closed form for $$\int_{0}^{\frac{\pi}{2}}\frac{x^m}{\sin x}\ dx$$
I slove it for some m but in general i failed.
I tried by part , by substitution,by using $\sin x =\frac{e^{ix}-e^{-ix}}{2i}$ .
I guess that $\sin x =\frac{e^{ix}-e^{-ix}}{2i}$ then using geometric series lead to the answer,I used it and got something it seems related to Zeta function.
can any one solve it using residue theorem or using real analysis ?
Well that's just a start, but I don't like this sin, so I tried this: I looked for a primitive of $\frac{1}{\sin(x)}$, which is $\ln(\tan(\frac{x}{2}))$ if $0 < x < \frac{\pi}{2}$, and I used the change of variable :
$$ u = \ln\left(\tan\left(\frac{x}{2}\right)\right)\rightarrow du = \frac{dx}{\sin(x)}$$
$$x = 0 \rightarrow u = -\infty ;\quad x=\frac{\pi}{2} \rightarrow u=0 $$
So, if I denote $I_m$ your integral, one gets:
$$ I_m = \int_{-\infty}^0 \left({2\arctan(e^u)}\right)^m du = 2^m \int_{-\infty}^0 \left({\arctan(e^u)}\right)^m du $$
If you let : $t= e^u \rightarrow dt = e^u du,\ u = -\infty \rightarrow t = 0 ; $ $ u=0 \rightarrow t=1 $ , you get:
$$I_m = 2^m\int_{0}^1 \frac{{\arctan^m(t)}}{t} dt $$
This one is easier to treat I think, maybe the last change of variable was not the best. I'll edit if I can finish this
