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$a,c \in \mathbb{R}$ 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.

LOREY CHU
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    Actually it seems the integral does not converge. Are you sure you wrote it well? – Enrico M. Jan 30 '22 at 16:21
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    Have you substituted $x=\left(\frac{cy}{1-y}\right)^{1/a}$? – J.G. Jan 30 '22 at 16:59
  • @J.G. I have not. What does that do? – LOREY CHU Jan 30 '22 at 18:00
  • @EllipticCurve Really? At least for the special case n=1 it is convergent, right? – LOREY CHU Jan 30 '22 at 18:02
  • "What does that do?" Try it and see. You know enough about Beta and Gamma functions. – J.G. Jan 30 '22 at 18:07
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    @LOREYCHU If $n = 1$ we obtain the following result: $$\fbox{$\frac{\pi \left(\frac{1}{c}\right)^{-1/a} \csc \left(\frac{\pi }{a}\right)}{a}\text{ if }a>1\land \left(\Re\left((-c)^{1/a}\right)\leq 0\lor (-c)^{1/a}\notin \mathbb{R}\right)$}$$ So it's subject to conditions on the parameters $a$ and $c$. You did not specify what $a$ and $c$ are... – Enrico M. Jan 30 '22 at 18:15
  • @EllipticCurve Sorry, you are right. $a$ is bigger than 1, and c is a positive real number. – LOREY CHU Jan 30 '22 at 18:33
  • @J.G. Thank you. With this substitution, I have something includes $\int^1_0(1-y)^{-1-1/a}\hspace{5pt}y^{-1+1/a}\hspace{5pt}(1-y^n)dy$ which unfortunately does not apply to any beta function formula to the best of my knowledge. – LOREY CHU Feb 02 '22 at 14:46
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    @LOREYCHU It's the difference of two Beta integrals, one from the $1$ factor in $1-y^n$, the other from the $y^n$ factor. (You forgot the $dy$, by the way.) – J.G. Feb 02 '22 at 14:48
  • @J.G. Wow! Thank you so much, I actually spend last two days on that ; ( – LOREY CHU Feb 02 '22 at 14:50
  • @J.G. Sorry, I have to bother you again. Actually the integral when you take 1 factor in $1-y^n$ seems ill. If we represent it by beta function it will be $B(1/a,-1/a) = \Gamma(1/a)\Gamma(-1/a)/\Gamma(0)$. while $\Gamma(0)$ is not defined. – LOREY CHU Feb 02 '22 at 18:48
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    That's why @EllipticCurve hinted you should delete the $1-$ at the start of the integrand. The resulting integral, say $J$, is finite. But then $I+J=\int_0^\infty dx=\infty$. – J.G. Feb 02 '22 at 18:59

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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 $\Gamma(n)n = \Gamma(n+1)$.

LOREY CHU
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