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 $x,y>0$.
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?
 A: Hint:
It is easy to just integrate $t^{x-1}$ for $x>0$, 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 $(1-t)^{y-1}$.
A: Yes, write $$B(x,y)=\int_{0}^{1}\frac{dt}{t^{1-x}(1-t)^{1-y}}$$ is an improper integral it will be definedconvergent (real and finite), if $1-x<1$ and $1-y<1 \Rightarrow x, y>0.$
Like $$\int_{0}^{1} \frac{dt}{\sqrt{t}} =2,~~ \int_{0}^{1} \frac{dt}{(1-t)^{1/3}}=3/4, ~ \int_{0}^{1} \frac{dt}{\sqrt{t(1-t}}=\pi $$
Recurrence relations 
https://en.wikipedia.org/wiki/Beta_function
like $B(x+1,y ) = B(x,y) \frac{x}{x+y}, B(x,y+1) = B(x,y) \frac{y}{x+y}, B(x,y)= B(x+1,y)+B(x,y+1)$, $B(1,x)=\frac{1}{x}, B(1,1-x) = \frac{\pi}{\sin \pi x}$ 
can be helpful if one does not want to compute them by integration or by the well known formula.
A: $\frac{\Gamma{(\alpha+\beta)}}{\Gamma{(\alpha)}\Gamma{(\beta)}}t^{x-1}(1-t)^{y-1}$ is the Beta density and the area under the curve is 1
Then $\int_{0}^{1}t^{x-1}(1-t)^{y-1}dt=\frac{\Gamma{(\alpha)}\Gamma{(\beta)}}{\Gamma{(\alpha+\beta)}}$
