Integral of Hypergeometric Function with polynomial and exponential

Question

I was working on some mathematical derivations and faced this integral:

$$I=\int_{0}^{\infty}x^{\alpha-1}e^{-\beta x}{_2F_1}{(a,b;c;1-hx)}\,\mathrm{d}x$$

how can I integrate it?

Answer 1

Since nobody has bothered and since it's been two years I presume that any homework has expired, here is a way . Anybody who uses this should really check it for typos. I am using DLMF (http://dlmf.nist.gov/) and EMOT (http://authors.library.caltech.edu/43489/1/Volume%201.pdf) for reference. Problem:
$I={\displaystyle \intop_{0}^{\infty}}x^{α-1}⋅e^{-βx}⋅_2F_1(a,b;c;1-h⋅x)dx$
Examining the expression we see that we can use standard Laplace or Mellin tables if we reshape the Hypergeometric function to a standard form.
As suggested, Let: $x'=h\cdot x$ thus $x=\frac{x'}{h},dx=\frac{dx'}{h}$
$I=\frac{1}{h^{\alpha-1}}\cdot{\displaystyle \intop_{0}^{\infty}}x'^{\alpha-1}\cdot e^{-\frac{\beta}{h}\cdot x'}\cdot_{2}F_{1}\left(a,b;c;1-x'\right)\frac{dx'}{h}$
$I=\frac{1}{h^{\alpha}}\cdot{\displaystyle \intop_{0}^{\infty}}x'^{\alpha-1}\cdot e^{-\frac{\beta}{h}\cdot x'}\cdot_{2}F_{1}\left(a,b;c;1-x'\right)dx'$

We use DLMF 15.8.4:

$\frac{\sin ( \pi (c - a - b))}{\pi} \mathbf{F} \left( {a, b};{c} ; z \right) = \frac{1}{\Gamma ( c - a) \Gamma ( c - b)} \mathbf{F} \left( {a, b};{a + b - c + 1} ; 1 - z \right) - \frac{(1 - z)^{c - a - b}}{\Gamma ( a) \Gamma ( b)} \mathbf{F} \left( {c - a, c - b};{c - a - b + 1} ; 1 - z \right),$

Modified: $x'=1-z$

$\mathbf{F} \left({a, b};{c} ; 1 - x') = \frac{\Gamma( c - a - b) \Gamma( 1-( c - a - b))}{\Gamma(c - a)\Gamma( c - b)} \mathbf{F}\left({a, b};{a + b - c + 1} ; x' \right) - \frac{\Gamma( c - a - b)\Gamma( 1 - ( c - a - b)) (x')^{c - a - b}}{\Gamma( a) \Gamma( b)} \mathbf{F}\left({c - a, c - b};{c - a - b + 1} ; x' \right) \right.$
So we have two terms:

$I_1 = \frac{1}{h^{\alpha}} \frac{\Gamma( c - a - b) \Gamma( 1 - ( c - a - b))}{\Gamma( c -a) \Gamma( c - b)}$

$ \cdot \int_0^{\infty} x'^{\alpha - 1} \cdot e^{- \frac{\beta}{h} \cdot x'} \cdot_2 F_1 ( a, b ; a + b - c + 1 ; x'){dx}'$

$ I_2 = \frac{-1}{h^{\alpha}} \frac{\Gamma( c - a - b) \Gamma( 1 - ( c - a - b))}{\Gamma( a)\Gamma( b)} $
$\cdot \int_0^{\infty} x'^{c - a - b + \alpha - 1} \cdot e^{- \frac{\beta}{h} \cdot x'} \cdot_2 F_1 ( c - a, c - b ; c - a - b + 1 ; x') {dx}' $

Setting: $s = \frac{\beta}{h}$ and using EMOT 4.23 (17)

$I_1 = \frac{1}{h^{\alpha}} \frac{\Gamma( c - a - b) \Gamma( 1 - ( c - a - b))}{\Gamma( c - a) \Gamma( c - b)} $
$\cdot \Gamma (\alpha) \cdot (s^{-{\alpha}}) \cdot_3 F_1 \left( a, b,\alpha; a + b - c + 1 ; \frac{1}{s} \right)$

$I_2 = \frac{-1}{h^{\alpha}} \frac{\Gamma( c - a - b) \Gamma( 1 - ( c - a - b))}{\Gamma( a) \Gamma( b)} $
$\cdot \Gamma ({\alpha}+ c - a - b) \cdot (s^{- ({\alpha}+ c - a - b)}) \cdot_3 F_1 \left( c - a, c - b, ({\alpha}+ c - a - b) ; c - a - b + 1 ; \frac{1}{s} \right)$