Behavior of the Gaussian Hypergeometric function when one of its arguments approaches $0$ or $1$

Question

For two positive integers $a,b$, denote by $_2F_1(a,1-b;a+1;z)$ the Gaussian Hypergeometric function whose first three parameters are fixed at $a,1-b$ and $a+1$, respectively. such function is linked to the regularized incomplete Beta function (namely the CDF of a Beta distribution) by the relation: $$ B(z;a,b)=\frac{z^a}{a} {}_2F_1(a,1-b;a+1;z). $$ I was wondering whether there exist functions $f,\, g$, such that $$ f(z)\sim {}_2F_1(a,1-b;a+1;z) \quad (z\downarrow 0),\\ g(z)\sim {}_2F_1(a,1-b;a+1;z) \quad (z\uparrow 1). $$ In particular, I would be interested in the case $a,b\geq 1$.

Answer 1

We have $$ \eqalign{ & F(z,a,b) = {}_2F_{\,1} \left( {\left. {\matrix{ {a,\;1 - b} \cr {a + 1} \cr } \;} \right|\;z} \right) = \cr & = \sum\limits_{0\, \le \,k} {{{a^{\,\overline {\,k\,} } \left( {1 - b} \right)^{\,\overline {\,k\,} } } \over {\left( {a + 1} \right)^{\,\overline {\,k\,} } }} {{z^{\,k} } \over {k!}}} = \cr & = a\sum\limits_{0\, \le \,k} {{{\left( {1 - b} \right)^{\,\overline {\,k\,} } } \over {\left( {a + k} \right)}}{{z^{\,k} } \over {k!}}} = \cr & = a\sum\limits_{0\, \le \,k} {{{\left( { - 1} \right)^{\,k} \left( {b - 1} \right)^{\,\underline {\,k\,} } } \over {\left( {a + k} \right)}}{{z^{\,k} } \over {k!}}} \cr & = a\sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k} \left( \matrix{ b - 1 \cr k \cr} \right){{z^{\,k} } \over {\left( {a + k} \right)}}} \cr} $$

Then, for the first limit you have $$ \mathop {\lim }\limits_{z\, \to \,0 + } F(z,a,b) = 1 $$

Continuing the development $$ \eqalign{ & F(z,a,b) = a\sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k} \left( \matrix{ b - 1 \cr k \cr} \right){{z^{\,k} } \over {\left( {a + k} \right)}}} = \cr & = a\,z^{\, - \,a} \sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k} \left( \matrix{ b - 1 \cr k \cr} \right){{z^{\,a + k} } \over {\left( {a + k} \right)}}} = \cr & = a\,z^{\, - \,a} \sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k} \left( \matrix{ b - 1 \cr k \cr} \right)\int_{t = 0}^z {t^{\,a + k - 1} } dt} = \cr & = a\,z^{\, - \,a} \int_{t = 0}^z {t^{\,a - 1} \sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k} \left( \matrix{ b - 1 \cr k \cr} \right)} \left( { - t} \right)^{\,k} } dt = \cr & = a\,z^{\, - \,a} \int_{t = 0}^z {t^{\,a - 1} \left( {1 - t} \right)^{\,b - 1} } dt \cr} $$ which gives the relation with the Incomplete Beta function you already know. $$ F(z,a,b) = a\,z^{\, - \,a} \,{\rm B}(z\,;a,b)\quad \left| {\;0 < a,b} \right. $$

From here we get $$ \eqalign{ & \mathop {\lim }\limits_{z\, \to \,1 - } F(z,a,b) = a\int_{t = 0}^1 {t^{\,a - 1} \left( {1 - t} \right)^{\,b - 1} } dt = \cr & = a\,{\rm B}(a,b) \cr} $$ and since (re. for instance to this) $$ {\rm B}(z\,;a,b) = {{z^{\,\,a} } \over a}\left( {1 + O(z)} \right) $$ also $$ \mathop {\lim }\limits_{z\, \to \,0 + } \left( {F(z,a,b) = a\,z^{\, - \,a} \,{\rm B}(z\,;a,b)} \right) = 1\quad \left| {\;0 < a,b} \right. $$ and you can use the above expression for the whole domain of your interest.