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

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

==Proof==

==References==
* {{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-t)^(y-1)(1+bt)^(-x-y)dt = (1+b)^(-x)B(x,y)|next=integral (t-b)^(x-1)(a-t)^(y-1)dt=(a-b)^(x+y-1)B(x,y)}}: $\S 1.5 (12)$

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