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I am interested in learning how to obtain the result

$$\int_0^{+\infty} z^{c-2} {}_2 F_1(a,b;c;-z) \mathrm{d}z = \frac{\Gamma(1 + a - c) \Gamma(1 + b - c) \Gamma(-1 + c) \Gamma(c)}{\Gamma(a) \Gamma(b)},$$

which I currently got from Mathematica. I was able to play around with the series expression for ${}_2 F_1$ to obtain $$\int_0^{x} z^{c-2} {}_2 F_1(a,b;c;-z) \mathrm{d}z = \frac{x^{c-1}}{c-1} {}_3 F_2(a,b,c-1;c,c;-x),$$ but I wasn't able to make any more progress, since I don't really know how to take the limit $x \to +\infty$ at this stage.

Níckolas Alves
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    See here. – Random Variable Mar 08 '23 at 04:21
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    Note: the series expansion for ${}_2 F_1(a,b;c;-z)$ usually has radius of convergence $1$. So you cannot use it to do this integral from $0$ to $\infty$. So you will need to use the analytic continuation (also a hypergeometric) to make sense of the integral. – GEdgar Mar 08 '23 at 13:28
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    @GEdgar True. But Ramanujan's Master Theorem is still applicable because the proof of the theorem doesn't assume that the series has an infinite radius of convergence. – Random Variable Mar 08 '23 at 15:12

1 Answers1

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You probably noticed that Mathematica returns an antiderivative without any restriction $$\int z^{c-2} \, _2F_1(a,b;c;-z)\,dz=\frac {z^{c-1}}{c-1}\, _3F_2(a,b,c-1;c,c;-z)$$ For the integral $$I=\int_0^x z^{c-2} \, _2F_1(a,b;c;-z)\,dz=\frac {x^{c-1}}{c-1}\, _3F_2(a,b,c-1;c,c;-x)$$ the result is valid if $$\Re(c)>1\land (\Re(x)>-1\lor x\notin \mathbb{R})$$ Using series expansion for large $x$ $$\, _3F_2(a,b,c-1;c,c;-x)=x^{1-c}\,\frac{ \Gamma (c)^2 \,\,\Gamma (a-c+1)\,\, \Gamma (b-c+1)}{\Gamma(a) \,\,\Gamma (b)}-$$ $$x^{c-2} \,\Gamma (c)\,(A+B)+O\left(\frac{1}{x^2}\right)$$ where $$A=x^{-a} \frac{\Gamma (-a+b-1) \left(a (a-c+1)^2-x (a-b+1) (a-c+2)\right)}{(a-c+1) (a-c+2) \Gamma (b) \Gamma (c-a)}$$ $$B=x^{-b} \frac{\Gamma (a-b-1) \left(x (a-b-1) (b-c+2)+b (b-c+1)^2\right)}{\Gamma (a) (b-c+1) (b-c+2) \Gamma (c-b)}$$

So, the result is for $x \to \infty$ $$\color{blue}{\frac{\Gamma(c)\,\Gamma (c-1) \,\Gamma (a-c+1) \,\Gamma (b-c+1)}{\Gamma (a)\,\Gamma (b)}}$$ which is your formula.

Claude Leibovici
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