Gamma Function of Zero

Theorem

Let $\Gamma$ denotes the Gamma function

Then:

$\map \Gamma 0$ is not defined.

Proof

\(\ds \map \Gamma 1\)

\(=\)

\(\ds 0 \, \map \Gamma 0\)

Gamma Difference Equation

\(\ds \leadsto \ \ \)

\(\ds \map \Gamma 0\)

\(=\)

\(\ds \dfrac {\map \Gamma 1} 0\)

\(\ds \)

\(=\)

\(\ds \dfrac 1 0\)

Gamma Function Extends Factorial

But $\dfrac 1 0$ is not defined.

Hence the result.

$\blacksquare$

Sources

• 1965: Murray R. Spiegel: Theory and Problems of Laplace Transforms ... (previous) ... (next): Chapter $1$: The Laplace Transform: Solved Problems: The Gamma Function: $33 \ \text{(d)}$