(m,n)=\frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}$$ hence, $$G(t)=\sum_{n=0}^\infty\binom{2n}{n}\frac{4^{-n}}{2(2n+1)}\frac{\Gamma\left(n+t/2+1\right)\Gamma(1/2-t/2)}{\Gamma(n+3/2)}.$$ Now I need to extract the coefficient of $t^n/n!$ as, $$I_n=[t^n/n!]G(t)$$ but I am having difficulties doing so. If I can get a closed form for $G(t)$ I may be able to take its logarithmic derivative and use the Cauchy product, but I don't have a closed form either.
Any ideas? Thanks in advance.