Euler's Integral Representation of Hypergeometric Function

Theorem
Let $a, b, c \in \C$.

Let $\size x < 1$

Let $\map \Re c > \map \Re b > 0$.

Then:
 * $\ds \map F {a, b; c; x} = \dfrac {\map \Gamma c } {\map \Gamma b \map \Gamma {c - b} } \int_0^1 t^{b - 1} \paren {1 - t}^{c - b - 1} \paren {1 - x t}^{- a} \rd t$

in the $x$ plane cut along the real axis from $1$ to $\infty$.

Here it is understood that:
 * $\map \arg t = \map \arg {1 - t} = 0$
 * $\paren {1 - x t}^{-a}$ has its principal argument


 * $\map F {a, b; c; x}$ is the Gaussian hypergeometric function of $x$:
 * $\map F {a, b; c; x} := \ds \sum_{k \mathop = 0}^\infty \dfrac {a^{\overline k} b^{\overline k} } {c^{\overline k} } \dfrac {x^k} {k!}$
 * $x^{\overline k}$ denotes the $k$th rising factorial power of $x$
 * $\map \Gamma {n + 1} = n!$ is the Gamma function.

Proof
Letting $\size x < 1$ and expanding the product of $\paren {1 - x t}^{-a}$:

Therefore:

Therefore:
 * $\ds \map F {a, b; c; x} = \dfrac {\map \Gamma c} {\map \Gamma b \map \Gamma {c - b} } \int_0^1 t^{b - 1} \paren {1 - t}^{c - b - 1} \paren {1 - x t}^{-a} \rd t$

Also see

 * Gauss's Hypergeometric Theorem