View source for Integral of (1+t)^(2x-1)(1-t)^(2y-1)(1+t^2)^(-x-y)dt=2^(x+y-2)B(x,y)

==Theorem==
The following formula holds for $\mathrm{Re}(x)>0$ and $\mathrm{Re}(y)>0$:
$$\displaystyle\int_{-1}^1 (1+t)^{2x-1}(1-t)^{2y-1}(1+t^2)^{-x-y} \mathrm{d}t= 2^{x+y-2}B(x,y),$$
where $B$ denotes the [[beta]] function.

==Proof==

==References==
* {{BookReference|Higher Transcendental Functions Volume I|1953|Harry Bateman|prev=Integral of t^(x-1)(1-t^z)^(y-1) dt=(1/z)B(x/z,y)|next=Beta in terms of sine and cosine}}: $\S 1.5 (18)$

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