Contour Integration $\int_0^1\frac1{\sqrt[n]{1-x^n}}dx$ I want to compute:
$$\int^{1}_{0}\frac{1}{\sqrt[n]{1-x^n}}dx$$
for natural $n>1$ using Residue Calculus.
I am thinking of using some kind of a keyhole or bone contour that could go around the $n$th roots of unity (singularities in this case). The problem is I believe it is not clear how to define a suitable branch (or branches) of $\log$ in this region for it to work, also considering we only care about the segment from $0$ to $1$.

From Omran Kouba's answer, we have
\begin{eqnarray}
I_n&=&\frac{1}{n}\int_{-\infty}^\infty \frac{e^{t/n}}{1+e^t}dt=\frac{1}{n}\int_{0}^\infty \frac{e^{t/n}}{1+e^t}dt+\frac{1}{n}\int_{-\infty}^0 \frac{e^{t/n}}{1+e^t}dt\\
&=&\frac{1}{n}\int_{0}^\infty \frac{e^{t/n-1}}{1+e^{-t}}dt+\frac{1}{n}\int_{-\infty}^0 \frac{e^{t/n}}{1+e^t}dt\\
&=&\frac{1}{n}\int_{0}^\infty \sum_{k=0}^\infty(-1)^ke^{t/n-t-kt}dt+\frac{1}{n}\int_{-\infty}^0 \sum_{k=0}^\infty(-1)^ke^{t/n+kt}dt\\
&=&\sum_{k=0}^\infty(-1)^k\frac{1}{nk+1}+\sum_{k=0}^\infty(-1)^k\frac{1}{nk+n-1}\\
&=&\sum_{k=-\infty}^\infty(-1)^k\frac{1}{nk+1}=\frac{\pi}{n\sin\frac{\pi}{n}}.
\end{eqnarray}
