Difference between revisions of "B(x,y)=2^(1-x-y)integral (1+t)^(x-1)(1-t)^(y-1)+(1+t)^(y-1)(1-t)^(x-1) dt"

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(Created page with "==Theorem== The following formula holds for $\mathrm{Re}(x)>0$ and $\mathrm{Re}(y)>0$: $$B(x,y) = 2^{1-x-y} \displaystyle\int_0^1 (1+t)^{x-1}(1-t)^{y-1} + (1+t)^{y-1}(1-t)^{x-...")
 
 
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==References==
 
==References==
* {{BookReference|Higher Transcendental Functions Volume I|1953|Harry Bateman|prev=1/B(n,m)=n((n+m-1) choose (m-1))|next=integral t^(x-1)(1-t)^(y-1)(1+bt)^(-x-y)dt = (1+b)^(-x)B(x,y)}}: $\S 1.5 (10)$
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* {{BookReference|Higher Transcendental Functions Volume I|1953|Arthur Erdélyi|author2=Wilhelm Magnus|author3=Fritz Oberhettinger|author4=Francesco G. Tricomi|prev=1/B(n,m)=n((n+m-1) choose (m-1))|next=integral t^(x-1)(1-t)^(y-1)(1+bt)^(-x-y)dt = (1+b)^(-x)B(x,y)}}: $\S 1.5 (10)$
  
 
[[Category:Theorem]]
 
[[Category:Theorem]]
 
[[Category:Unproven]]
 
[[Category:Unproven]]

Latest revision as of 21:02, 3 March 2018

Theorem

The following formula holds for $\mathrm{Re}(x)>0$ and $\mathrm{Re}(y)>0$: $$B(x,y) = 2^{1-x-y} \displaystyle\int_0^1 (1+t)^{x-1}(1-t)^{y-1} + (1+t)^{y-1}(1-t)^{x-1} \mathrm{d}t,$$ where $B$ denotes the beta function.

Proof

References