Difference between revisions of "Gamma(z+1)=zGamma(z)"

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m (Tom moved page Factorial property of gamma to Gamma(z+1)=zGamma(z): Gamma(z+1)=zGamma(z))
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Revision as of 19:42, 15 March 2018

Theorem: $$\Gamma(x+1)=x\Gamma(x), \quad x>0,$$ where $\Gamma$ denotes the gamma function.

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