MHB Solving the Hypergeometric Function Integral Representation

AI Thread Summary
The discussion revolves around proving the integral representation of the hypergeometric function, specifically the equation involving the Gamma function and an integral from 0 to 1. Participants suggest that demonstrating this representation solves the hypergeometric differential equation could be a valid approach. One user expresses difficulty in simplifying the integrals that arise when substituting the integral representation into the differential equation. There is acknowledgment of the singular points of the hypergeometric function, which do not appear in the integral representation. The conversation highlights the challenge of the task and the need for further exploration of the integral's properties.
alyafey22
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Prove the following

$$ {}_2 F_1 \left( a,b; c ; x \right) = \frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}\int^1_0 t^{b-1}(1-t)^{c-b-1} (1-xt)^{-a} \, dt$$​

Hypergeometric function .
 
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Hint

$$(a)_k = \frac{\Gamma(k+a)}{\Gamma(a)}$$
 
Would it be sufficient to show that your version of _2 F _1 solves the hypergeometric differential equation? I haven't tried it yet, and it looks like it might be a bit on the nasty side of tedious, but it should be a tolerable exercise.

-Dan
 
topsquark said:
Would it be sufficient to show that your version of _2 F _1 solves the hypergeometric differential equation? I haven't tried it yet, and it looks like it might be a bit on the nasty side of tedious, but it should be a tolerable exercise.

-Dan

Not a clue . I skipped that part when I read about hypergeomtric function (Emo)
 
ZaidAlyafey said:
Prove the following

$$ {}_2 F_1 \left( a,b; c ; x \right) = \frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}\int^1_0 t^{b-1}(1-t)^{c-b-1} (1-xt)^{-a} \, dt$$​

Hypergeometric function .
This is more or less an exercise in Gamma functions.

The series solution for the hypergeometric function is:
[math]_2 F _1 (a, b, c; x) = \sum_{n = 0}^{\infty} \frac{(a)_n (b)_n}{(c)_n} \frac{x^n}{n!}[/math]

where [math](a)_n \equiv \frac{ \Gamma (a + n)}{ \Gamma (a) }[/math]

We start with the integral formula:
$$_2 F_1 (a, b, c; x) = \frac{\Gamma (c)}{\Gamma (b)\Gamma (c-b)}\int_0^1 t^{b-1} (1-t)^{c-b-1} (1-xt)^{-a}~dt$$

Expanding the [math](1 - xt)^{-a}[/math] and inserting it into the integral formula gives
$$_2 F_1 (a, b, c; x) = \frac{\Gamma (c)}{\Gamma (b)\Gamma (c-b)}\int_0^1 t^{b-1} (1-t)^{c-b-1} \left ( \sum_{n = 0}^{\infty} \frac{ \Gamma (a + n) }{ \Gamma (a) } \frac{(tx)^n}{n!} \right )~dt$$

After a bit of "massaging":
$$_2 F_1 (a, b, c; x) = \sum_{n = 0}^{\infty} (a)_n \frac{\Gamma (c)}{\Gamma (b)\Gamma (c-b)} \left ( \int_0^1 t^{b + n -1} (1-t)^{c-b-1} dt \right ) \frac{x^n}{n!}$$

Now,
[math]\int_0^1 t^{b + n -1} (1-t)^{c-b-1} dt = B(b + n, c- b) = \frac{ \Gamma (b + n) \Gamma (c - b)}{\Gamma (c + n)}[/math]
where B is the beta function.

Plugging this in gives
[math]_2 F_1 (a, b, c; x) = \sum_{n = 0}^{\infty} \frac{\Gamma (c)}{\Gamma (b)\Gamma (c-b)} (a)_n \left ( \frac{ \Gamma (b + n) \Gamma (c - b)}{\Gamma (c + n)} \right ) \frac{x^n}{n!}[/math]

Simplifying gives
[math]_2 F_1 (a, b, c; x) = \sum_{n = 0}^{\infty} (a)_n \left ( \frac{\Gamma (b) }{ \Gamma (b + n)} \frac{\Gamma (c)}{\Gamma (c + n)} \right ) \frac{x^n}{n!} = \sum_{n = 0}^{\infty} \frac{(a)_n (b)_n}{(c)_n} \frac{x^n}{n!}[/math]

-Dan

Edit: Okay it's 1:57 AM (Eastern) and I think I got all the typos fixed.

Edit 2: I forgot to mention that the hypergeometric function has singular points at [math]x = 0, 1, \infty[/math]. None of these singularities appear in the given integral representation. Nor does the series representation take into account the x = 0 and x = 1 singularities.
 
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Way to go topsquark (Clapping)
 
It would be cool if someone could input the integral rep into the differential equation and show that it's a solution. I wasn't able to do it.

-Dan
 
topsquark said:
It would be cool if someone could input the integral rep into the differential equation and show that it's a solution. I wasn't able to do it.

-Dan

Hey Dan , can you post your attempt . I need to know what you are thinking about ?
 
I'm really not able to do much more than putting the integral form into the differential equation and a little bit of simplifying, but here it is.

First we are given:
[math]_2 F _1 (a, b, c; x) = \frac{ \Gamma (c) }{ \Gamma (b) \Gamma (c - b) } \int_0^1 t^{b - 1} (1 - t)^{c - b - 1} (1 - xt)^{-a} dt[/math]

The hypergeometric differential equation is:
[math]x(1 - x) y''(x) + (c - (a + b + 1)x) y'(x) - ab y(x) = 0[/math]

One of the solutions of this equation is [math]y(x) =~_2 F _1 (a, b, c; x)[/math]

Inserting the integral representation gives:
[math]x(1 - x) \left [ \frac{ \Gamma (c) }{ \Gamma (b) \Gamma (c - b) } \int_0^1 t^{b - 1} (1 - t)^{c - b - 1} a(a + 1)t^2 (1 - xt)^{-(a + 2)} dt \right ] [/math]
[math]+ \left [ ( c - (a + b + 1)x ) \frac{ \Gamma (c) }{ \Gamma (b) \Gamma (c - b) } \int_0^1 t^{b - 1} (1 - t)^{c - b - 1} at (1 - xt)^{-(a + 1)} dt \right ] [/math]
[math]- ab \left [ \frac{ \Gamma (c) }{ \Gamma (b) \Gamma (c - b) } \int_0^1 t^{b - 1} (1 - t)^{c - b - 1} (1 - xt)^{-a} dt \right ] = 0[/math]

Simplifying:
[math]x(1 - x) (a + 1) \int_0^1 t^{b + 1} (1 - t)^{c - b - 1} (1 - xt)^{-(a + 2)} dt [/math]
[math]+ ( c - (a + b + 1)x ) \int_0^1 t^b (1 - t)^{c - b - 1} (1 - xt)^{-(a + 1)} dt - b \int_0^1 t^{b - 1} (1 - t)^{c - b - 1} (1 - xt)^{-a} dt = 0[/math]

There is a way, obviously, to put all the x's and constants inside the integrals and then put this all under one integral but I am unable to see any value to this approach.

-Dan
 
  • #10
So your attempt is to prove that the hypergeometric function satisfies the differential equation using the integral representation ?
 
  • #11
ZaidAlyafey said:
So your attempt is to prove that the hypergeometric function satisfies the differential equation using the integral representation ?
Yup. It should be workable, but I'm out of tricks on this one. I can't figure out how to solve the integrals that come up.

-Dan
 

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