- The hypergeometric series \(F ( \alpha , \beta , \gamma ; z )\) was defined in Exercise 16 of Chapter 1. Show that
\[ F (\alpha , \beta , \gamma ; z) = \frac {\Gamma (\gamma)}{\Gamma (\beta) \Gamma (\gamma - \beta)} \int_ {0} ^ {1} t ^ {\beta - 1} (1 - t) ^ {\gamma - \beta - 1} (1 - z t) ^ {- \alpha} d t. \]
Here \(\alpha > 0 , \beta > 0 , \gamma > \beta\) , and \(| z | < 1\)
Show as a result that the hypergeometric function, initially defined by a power series convergent in the unit disc, can be continued analytically to the complex plane slit along the half-line \(\lbrack 1 , \infty )\)
Note that
\[ \log (1 - z) = - z F (1, 1, 2; z), \]
\[ e ^ {z} = \lim _ {\beta \rightarrow \infty} F (1, \beta , 1; z / \beta), \]
\[ (1 - z) ^ {- \alpha} = F (\alpha , 1, 1; z). \]
[Hint: To prove the integral identity, expand \(( 1 - z t ) ^ { - \alpha }\) as a power series.]
Solution
Recall \[ F(\alpha,\beta,\gamma;z) =\sum_{n=0}^\infty \frac{(\alpha)_n(\beta)_n}{(\gamma)_n\,n!}z^n. \] For \(|z|<1\), \[ (1-zt)^{-\alpha} =\sum_{n=0}^\infty\frac{(\alpha)_n}{n!}(zt)^n, \] uniformly for \(0\le t\le1\) on compact subsets of the unit disc. Hence termwise integration gives \[ \int_0^1 t^{\beta-1}(1-t)^{\gamma-\beta-1}(1-zt)^{-\alpha}\,dt =\sum_{n=0}^\infty\frac{(\alpha)_n z^n}{n!} B(\beta+n,\gamma-\beta). \] Using the beta-gamma identity, \[ B(\beta+n,\gamma-\beta) =\frac{\Gamma(\beta+n)\Gamma(\gamma-\beta)}{\Gamma(\gamma+n)} =\frac{\Gamma(\beta)\Gamma(\gamma-\beta)}{\Gamma(\gamma)} \frac{(\beta)_n}{(\gamma)_n}. \] Multiplying by the prefactor yields exactly the hypergeometric series, proving the integral formula.
Now let \[ \Omega=\mathbb C\setminus[1,\infty). \] For \(z\in\Omega\) and \(0\le t\le1\), the number \(1-zt\) never lies on the nonpositive real axis, so the principal branch of \((1-zt)^{-\alpha}\) is holomorphic in \(z\). On every compact \(K\subset\Omega\) it is uniformly bounded for \(0\le t\le1\). Since \[ t^{\beta-1}(1-t)^{\gamma-\beta-1} \] is integrable, differentiation under the integral is valid locally uniformly on \(\Omega\). Thus the integral defines a holomorphic function on \(\Omega\) agreeing with the original power series on \(|z|<1\). It is therefore the desired analytic continuation to the slit plane.