Difference between revisions of "B(x,y)B(x+y,z)B(x+y+z,u)=Gamma(x)Gamma(y)Gamma(z)Gamma(u)/Gamma(x+y+z+u)"
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(Created page with "==Theorem== The following formula holds: $$B(x,y)B(x+y,z)B(x+y+z,u)=\dfrac{\Gamma(x)\Gamma(y)\Gamma(z)\Gamma(u)}{\Gamma(x+y+z+u)},$$ where $B$ denotes the beta function an...") |
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Revision as of 23:02, 24 June 2017
Theorem
The following formula holds: $$B(x,y)B(x+y,z)B(x+y+z,u)=\dfrac{\Gamma(x)\Gamma(y)\Gamma(z)\Gamma(u)}{\Gamma(x+y+z+u)},$$ where $B$ denotes the beta function and $\Gamma$ denotes the gamma function.
Proof
References
- 1953: Harry Bateman: Higher Transcendental Functions Volume I ... (previous) ... (next): $\S 1.5 (8)$
