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

From specialfunctionswiki
Jump to: navigation, search
(Created page with "==Theorem== The following holds for $b>-1$, $\mathrm{Re}(x)>0$, and $\mathrm{Re}(y)>0$: $$\displaystyle\int_0^1 t^{x-1}(1-t)^{y-1}(1+bt)^{-x-y} \mathrm{d}t = (1+b)^{-x} B(x,y)...")
 
 
Line 7: Line 7:
  
 
==References==
 
==References==
* {{BookReference|Higher Transcendental Functions Volume I|1953|Harry Bateman|prev=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|next=integral t^(x-1)(1+bt)^(-x-y) dt = b^(-x) B(x,y)}}: $\S 1.5 (11)$
+
* {{BookReference|Higher Transcendental Functions Volume I|1953|Arthur Erdélyi|author2=Wilhelm Magnus|author3=Fritz Oberhettinger|author4=Francesco G. Tricomi|prev=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|next=integral t^(x-1)(1+bt)^(-x-y) dt = b^(-x) B(x,y)}}: $\S 1.5 (11)$
  
 
[[Category:Theorem]]
 
[[Category:Theorem]]
 
[[Category:Unproven]]
 
[[Category:Unproven]]

Latest revision as of 21:02, 3 March 2018

Theorem

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

Proof

References