The classical hypergeometric function ${}_{2}F_1$ is defined by
$$
{}_{2}F_1(a,b;c;z)
=
\sum_{k=0}^\infty \frac{(a)_k(b)_k}{(c)_k} \frac{z^k}{k!}
$$
where $(\cdot)_k$ is Pochhammer symbol, that is,
$$
(q)_k = \frac{\Gamma(q+k)}{\Gamma(q)}
$$
provided that $q+k$ is not a negative integer, with the convention $1/{\pm\infty} = 0$. Note that we may naturally generalize the classical hypergeometric function to the generalized hypergeometric series:
$$
{}_p F_q \begin{bmatrix}a_1, a_2, \cdots, a_p \\ b_1, b_2, \cdots, b_q;z \end{bmatrix} = \sum_{k=0}^\infty \frac{(a_1)_k (a_2)_k \cdots (a_p)_k}{(b_1)_k (b_2)_k \cdots (b_q)_k} \frac{z^k}{k!}
$$
for $p,q\in \mathbb N$. Note that we usually set $p=q+1$ so that the series converges when $|z|<1$ for all possible choices of parameters. Of course, there are lots of formulas, which expand the domain of the function through the analytic continuation. For the simplicity, we frequently use the notation
$$
{}_p F_q (a_1, \cdots, a_p; b_1, \cdots, b_q; z) = {}_p F_q \begin{bmatrix}a_1, \cdots, a_p \\ b_1, \cdots, b_q;z \end{bmatrix}.
$$
Theorem 1. (Euler’s Integral Representation) Let $\text{Re} c> \text{Re} b>0$, then $$ _2 F_1 (a,b;c;z) = \frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)} \int_0^1 \frac{t^{b-1} (1-t)^{c-b-1}}{(1-tz)^a}\,dt $$ holds in the $z$ plane cut along the real axis from $1$ to $\infty$. |
Proof. By the binomial theorem, we have
$$
\frac{1}{(1-tz)^{a}} = \sum_{k=0}^\infty \frac{(a)_k}{k!}(tz)^k
$$
so that
\begin{eqnarray*}
\int_0^1 \frac{t^{b-1}(1-t)^{c-b-1}}{(1-tz)^a}\,dt
&=&
\int_0^1 t^{b-1}(1-t)^{c-b-1}\sum_{k=0}^\infty \frac{(a)_k}{k!}(tz)^k\,dt\\
&=&
\sum_{k=0}^\infty \frac{(a)_k}{k!} z^k \int_0^1 t^{k+b-1} (1-t)^{c-b-1}\,dt\\
&=&
\sum_{k=0}^\infty \frac{(a)_k}{k!} z^k \frac{\Gamma(k+b)\Gamma(c-b)}{\Gamma(k+c)}.
\end{eqnarray*}
Therefore we obtain
\begin{eqnarray*}
\frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}
\int_0^1 \frac{t^{b-1}(1-t)^{c-b-1}}{(1-tz)^a}\,dt
&=&
\sum_{k=0}^\infty \frac{(a)_k}{k!} z^k \frac{\Gamma(k+b)}{\Gamma(b)} \frac{\Gamma(c)}{\Gamma(k+c)}\\
&=&
\sum_{k=0}^\infty \frac{(a)_k (b)_k}{(c)_k k!} z^k\\
&=&
_2 F_1 (a,b;c;z)
\end{eqnarray*}
which proves Theorem 1 for $|z|<1$. Since the integral is analytic in the cut plane, the theorem holds.$\square$
References.
[1] en.wikipedia.org/wiki/Hypergeometric_function
[2] aw.twi.tudelft.nl/~koekoek/onderw1112/specfunc_en.html
Leave A Reply
[…] ← Previous […]
[…] Proof. We remember that the Euler Integral Representation for the hypergeometric function […]