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?
 A: The first thing to notice is that the equation is symmetric in $\, a,b \,$ which comes from changing $\, k \,$ to $\, -k. \,$ Thus, without loss of generality suppose $\, k\ge 0. \,$ Let us define some notation:
 $\, f_3 := (\Gamma(c)\Gamma(c-a-b))/(\Gamma(c-a)\Gamma(c-b)), \quad
 f_4 := (\Gamma(c)\Gamma(a+b-c))/(\Gamma(a)\Gamma(b)) $
which are multiplied by $\, w_3(z) \,$ and $\, w_4(z) \,$ respectively.
The equation is $\, w_1(z) = w_3(z) f_3 + w_4(z) f_4. \,$ 
Note that we will also add $\,\epsilon\,$ to $\,a\,$ to probe sensitivity to parameters. We are interested in the values of the functions at $\, z=1. \,$
For all integer $\, k, \,$ we have $\, w_3(1) = 1. \,$  We have the cases:


*

*If $\, k \,$ is even, $\, w_1 \,$ has a pole at $\,z=1,\,$ but the factor $\, f_3 \,$ has an $\, \epsilon \,$ pole, and 
$\, w_4 \,$ has an $\, \epsilon \,$ pole, but its factor $\, f_4 \,$ is finite.

*If $\, k=1 \,$ then both $\, w_1 \,$ and $\, w_4 \,$ have a $\, z=1 \,$ pole while $\, f_3 = 0. \,$

*If $\, k>1 \,$ and $\, k=4n+1 \,$ then $\, w_1 \,$ has a pole at $\,z=1,\,$
while $\, f_3 \,$ has  an $\, \epsilon \,$ pole, and 
$\, w_4 \,$ has an $\, \epsilon \,$ pole while $\, f_4 \,$ is finite.

*If $\, k = 4n+3, \,$ then $\, w_1, w_3, f_3 \,$ are all finite, while $\, f_4 = 0 \,$ but $\, w_4 \,$ has both a $\, z=1 \,$ pole and an $\, \epsilon \,$ pole.
Note that $\, w_1(z) \,$ is a polynomial in $\, z \,$ with $\, w_1(1) = 1. \,$
A: 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
$$F(x) = \frac {\sqrt \pi}
  {\Gamma\left( \frac {3+k} 4 \right)
   \Gamma\left( \frac {3-k} 4 \right)
   (1-x)} +
 O(|\!\ln(1-x)|).$$
