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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==References==
 
==References==
* {{BookReference|Higher Transcendental Functions Volume I|1953|Harry Bateman|prev=B(x,y)B(x+y,z)=B(z,x)B(x+z,y)|next=1/B(n,m)=m((n+m-1) choose (n-1))}}: $\S 1.5 (8)$
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* {{BookReference|Higher Transcendental Functions Volume I|1953|Arthur Erdélyi|author2=Wilhelm Magnus|author3=Fritz Oberhettinger|author4=Francesco G. Tricomi|prev=B(x,y)B(x+y,z)=B(z,x)B(x+z,y)|next=1/B(n,m)=m((n+m-1) choose (n-1))}}: $\S 1.5 (8)$
  
 
[[Category:Theorem]]
 
[[Category:Theorem]]
 
[[Category:Unproven]]
 
[[Category:Unproven]]

Latest revision as of 21:01, 3 March 2018

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