<s>[INST] Connection Formula for Hypergeometric Function 2F1 Suppose I have the function $_2F_1\left(a,b;c;x^2\right)$ with $a=\frac{3}{4}+\frac{k}{4}$, $b=\frac{3}{4}-\frac{k}{4}$ and $c=\frac{1}{2}$.
I want to know the behaviour about $x=1.\,$
I go to DLMF equation 15.10.21 and choose
$$
w_1\left(x^2\right) = {\frac {\Gamma \left( c \right) \Gamma \left( c-a-b \right)}{
\Gamma \left( c-a \right) \Gamma \left( c-b \right) }} \, w_3\left(x^2\right) +{\frac {\Gamma
 \left( c \right) \Gamma \left( a+b-c \right)}{\Gamma \left( 
a \right) \Gamma \left( b \right) }} \, w_4\left(x^2\right).
$$
Since $w_4$ is singular at $x=1$ (and it should be finite) I expected this typical constraint that $a$ or $b$ must be some $n\leq 0 \in \mathbb{Z}$ for the second term to vanish. Now the first term has $\Gamma(-1)$. Is that a problem or can I absorb this into a constant? However if I do so, then the original function $w_1$ is not really defined. Does this mean the solution is not valid unless $k=3$?
PS: Actually if $k=4n+3$ $(n\geq 0)$ then the second term vanishes and in the first term the poles cancel?
 [/INST] Let
$$F(x) = {_2F_1}\left( \frac {3+k} 4, \frac {3-k} 4; \frac 1 2; x \right),
\quad k \geq 0.$$
If $(3-k)/4$ is an integer, $F(x)$ becomes a polynomial and we have
$$F(x) = F(1) + O(|1-x|) =
\frac {(-1)^{(k-3)/4} \sqrt \pi \,\Gamma \left( \frac {k+5} 4 \right)}
{\Gamma \left( \frac {k-1} 4 \right)} + O(|1-x|).$$
Otherwise this is the logarithmic case, the formulas for which can be found here. The leading term is
$$F(x) = \frac {\sqrt \pi}
  {\Gamma\left( \frac {3+k} 4 \right)
   \Gamma\left( \frac {3-k} 4 \right)
   (1-x)} +
 O(|\!\ln(1-x)|).$$
</s>