Difference between revisions of "Gamma(z+1)=zGamma(z)"
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(Created page with "<div class="toccolours mw-collapsible mw-collapsed" style="width:800px"> <strong>Theorem:</strong> $$\Gamma(x+1)=x\Gamma(x), \quad x>0,$$ wher...") |
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<strong>Proof:</strong> Use [[integration by parts]] to compute | <strong>Proof:</strong> Use [[integration by parts]] to compute | ||
$$\begin{array}{ll} | $$\begin{array}{ll} | ||
| − | \Gamma(x+1) &= \displaystyle\int_0^{\infty} \xi^x e^{-\xi} d\xi \\ | + | \Gamma(x+1) &= \displaystyle\int_0^{\infty} \xi^x e^{-\xi} \mathrm{d}\xi \\ |
| − | &= -\xi^x e^{-\xi}|_0^{\infty} \displaystyle\int_0^{\infty} x \xi^{x-1} e^{-\xi} d\xi \\ | + | &= -\xi^x e^{-\xi}|_0^{\infty} \displaystyle\int_0^{\infty} x \xi^{x-1} e^{-\xi} \mathrm{d}\xi \\ |
&= x\Gamma(x), | &= x\Gamma(x), | ||
\end{array}$$ | \end{array}$$ | ||
Revision as of 05:44, 16 May 2016
Theorem: $$\Gamma(x+1)=x\Gamma(x), \quad x>0,$$ where $\Gamma$ denotes the gamma function.
Proof: Use integration by parts to compute $$\begin{array}{ll} \Gamma(x+1) &= \displaystyle\int_0^{\infty} \xi^x e^{-\xi} \mathrm{d}\xi \\ &= -\xi^x e^{-\xi}|_0^{\infty} \displaystyle\int_0^{\infty} x \xi^{x-1} e^{-\xi} \mathrm{d}\xi \\ &= x\Gamma(x), \end{array}$$ as was to be shown. █
