# Hypergeometric Challenge

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Prove the following

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

Hypergeometric function .

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Hint

$$\displaystyle (a)_k = \frac{\Gamma(k+a)}{\Gamma(a)}$$

#### topsquark

##### Well-known member
MHB Math Helper
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

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
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

#### topsquark

##### Well-known member
MHB Math Helper
Prove the following

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

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

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

where $$\displaystyle (a)_n \equiv \frac{ \Gamma (a + n)}{ \Gamma (a) }$$

$$\displaystyle _2 F_1 (a, b, c; x) = \frac{\Gamma (c)}{\Gamma (b)\Gamma (c-b)}\int_0^1 t^{b-1} (1-t)^{c-b-1} (1-xt)^{-a}~dt$$

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

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

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

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

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

-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 $$\displaystyle x = 0, 1, \infty$$. 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.

Last edited:

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Way to go topsquark

#### topsquark

##### Well-known member
MHB Math Helper
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

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
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 ?

#### topsquark

##### Well-known member
MHB Math Helper
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:
$$\displaystyle _2 F _1 (a, b, c; x) = \frac{ \Gamma (c) }{ \Gamma (b) \Gamma (c - b) } \int_0^1 t^{b - 1} (1 - t)^{c - b - 1} (1 - xt)^{-a} dt$$

The hypergeometric differential equation is:
$$\displaystyle x(1 - x) y''(x) + (c - (a + b + 1)x) y'(x) - ab y(x) = 0$$

One of the solutions of this equation is $$\displaystyle y(x) =~_2 F _1 (a, b, c; x)$$

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

Simplifying:
$$\displaystyle x(1 - x) (a + 1) \int_0^1 t^{b + 1} (1 - t)^{c - b - 1} (1 - xt)^{-(a + 2)} dt$$
$$\displaystyle + ( c - (a + b + 1)x ) \int_0^1 t^b (1 - t)^{c - b - 1} (1 - xt)^{-(a + 1)} dt - b \int_0^1 t^{b - 1} (1 - t)^{c - b - 1} (1 - xt)^{-a} dt = 0$$

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

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
So your attempt is to prove that the hypergeometric function satisfies the differential equation using the integral representation ?

#### topsquark

##### Well-known member
MHB Math Helper
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