How to derive the integral $ I = \int^{\infty}_{0} 1-\left(\frac{x^a}{c + x^a}\right)^ndx, n\in \mathbb{Z}^+$ $a,c \in \mathbb{R}$ and $a > 1, c > 0$.
I already know the answer includes $\Gamma(n+1/a)\Gamma(1-1/a)/\Gamma(n)$. But I couldn't get how integral like this can be expressed by gamma function since gamma function includes exponential function ($\int_0^\infty u^a e^{-u}\:du=\Gamma(a+1)$). What is the methodology to deal with integral like that and express it in gamma function?
I have checked answers in Expressing Integrals in terms of gamma functions however, does not apply to my problem perfectly.
 A: I finally crack this. Thanks for the help from J.G. and Elliptic Curve.
With substitution $t = cx^{-a}$, we can have
\begin{align}
I 
&= \frac{c^\frac{1}{a}}{a}\int^{\infty}_{0} \left(1 - \left(\frac{1}{t+1}\right)^n\right) t^{-\frac{1}{a}-1} dt\\
&= \frac{c^\frac{1}{a}}{a}\int^{\infty}_{0} \left(1 - \left(\frac{1}{t+1}\right)^n\right) \left(-at^{-\frac{1}{a}}\right)^{'} dt\\
&\overset{(a)}= \frac{c^\frac{1}{a}}{a} (-at^{-\frac{1}{a}})\frac{(1+t)^n-1}{(1+t)^n}\Bigg\vert_{0}^{\infty} - \frac{c^\frac{1}{a}}{a}\int^\infty_0 n(1+t)^{-n-1}(-at^{-\frac{1}{a}})dt\\
\end{align}
(a) is given by integral by parts. Next, we have
$$
\lim_{t \to 0} \frac{(1+t)^n-1}{t^{1/a}(1+t)^n} =  \lim_{t \to 0} \frac{n(1+t)^{n-1}}{\frac{1}{a}t^{1/a-1}(1+t)^n + n(1+t)^{n-1}t^{1/a}} = 0
$$
The equivalence is based on L'Hôpital's rule and the fact that $a>1$ respectively. And it is easy to see
$$
\lim_{t \to \infty} \frac{(1+t)^n-1}{t^{1/a}(1+t)^n} = 0
$$
Then we can have
$$
I = \frac{c^\frac{1}{a}}{a}\int^\infty_0 n(1+t)^{-n-1}(at^{-\frac{1}{a}})dt
$$
According to the formula of Beta function
$$
B(x,y) = \int^{\infty}_{0} \frac{t^{x-1}}{(1+t)^{x+y}}dt
$$
We have $x= 1-\frac{1}{a}$ and $y = n-1+\frac{1}{a}$. Thus
\begin{align}
I &= nc^{\frac{1}{a}}B(1-\frac{1}{a},n + \frac{1}{a})\\
&\overset{(b)}= nc^{\frac{1}{a}}\frac{\Gamma(1-\frac{1}{a})\Gamma(n +\frac{1}{a})}{\Gamma(n+1)}\\
&\overset{(c)}=c^{\frac{1}{a}}\frac{\Gamma(1-\frac{1}{a})\Gamma(n +\frac{1}{a})}{\Gamma(n)}
\end{align}
where (b) is given by $B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$ and (c) is given by $\Gamma(n)n = \Gamma(n+1)$.
