TITLE: Generalised Integral $I_n=\int_0^{\pi/2} \frac{x^n}{\sin ^n x} \ \mathrm{d}x, \quad n\in \mathbb{Z}^+.$
QUESTION [22 upvotes]: I have this integral,  $$I_n=\displaystyle \int_0^{\pi/2} \frac{x^n}{\sin ^n x} \ \mathrm{d}x, \qquad n\in \mathbb{Z}^+.$$
We have the results
$$
\begin{align}
I_1 & = 2C, \\
I_2 &= \pi\log 2, \\
I_4 & = -\frac{\pi^3}{12} + 2\pi\log 2 + \frac{\pi^3}{3}\log 2-\frac{3\pi}{2}\zeta(3),
\end{align} 
$$
where $C$ is Catalan's constant. Can we prove any of these results, or make any progress on $I_3$, or the general case?

REPLY [14 votes]: For $n\in\mathbb{N}$ we have: 

$$\int\limits_0^{\pi/2}\bigg(\frac{x}{\sin x}\bigg)^ndx=\sum\limits_{j=0}^{n-1}2^{n-j}\begin{bmatrix}n\\j+1\end{bmatrix}\sum\limits_{v=0}^j\binom{j}{v}(-n)^{j-v+1}\sum\limits_{l=1}^{\big\lfloor\frac{n+1}{2}\big\rfloor}\frac{(-1)^l\Big(\tfrac{\pi}{2}\Big)^{n-2l+1}}{(n-2l+1)!}f_n(2l-v),$$

with the Stirling numbers of the first kind $\begin{bmatrix}n\\k\end{bmatrix}$ defined by $\displaystyle\sum\limits_{k=0}^n\begin{bmatrix}n\\k\end{bmatrix}x^k:=\prod\limits_{k=0}^{n-1}(x+k),$ 
with $f_{2m-1}(s):=(-1)^{m-1}\beta(s),$ where $\beta(s)$ := Dirichlet $\beta$ function, 
and $f_{2m}(s):=(-1)^{m-1}2^{-s}\eta(s),$ where $\eta(s)$ := Dirichlet $\eta$ function, 
for $m\in\mathbb{N}$, with the analytical extensions $(s\in\mathbb{C})$ 

$$\begin{align}
\beta(1-s)&=\bigg(\dfrac{2}{\pi}\bigg)^s\sin\bigg(\dfrac{\pi s}{2}\bigg)~\Gamma(s)~\beta(s)
\\\\
\eta(1-s)&=\dfrac{2^s-1}{1-2^{s-1}}~\pi^{-s}\cos\bigg(\dfrac{\pi s}{2}\bigg)~\Gamma(s)~\eta(s)
\end{align}$$

and with the simplifications $(k\in\mathbb{N}_0)$ 

$$\begin{align}
\beta(-2k-1)~&=~0
\\\\
\beta(-2k)~&=~-\frac{2^{4k+1}}{2k+1}~B_{2k+1}\bigg(\frac{1}{4}\bigg)
\\\\
\beta(2k+1)~&=~(-1)^{k-1}~\frac{(2\pi)^{2k+1}}{2(2k+1)!}~B_{2k+1}\bigg(\frac{1}{4}\bigg)
\\\\
\eta(1-k)~&=~\frac{2^k-1}{k}~B_k
\\\\
\eta(2k)~&=~(-1)^{k-1}~\frac{2^{2k-1}-1}{(2k)!}~B_{2k}~\pi^{2k}
\\\\
\eta(2k+1)~&=~\bigg(1-\frac{1}{2^{2k}}\bigg)\zeta(2k+1)
\end{align}$$


Examples include: 

$$\begin{align}
\int\limits_0^{\pi/2}\bigg(\frac{x}{\sin x}\bigg)^1dx~&=~2\beta(2)\approx1.83
\\\\
\int\limits_0^{\pi/2}\bigg(\frac{x}{\sin x}\bigg)^2dx~&=~\pi\ln2\approx2.178
\\\\
\int\limits_0^{\pi/2}\bigg(\frac{x}{\sin x}\bigg)^3dx~&=~-6\beta(4)+\bigg(\frac{3}{4}\pi^2+6\bigg)\beta(2)-\frac{3}{8}\pi^2\approx2.64
\\\\
\int\limits_0^{\pi/2}\bigg(\frac{x}{\sin x}\bigg)^4dx~&=~-\frac{3}{2}\pi\zeta(3)+\bigg(\frac{\pi^3}{3}+2\pi\bigg)\ln2-\frac{\pi^3}{12}\approx3.27
\\\\
\int\limits_0^{\pi/2}\bigg(\frac{x}{\sin x}\bigg)^5dx~&=~90\beta(6)-\bigg(\frac{45}{4}\pi^2+100\bigg)\beta(4)+\bigg(\frac{45}{192}\pi^4+\frac{25}{4}\pi^2+10\bigg)\beta(2)-
\\
&-\bigg(\frac{55}{384}\pi^4+\frac{5}{8}\pi^2\bigg)\approx4.135
\end{align}$$