Integral of the nth root of csc(x)

We have (the integral only exists for $n>1$ ) $\int_0^{\frac12\pi}\csc^{\frac1n}\theta\,d\theta \; = \; \tfrac12B(\tfrac12 - \tfrac{1}{2n},\tfrac12) \; = \; \frac{\Gamma(\frac12-\frac{1}{2n})\sqrt{\pi}}{2\Gamma(1 - \frac{1}{2n})} \; = \; -\frac{\sqrt{\pi} n \Gamma(\frac12 - \frac{1}{2n})}{\Gamma(-\frac{1}{2n})}$ and since $\Gamma(-\tfrac1n) = \tfrac{1}{\sqrt{2\pi}}2^{-\frac12-\frac1n}\Gamma(-\tfrac{1}{2n})\Gamma(\tfrac12 - \tfrac{1}{2n})$ we deduce that the integral is equal to $- \frac{n\pi 2^{1 + \frac1n}\Gamma(-\frac1n)}{\Gamma(-\frac{1}{2n})^2} \; = \; \frac{n^2\pi 2^{1+\frac1n}\Gamma(1-\frac1n)}{\Gamma(-\frac{1}{2n})^2}$ Thus $\alpha=\pi$ , $\beta=2$ , $\gamma=-1$ , $\delta=2$ , $\epsilon=-2$ , giving the answer $\boxed{6\pi+2}$ .