# Beta function with one negative and one positive arguments

Beta function defined as $B(x,y) = \int_0^1 t^{x-1} (1-t)^{y-1} dt$ is only well defined when Re $x,$ Re $y >0$. However, according to "http://www.efunda.com/math/beta/," we can use the fact that Beta function is expressed in terms of Gamma functions to define beta function at negative arguments.

I am questioning the logic of this page, since identification of beta function to gamma function is probably valid at non-negative arguments to begin with. Can we actually say that the result of this page at negative arguments of (x,y) is true at the above integration level?

Also, if this is true, then can we also do this for negative 'x' and positive 'y', for instance?

Let $f:G\rightarrow\mathbb{C}$ be defined by $f(z)=\int_{0}^{\infty}e^{-s}s^{z-1}\mathrm{d}s$, where $G=\left\{z\in\mathbb{C}:\Re z>0\right\}$. It can be shown that $f$ is an analytic function on $G$.
Then we define $g:\mathbb{C}\setminus\mathbb{Z}_{\leq0}\rightarrow\mathbb{C}$ as $$g(z) =\begin{cases} f(z)&\mbox{ if }\Re z>0\\ \frac{f(z+n)}{z(z+1)\cdots(z+n-1)}&\mbox{ if }\Re z+n\in(0,1] \end{cases}$$ So that $g$ is analytic in its domain, effectively an analytic extension of $f$. Note crucially that $g$ is not given by a divergent integral, the shift of the argument in $f(z+n)$ precisely avoids this.