View source for Integral of (t-b)^(x-1)(a-t)^(y-1)/(c-t)^(x+y) dt = (a-b)^(x+y-1)/((c-a)^x (c-b)^y) B(x,y)

==Theorem==
The following formula holds for $\mathrm{Re}(x)>0$, $\mathrm{Re}(y)>0$, and $b<a<c$:
$$\displaystyle\int_a^b \dfrac{(t-b)^{x-1}(a-t)^{y-1}}{(c-t)^{x+y}} \mathrm{d}t = \dfrac{(a-b)^{x+y-1}}{(c-a)^x (c-b)^y} B(x,y),$$
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 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)|next=integral of (1+bt^z)^(-y)t^x dt = (1/z)*b^(-(x+1)/z) B((x+1)/z,y-(x+1)/z)}}: $\S 1.5 (15)$

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