Gamma(z+1)=zGamma(z)

Theorem

The following formula holds: $$\Gamma(z+1)=z\Gamma(z),$$ where $\Gamma$ denotes gamma.

Proof

Use integration by parts to compute $$\begin{array}{ll} \Gamma(z+1) &= \displaystyle\int_0^{\infty} \xi^z e^{-\xi} \mathrm{d}\xi \\ &= -\xi^z e^{-\xi}\Bigg|_0^{\infty}- \displaystyle\int_0^{\infty} z \xi^{z-1} e^{-\xi} \mathrm{d}\xi \\ &= z\Gamma(z), \end{array}$$ as was to be shown. █

References

• 1968: W.W. Bell: Special Functions for Scientists and Engineers ... (previous) ... (next): Theorem 2.2
• 2010: Roelof Koekoek, Peter A. Lesky and René F. Swarttouw: Hypergeometric Orthogonal Polynomials and Their q-Analogues ... (previous) ... (next): $(1.2.2)$
• 2010: Richard Beals and Roderick Wong: Special functions, a graduate text ... (previous) ... (next): $(2.1.2)$