First Look into Hypergeometric Functions (involving the gamma and beta functions)

We look at the following hypergeometric series:

$F(a, b, c; u) = 1 + \frac{a \cdot b}{1 \cdot c} \frac{u}{1!} + \frac{a(a+1)b(b+1)}{c(c+1)0} \frac{u^2}{2!} + ...$

$\equiv \sum_{n=0}^{\infty} \frac{(a)_n (b)_n}{(c)_n} \frac{z^n}{n!}$ (*)

where $(a)_n = a(a+1) ... (a+n-1)$ ( $(b)_n$ , and $(c)_n$ are defined similarly) and $a, b, c \in \mathbb{C}$ .

We now find the integral representation of this series.

Key: $(a)_n = \frac{\Gamma(a+n)}{\Gamma(a)}$ , $(b)_n = \frac{\Gamma(b+n)}{\Gamma(b)}$ , and $(c)_n = \frac{\Gamma(c+n)}{\Gamma(c)}$ .

It follows that (*) becomes:

$\sum_{n=0}^{\infty} \frac{\Gamma(a+n) \Gamma(b+n) \Gamma(c)}{\Gamma(a) \Gamma(b) \Gamma(c+n)} \frac{z^n}{n!}$ (**).

Note: the Beta function is defined as

$B(x, y) = \int_0^1 t^{x-1} (1-t)^{y-1} dt = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}$ .

Thus, we have:

(**) = $\frac{\Gamma(c)}{\Gamma(a) \Gamma(b)} \Sigma_{n=0}^{\infty} \frac{\Gamma(a+n) z^n}{n!} \frac{\Gamma(b+n)}{\Gamma(c+n)}$

$= \frac{\Gamma(c)}{\Gamma(a) \Gamma(b) \Gamma(c-b)} \Sigma_{n=0}^{\infty} \frac{\Gamma(a+n) z^n}{n!} \int_0^1 t^{b+n-1} (1-t)^{c-b-1}dt$

Here, without considering much about convergence, we move the infinite sum into the integral:

$= \frac{\Gamma(c)}{\Gamma(a) \Gamma(b) \Gamma(c-b)} \int_0^1 t^{b-1} (1-t)^{c-b-1} [\Sigma_{n=0}^{\infty} \frac{\Gamma(a+n)}{n!}(zt)^n] dt$

But notice:

$\Sigma_{n=0}^{\infty} \frac{\Gamma(a+n)}{n!}(zt)^n = \Gamma(a) \Sigma_{n=0}^{\infty} \frac{(a)_n}{n!}(zt)^n = \Gamma(a) (1-zt)^{-a}$

we use the generalized binomial series.

Hence, we obtain:

$F(a, b, c; u) = \frac{\Gamma(c)}{\Gamma(a) \Gamma(b) \Gamma(c-b)} \int_0^1 t^{b-1} (1-t)^{c-b-1} [\Gamma(a) (1-zt)^{-a}]dt$

$= \frac{\Gamma(c)}{\Gamma(b) \Gamma(c-b)} \int_0^1 t^{b-1} (1-t)^{c-b-1} (1-zt)^{-a} dt$

which gives us an integral representation of our series.

Part 2: the corresponding differential equation

Notice that for fixed a, b, c:

$F'(u) = \frac{a \cdot b}{c \cdot 0!} + \frac{a(a+1)b(b+1)}{c(c+1)} \frac{u}{1!} + ...$

$F''(u) = \frac{a(a+1)b(b+1)}{c(c+1)} + \frac{a(a+1)(a+2)b(b+1)(b+2)}{c(c+1)(c+2)} \frac{u}{1!} + ...!$

and so it is not difficult to check that:

$u(1-u)F'' + (c-(a+b+1)u)F' - abF = 0$ (*)

i.e. the hypergeometric series is a solution to the above second-order differential equation.

We can rewrite this equation as:

$F'' + \frac{c-(a+b+1)u}{u(1-u)}F' - \frac{ab}{u(1-u)}F = 0$ (**)

Interesting Question: what is the other independent solution corresponding to (**)?

In the appendix of Klein’s The Icosahedron, it is mentioned that the more general hypergeometric differential equation is:

$z'' + \frac{(\alpha + \alpha' - 1) + (\beta + \beta' + 1)u}{u(u-1)} z' + \frac{\alpha \alpha' - (\alpha \alpha' + \beta \beta' - \gamma \gamma')u + \beta \beta' u^2}{u^2 (u-1)^2} z = 0$

where $\alpha' = 1 - c + \gamma$ , $\gamma' = c - a - b + \gamma$ , $\beta = a - \alpha - \gamma$ , $\beta' = b - \alpha - \gamma$

and one corresponding solution (known as the hypergeometric functions):

$u^{\alpha}(1-u)^{\gamma}F(a,b,c;u)$

where $\alpha, \gamma \in \mathbb{C}$ .

First Look into Hypergeometric Functions (involving the gamma and beta functions)

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