Kummer's Hypergeometric Theorem/Proof 2

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Theorem

$\map F {n, -x; x + n + 1; -1} = \dfrac {\map \Gamma {x + n + 1} \map \Gamma {\dfrac n 2 + 1} } {\map \Gamma {x + \dfrac n 2 + 1} \map \Gamma {n + 1} }$


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 $x \to -1$.


By symmetry, we have:

$\ds \map F {n, -x; x + n + 1; -1} = \ds \map F {-x, n; x + n + 1; -1}$

Therefore:

\(\ds \map F {-x, n; x + n + 1; -1}\) \(=\) \(\ds \dfrac {\map \Gamma {x + n + 1} } {\map \Gamma n \map \Gamma {x + n + 1 - n } } \int_0^1 t^{n - 1} \paren {1 - t}^{x + n + 1 - n - 1} \paren {1 - \paren {-1} t}^{- \paren {-x} } \rd t\)
\(\ds \leadsto \ \ \) \(\ds \) \(=\) \(\ds \dfrac {\map \Gamma {x + n + 1} } {\map \Gamma n \map \Gamma {x + 1 } } \int_0^1 t^{n - 1} \paren {1 - t}^x \paren {1 + t}^x \rd t\) simplifying
\(\ds \leadsto \ \ \) \(\ds \) \(=\) \(\ds \dfrac {\map \Gamma {x + n + 1} } {\map \Gamma n \map \Gamma {x + 1 } } \int_0^1 t^{n - 1} \paren {1 - t^2}^x \rd t\) simplifying further: $\paren {1 - t^2} = \paren {1 - t}\paren {1 + t}$

We now apply a u-substitution: Let $u = t^2$

\(\ds u\) \(=\) \(\ds t^2\)
\(\ds \leadsto \ \ \) \(\ds t\) \(=\) \(\ds \sqrt u\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d u} {\d t}\) \(=\) \(\ds 2 t\) Power Rule for Derivatives
\(\ds \leadsto \ \ \) \(\ds \d t\) \(=\) \(\ds \frac {\d u} {2 t}\)
\(\ds \leadsto \ \ \) \(\ds \) \(=\) \(\ds \frac {\d u} {2 \sqrt u}\)

Substituting back into our equation, we have:

\(\ds \map F {-x, n; x + n + 1; -1}\) \(=\) \(\ds \dfrac {\map \Gamma {x + n + 1} } {\map \Gamma n \map \Gamma {x + 1 } } \int_0^1 \paren {\sqrt u}^{n - 1} \paren {1 - u}^x \frac {\d u} {2 \sqrt u}\)
\(\ds \) \(=\) \(\ds \dfrac 1 2 \dfrac {\map \Gamma {x + n + 1} } {\map \Gamma n \map \Gamma {x + 1 } } \int_0^1 u^{\frac n 2 - 1} \paren {1 - u}^x \d u\)
\(\ds \leadsto \ \ \) \(\ds \) \(=\) \(\ds \dfrac 1 2 \dfrac {\map \Gamma {x + n + 1} } {\map \Gamma n \map \Gamma {x + 1 } } \dfrac {\map \Gamma {\dfrac n 2 } \map \Gamma {x + 1 } } {\map \Gamma {\dfrac n 2 + x + 1 } }\) Definition of Beta Function
\(\ds \leadsto \ \ \) \(\ds \) \(=\) \(\ds \dfrac {\dfrac 1 2 \map \Gamma {x + n + 1} } {\map \Gamma n } \dfrac {\map \Gamma {\dfrac n 2 } } {\map \Gamma {\dfrac n 2 + x + 1 } }\) simplifying and canceling $\map \Gamma {x + 1 }$
\(\ds \leadsto \ \ \) \(\ds \) \(=\) \(\ds \dfrac {\dfrac n 2 \map \Gamma {x + n + 1} } {n \map \Gamma n } \dfrac {\map \Gamma {\dfrac n 2 } } {\map \Gamma {\dfrac n 2 + x + 1 } }\) multiplying by $1$
\(\ds \leadsto \ \ \) \(\ds \) \(=\) \(\ds \dfrac {\map \Gamma {x + n + 1} \map \Gamma {\dfrac n 2 + 1} } {\map \Gamma {x + \dfrac n 2 + 1} \map \Gamma {n + 1} }\) Definition of Gamma Function

$\blacksquare$


Source of Name

This entry was named for Ernst Eduard Kummer.