Hypergeometric function discontinuity The hypergeometric function$\ _2F_1(a,b;c;z)$ has a branch point at $z=1$. How do I compute the discontinuity around the point? In particular, how do I compute the following?
$$\lim_{\epsilon\rightarrow 0^+} \ _2F_1(1,b;b+1;1+\epsilon)-\ _2F_1(1,b;b+1;1-\epsilon)$$
Mathematica says that it is $-ib\pi$.
 A: Here is Luke's “The Special Functions and Their Approximations: Volume 1”
https://www.amazon.com/Special-Functions-Their-Approximations/dp/0124110371
Section 3.4, Eq:15
answer with her variables
$A\left(z\right)=\,_{2}F_{1}\left(\begin{array}{c}
\nu+1,\nu+\mu+1\\
\nu+\mu+\lambda+2
\end{array};z\right)=\frac{\left(-1\right)^{\mu+1}\left(\nu+\mu+\lambda+1\right)!}{\lambda!\cdot\nu!\cdot\left(\nu+\mu\right)!\cdot\left(\mu+\lambda\right)!}\cdot\frac{d^{\nu+\mu}}{dz^{\nu+\mu}}\left[\left(1-z\right)^{\mu+\lambda}\cdot\frac{d^{\lambda}}{dz^{\lambda}}\left\{ z^{-1}\cdot ln\left(1-z\right)\right\} \right]$
Here is your example with $1\leq a\leq b < c;a,b,c:integer$
$\nu=0,a=1,b=\mu+1,\lambda=0,c=b+1,\lambda=0$
$A\left(1-z\right)=\,_{2}F_{1}\left(\begin{array}{c}
1,b\\
b+1
\end{array};1-z\right)=\frac{\left(-1\right)^{b}\left(b\right)!}{0!\cdot 0!\cdot\left(b-1\right)!\cdot\left(b-1\right)!}\cdot\frac{d^{b-1}}{dz^{b-1}}\left[\left(z\right)^{b-1}\cdot\frac{d^{0}}{dz^{0}}\left\{ \left(1-z\right)^{-1}\cdot ln\left(z\right)\right\} \right]$
For instance, for $b=3$:
Sagemath returns
$\frac{3\,\left(x^{2}-4\,x+2\,\log\left(x\right)+3\right)}{2\,\left(x-1\right)^{3}}$
Which matches Sagemath straight up result. The answer to the difference/shape around 1 is:
$-\frac{3\,\left(x^{2}+4\,x+2\,\log\left(-x\right)+3\right)}{2\,\left(x+1\right)^{3}}-\frac{3\,\left(x^{2}-4\,x+2\,\log\left(x\right)+3\right)}{2\,\left(x-1\right)^{3}}$
A: For the particular problem we can refer to Lerch's Function at:
Bateman Higher Transcendental Functions sections 1.11
and 6.11.1 and
DLMF  25.14
How do we get there?
We look at a term of the question;
say $x^{n}$ and set $x=1-\epsilon$
$\left[x^{n}\right]\,_{2}F_{1}\left(1,b;b+1;1-\epsilon\right)=\frac{\left(1\right)_{n}\left(b\right)_{n}}{\left(b+1\right)_{n}}\cdot\frac{x^{n}}{n!} =\frac{\Gamma\left(b+1\right)}{\Gamma\left(b\right)}\cdot\frac{\Gamma\left(b+n\right)}{\Gamma\left(b+1+n\right)}\cdot x^{n}=b\cdot\frac{1}{b+n}\cdot x^{n}$
And examining the terms involving n we get Lerch's Function
$=b\cdot\Phi\left(x,1,b\right)$
$\Phi\left(x,s,b\right)={\displaystyle \sum_{n=0}^{\infty}}\frac{x^{n}}{\left(b+n\right)^{s}}$
To me the most relevant part of section 1.11 starts at Eq-14 and winds up in the comments after Eq-18; but there are a lot of other goodies.
