Show the Beta function $B(x,y)=\int_{0}^{1}t^{x-1}(1-t)^{y-1}dt$ is defined for $x,y > 0$ without actually integrating An old exam question: Show that 
$$B(x,y)=\int_{0}^{1}t^{x-1}(1-t)^{y-1}dt$$ exists for all $x,y&gt;0$.

I'm sure because of time allotment that it's not in the scope of the question to actually integrate.

Is there an elegant way to show that the integral exists without calculating it explicitely?
Hint:

It is easy to just integrate $t^{x-1}$ for $x&gt;0$, and since $(1-t)^{y-1}\approx 1$ in a neighborhood of $t=0$, the integral converges near $t=0$.

Likewise, the same holds for $t=1$ and $(1-t)^{y-1}$.
$\frac{\Gamma{(\alpha+\beta)}}{\Gamma{(\alpha)}\Gamma{(\beta)}}t^{x-1}(1-t)^{y-1}$ is the Beta density and the area under the curve is 1

Then $\int_{0}^{1}t^{x-1}(1-t)^{y-1}dt=\frac{\Gamma{(\alpha)}\Gamma{(\beta)}}{\Gamma{(\alpha+\beta)}}$