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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?

Quantization
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It is true in the sense of analytic continuation. The integral you wrote down does not converge anymore. To clear it up a bit, this is an example of how analytic continuation works:

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.

There are, however, other integral representations, usually over a certain contour, that continue to converge in a larger set. One such example for the Beta function is this, you can read more about it here, where it is stated that This Pochhammer contour integral converges for all values of α and β and so gives the analytic continuation of the beta function.

shdp
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