Special Functions defined by Integrals.

Question

There was this integral which caught my attention, when I was checking out The Applications of Beta and Gamma Functions. So, how can i prove the below, using change of variable?

$$\int_{0}^{1}\frac{dx}{\sqrt{1-x^n}} = \frac{\sqrt{\pi}}{n}\frac{\Gamma(n^{-1})}{\Gamma(n^{-1}+\frac{1}{2})}$$

Thanks a lot, in advance.

Answer 1

Start with $$I(a,b)=\int_0^1 \frac{dx}{(1-x^a)^b}$$ Then let $x=u^{1/a}$: $$I(a,b)=\frac1a\int_0^1 u^{1/a-1}(1-u)^{-b}du$$ Which we immediately recognize as $$I(a,b)=\frac{\Gamma(\frac1a)\Gamma(1-b)}{a\Gamma(\frac1a-b)}$$ The result in question follows from the choice of $b=\frac12$ and $a=n$.

Answer 2

Corrected Answer:

\begin{align*} \int_0^1 \frac{dx}{(1-x^a)^b} &= \int_0^1 \frac{u^{\frac{1}{a}-1}}{a (1-u)^b} \, du\\ &=\frac{1}{a}\int_0^1 u^{\frac{1}{a}-1} (1-u)^{(1-b)-1}du\\ &=\frac{1}{a}B\left(\frac{1}{a},1-b\right),\qquad a>0,~~b<1 \end{align*}