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# integral representation of the hypergeometric function

When $\Re c>\Re b>0$, one has the representation

$F(a,b;c;z)={\Gamma(c)\over\Gamma(b)\Gamma(c-b)}\int_{0}^{1}t^{{b-1}}(1-t)^{{c-% b-1}}(1-tz)^{{-a}}\,dt$ |

Note that the conditions on $b$ and $c$ are necessary for the integral to be convergent at the endpoints $0$ and $1$. To see that this integral indeed equals the hypergeometric function, it suffices to consider the case $|z|<1$ since both sides of the equation are analytic functions of $z$. (This follows from the rigidity theorem for analytic functions although some care is required because the function is multiply-valued.) With this assumption, $|tz|<1$ if $t$ is a real number in the interval $[0,1]$ and hence, $(1-tz)^{{-a}}$ may be expanded in a power series. Substituting this series in the right hand side of the formula above gives

${\Gamma(c)\over\Gamma(b)\Gamma(c-b)}\int_{0}^{1}\sum_{{k=0}}^{\infty}t^{{b-1}}% (1-t)^{{c-b-1}}{\Gamma(k-a+1)\over\Gamma(1-a)\Gamma(k+1)}(-tz)^{k}\,dt$ |

Since the series is uniformly convergent, it is permissible to integrate term-by-term. Interchanging integration and summation and pulling constants outside the integral sign, one obtains

${\Gamma(c)\over\Gamma(b)\Gamma(c-b)}\sum_{{k=0}}^{\infty}{\Gamma(k-a+1)\over% \Gamma(1-a)\Gamma(k+1)}(-z)^{k}\int_{0}^{1}(1-t)^{{c-b-1}}t^{{b+k-1}}dt$ |

The integrals appearing inside the sum are Euler beta functions. Expressing them in terms of gamma functions and simplifying, one sees that this integral indeed equals the hypergeometric function.

## Mathematics Subject Classification

33C05*no label found*

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