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__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 .

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

Hypergeometric function .

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Hint

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

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-Dan

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Not a clue . I skipped that part when I read about hypergeomtric function

-Dan

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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 .

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

We start with the integral formula:

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

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:

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Way to go topsquark

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-Dan

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Hey Dan , can you post your attempt . I need to know what you are thinking about ?

-Dan

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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

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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