How do you evaluate this integral $I =\int_{0}^{1}x^{m-n}(x^2-1)^n dx$ Assuming $m \geq n$   (not necessary integers)
How do I calculate this integral $I =\int_{0}^{1}x^{m-n}(x^2-1)^n dx$ ?
I have tried using binomial series and then integrated each term. 
I got $\sum_{k=0}^{\infty}\frac{1}{m+n+1-2k}\binom{n}{k}$,
but I don't know how to tackle series of this kind. 
 A: With substitution $x^2=t$ :
\begin{align}
I 
&= \int_{0}^{1}x^{m-n}(x^2-1)^n dx \\
&= \dfrac12(-1)^n\int_{0}^{1}t^{\frac12(m-n-1)}(1-t)^n dx \\
&= \dfrac12(-1)^n\beta(\dfrac{m-n+1}{2},n+1) \\
&= \dfrac12(-1)^n\dfrac{\Gamma(\dfrac{m-n+1}{2})\Gamma(n+1)}{\Gamma(\dfrac{m+n+3}{2})}
\end{align}
where $\beta(x,y)$ is Beta function.
A: Let $x=\sin(t)$ to get
$$I =\int_{0}^{1}x^{m-n}(x^2-1)^n\, dx=(-1)^n\int_{0}^{\frac \pi 2}\cos^{2n+1}(t)\sin^{m-n}(t)\,dt$$ and try reduction formulae.
You must take care that, for getting a real result, $n$ must be an integer; otherwise, the result will be a complex number (which perfectly evaluate). Fo example, using $m=\pi$ and $n=e$, using 
$$I_{m,n}=(-1)^n\frac{ \Gamma (n+1) \Gamma \left(\frac{m-n+1}{2} \right)}{2\, \Gamma
   \left(\frac{m+n+3}{2} \right)}$$ as given by MyGlasses in his/her answer, we should get 
$$I_{\pi,e}=(-1)^e \frac{ \Gamma (e+1) \Gamma \left(\frac{\pi-e+1}{2} \right)}{2\, \Gamma
   \left(\frac{\pi+e+3}{2} \right)}=(-0.633256+0.773943\, i)\times 0.258234$$
