Difference between revisions of "Integral (t-b)^(x-1)(a-t)^(y-1)dt=(a-b)^(x+y-1)B(x,y)"

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(Created page with "==Theorem== The following formula holds for $\mathrm{Re}(x)>0$, $\mathrm{Re}(y)>0$, and $b<a$: $$\displaystyle\int_a^b (t-b)^{x-1}(a-t)^{y-1} \mathrm{d}t=(a-b)^{x+y-1}B(x,y),$...")
 
 
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==References==
 
==References==
* {{BookReference|Higher Transcendental Functions Volume I|1953|Harry Bateman|prev=Integral t^(x-1)(1+bt)^(-x-y) dt = b^(-x) B(x,y)|next=integral of (t-b)^(x-1)(a-t)^(y-1)/(t-x)^(x+y) dt=(a-b)^(x+y-1)/((a-c)^x(b-c)^y) B(x,y)}}: $\S 1.5 (13)$
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* {{BookReference|Higher Transcendental Functions Volume I|1953|Arthur Erdélyi|author2=Wilhelm Magnus|author3=Fritz Oberhettinger|author4=Francesco G. Tricomi|prev=Integral t^(x-1)(1+bt)^(-x-y) dt = b^(-x) B(x,y)|next=integral of (t-b)^(x-1)(a-t)^(y-1)/(t-x)^(x+y) dt=(a-b)^(x+y-1)/((a-c)^x(b-c)^y) B(x,y)}}: $\S 1.5 (13)$
  
 
[[Category:Theorem]]
 
[[Category:Theorem]]
 
[[Category:Unproven]]
 
[[Category:Unproven]]

Latest revision as of 21:03, 3 March 2018

Theorem

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

Proof

References