I am only interested in real numbers, so here $x,y>0$ real numbers. The Beta function is defined as $$B(x,y)= \int_0^1 w^{x-1}(1-w)^{y-1}dw.$$
The above integral is defined for every $x,y>0$. Now, take $t>0$ fixed and observe the following: $$\int_0^t w^{x-1}(t-w)^{y-1}dw = t^{x+y-1} \int_0^1 z^{x-1}(1-z)^{y-1}dw=t^{x+y-1} B(x,y),$$
where I used the change of variables $z=w/t$. Hence, $$\int_0^t w^{x-1}(t-w)^{y-1}dw = t^{x+y-1} B(x,y).$$
But, now, the above seems weird to me. In principle, $B$ is well-defined and convergent for any $x,y>0$. Nevertheless if, e.g. $x=y=1/4$ then $$\int_0^t w^{1/4-1}(t-w)^{1/4-1}dw = t^{-1/2} B(1/4,1/4),$$ and hence I get a negative exponent in $t$ which then explodes when $t\to 0$. What is wrong here? :S
Thanks for any ideas.
Indeed, there is no contradiction. The issue here is that $t$ appears in the integrand as well as in the upper limit of integration, so as $t \to 0^+$, the minimum value for $0 < x, y < 1$ of $$f(w;t) = w^{x-1} (t-w)^{y-1}$$ increases without bound. You can see this by setting $x = y = 1/4$ and noting the minimum is attained when $w = t/2$, giving the crude lower bound for the integral as $$\int_{w=0}^t w^{-3/4} (t-w)^{-3/4} \, dw > \frac{2\sqrt{2}}{\sqrt{t}}.$$ Then as $t \to 0^+$ the divergence is clear.
Geometrically, your thinking is a bit like saying, "if I take a unit square and I perform a sequence of transformations that scales its width by $t$ while scaling its height by $1/t^2$ as $t \to 0^+$, why would the limiting area become infinite?"