View source for Integral t^(x-1)(1-t)^(y-1)(1+bt)^(-x-y)dt = (1+b)^(-x)B(x,y)

==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==
* {{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)}}: $\S 1.5 (11)$

[[Category:Theorem]]
[[Category:Unproven]]