Gauss's Hypergeometric Theorem/Proof 2

Theorem

$\map F {a, b; c; 1} = \dfrac {\map \Gamma c \map \Gamma {c - a - b} } {\map \Gamma {c - a} \map \Gamma {c - b} }$

Proof

From Euler's Integral Representation of Hypergeometric Function, we have:

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

Where $a, b, c \in \C$.

and $\size x < 1$

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

Since Euler's Integral Representation only applies where $\size x < 1$, we will determine the limit of the integral as $x \to 1$.

Therefore:

\(\ds \map F {a, b; c; x}\)

\(=\)

\(\ds \dfrac {\map \Gamma c } {\map \Gamma b \map \Gamma {c - b} } \int_0^1 t^{b - 1} \paren {1 - t}^{c - b - 1} \paren {1 - \paren {1} t}^{- a} \rd t\)

\(\ds \leadsto \ \ \)

\(\ds \)

\(=\)

\(\ds \dfrac {\map \Gamma c } {\map \Gamma b \map \Gamma {c - b} } \int_0^1 t^{b - 1} \paren {1 - t}^{c - b - a - 1} \rd t\)

simplifying and Product of Powers

\(\ds \leadsto \ \ \)

\(\ds \)

\(=\)

\(\ds \dfrac {\map \Gamma c } {\map \Gamma b \map \Gamma {c - b} } \dfrac {\map \Gamma b \map \Gamma {c - a - b } } {\map \Gamma {c - a } }\)

Definition of Beta Function

\(\ds \leadsto \ \ \)

\(\ds \)

\(=\)

\(\ds \dfrac {\map \Gamma c \map \Gamma {c - a - b} } {\map \Gamma {c - a} \map \Gamma {c - b} }\)

simplifying and canceling $\map \Gamma b$

$\blacksquare$

Source of Name

This entry was named for Carl Friedrich Gauss.