Difference between revisions of "Gamma(z+1)=zGamma(z)"
From specialfunctionswiki
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* {{BookReference|Special Functions for Scientists and Engineers|1968|W.W. Bell|prev=Gamma(1)=1|next=Gamma(n+1)=n!}}: Theorem 2.2 | * {{BookReference|Special Functions for Scientists and Engineers|1968|W.W. Bell|prev=Gamma(1)=1|next=Gamma(n+1)=n!}}: Theorem 2.2 | ||
* {{BookReference|Hypergeometric Orthogonal Polynomials and Their q-Analogues|2010|Roelof Koekoek|author2=Peter A. Lesky|author3=René F. Swarttouw|prev=Gamma|next=findme}}: $(1.2.2)$ | * {{BookReference|Hypergeometric Orthogonal Polynomials and Their q-Analogues|2010|Roelof Koekoek|author2=Peter A. Lesky|author3=René F. Swarttouw|prev=Gamma|next=findme}}: $(1.2.2)$ | ||
| + | * {{BookReference|Special functions, a graduate text|2010|Richard Beals|author2=Roderick Wong|prev=Gamma|next=findme}}: $(2.1.2)$ | ||
[[Category:Theorem]] | [[Category:Theorem]] | ||
[[Category:Proven]] | [[Category:Proven]] | ||
[[Category:Justify]] | [[Category:Justify]] | ||
Latest revision as of 18:12, 16 June 2018
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): $(2.1.2)$
