I want to expand $_2F_1(\frac{1}{2},a,\frac{3}{2},1-x)$ around $x=0$, and then take $a \to 0$.
I know that $_2F_1(\frac{1}{2},0,\frac{3}{2},1-x)=1$ for any $x$, Which means only the zeroth value of this expansion will survive when $a \to 0$.
But it seems I have a problem. All formulas (in mathematica website) include the condition of $c-b-a \notin \mathbin{Z}$. Before I take a to zero, the expansion obeys this condition, but the limit of $a \to 0$ diverges. This condition originated from the derivative of $_2F_1$ around $z=1$ when we write the Taylor expansion.
What did I do wrong? Or maybe there is no formula in my case?
Stricctly speaking, it won't be Taylor expansion, because this function for $a\neq 0$ will have a part that behaves $\sim z^{-a}$, but there's a formula you can use: \begin{align} _2F_1(a,b;c;1-z) &= \frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)} \,_2F_1(a,b;1+a+b-c;z) + \\ &\qquad +\frac{\Gamma(c)\Gamma(a+b-c)}{\Gamma(a)\Gamma(b)} z^{c-a-b}\,_2F_1(c-a,c-b;1-a-b+c;z) \end{align} so \begin{align} _2F_1(\frac12,a;\frac32;1-z) &= \frac{\Gamma(\frac32)\Gamma(1-a)}{\Gamma(1)\Gamma(\frac32-a)} \,_2F_1(\frac12,a;a;z) + \\ &\qquad +\frac{\Gamma(\frac32)\Gamma(a-1)}{\Gamma(\frac12)\Gamma(a)} z^{1-a}\,_2F_1(1,\frac32-a;2-a;z) = \\ &= \frac{\sqrt{\pi}\Gamma(1-a)}{2\Gamma(\frac32-a)} \,_2F_1(\frac12,a;a;z) + \\ &\qquad - \frac{1}{2(1-a)} z^{1-a}\,_2F_1(1,\frac32-a;2-a;z)\end{align} This formula has a perfectly fine limit $a\rightarrow 0$: \begin{align} \,_2F_1(\frac12,0;\frac32;1-z) &= \Big(\lim_{a\rightarrow 0}\,_2F_1(\frac12,a;a;z) \Big) - \frac{z}{2} \,_2F_1(1,\frac32;2;z)\end{align} Note that we cannot simply write $_2F_1(\frac12,0;0,z)$ because it's not well defined (we'd get $\frac{(0)_n}{(0)_n}$ in the expansion). However we have $$ _2F_1(\frac12,a;a;z) = \frac{1}{\sqrt{1-z}}$$ $$ \,_2F_1(1,\frac32;2;z) = \frac{2(1-\sqrt{1-z})}{z\sqrt{1-z}}$$ and we get a correct formula $$ 1 = \frac{1}{\sqrt{1-z}} - \frac{z}{2}\cdot\frac{2(1-\sqrt{1-z})}{z\sqrt{1-z}} $$ On the level of expansions you have \begin{align} _2F_1(\frac12,a;\frac32;1-z) &= \sum_{n=0}^\infty \frac{\Gamma(1-a)\Gamma(\frac12+n)}{2\Gamma(\frac32-a)n!} z^n - z^{1-a} \sum_{n=0}^\infty \frac{\Gamma(1-a)\Gamma(\frac32-a+n)}{2\Gamma(\frac32-a)\Gamma(2-a+n)} z^n = \\ &= \frac{\Gamma(1-a)}{2\Gamma(\frac32-a)}\Big(\sum_{n=0}^\infty \frac{\Gamma(\frac12+n)}{\Gamma(1+n)} z^n - \sum_{n=1}^\infty \frac{\Gamma(\frac12+n-a)}{\Gamma(1+n-a)} z^{n-a}\Big)\end{align} and you can also take a limit $a\rightarrow 0$ here: \begin{align} _2F_1(\frac12,0;\frac32;1-z) &= \frac{\Gamma(1)}{2\Gamma(\frac32)}\Big(\sum_{n=0}^\infty \frac{\Gamma(\frac12+n)}{\Gamma(1+n)} z^n - \sum_{n=1}^\infty \frac{\Gamma(\frac12+n)}{\Gamma(1+n)} z^n\Big) = \\ &= \frac{\Gamma(1)}{2\Gamma(\frac32)} \frac{\Gamma(\frac12)}{\Gamma(1)} = 1\end{align}