View source for Integral of t^(x-1)(1-t^z)^(y-1) dt=(1/z)B(x/z,y)

==Theorem==
The following formula holds for $z>0$, $\mathrm{Re}(y)>0$, and $\mathrm{Re}(x)>0$:
$$\displaystyle\int_0^1 t^{x-1} (1-t^z)^{y-1} \mathrm{d}t = \dfrac{1}{z} B \left( \dfrac{x}{z}, y \right),$$
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 of (1+bt^z)^(-y)t^x dt = (1/z)*b^(-(x+1)/z) B((x+1)/z,y-(x+1)/z)|next=Integral of (1+t)^(2x-1)(1-t)^(2y-1)(1+t^2)^(-x-y)dt=2^(x+y-2)B(x,y)}}: $\S 1.5 (17)$

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