View source for B(x,y)B(x+y,z)=B(y,z)B(y+z,x)

==Theorem==
The following formula holds:
$$B(x,y)B(x+y,z)=B(y,z)B(y+z,x),$$
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=B(x,y+1)=(y/(x+y))B(x,y)|next=B(x,y)B(x+y,z)=B(z,x)B(x+z,y)}}: $\S 1.5 (7)$

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