Convergence of the beta function $\beta(x,y)$

Question

I have a question regarding the beta function $\beta(x,y)=\int_{0}^{1}t^{x-1}(1-t)^{y-1}dt$ which has been asked here (Show that the Beta Function $\beta (x,y)$ Converges When $x \gt 0, \space y \gt 0$), and an answer supplied, however to my mind there are still some questions which I think need clarifying, or at least a fuller answer provided.

The result I would like explained/shown is why $\beta(x,y)$ converges for any $x,y>0$, whereas it appears that in fact this convergence only occurs for $x,y>1$, as the OP in the question I have referenced here has also concluded.

It appears the strategy is to split the integral into two sections, one on $(0,0.5]$ and the other on $[0.5,1)$. The logic for this as far as I can tell is as follows. For $y-1<0$ the term $(1-t)^{y-1}$ is undefined at $t=1$, and so by restricting $t$ away from $1$, say $0<t\leq c<1$ then $(1-t)^{y-1}$ is defined for all $y$, and furthermore it is bounded on $(0,c]$. Now since $t^{x-1}(1-t)^{y-1}\leq(1-t)^{y-1}$ when $x-1\geq 0$ we have that

$\int_{0}^{c}t^{x-1}(1-t)^{y-1}dt\leq\int_{0}^{c}(1-t)^{y-1}dt\hspace{20pt}$ for all $y$,$\hspace{5pt}x-1\geq0$,

and furthermore that the above integral is finite.

Similarly for $x-1<0$ the term $t^{x-1}$ is undefined at $t=0$, and so by restricting $t$ away from $0$, say $0<c\leq t<1$ the term $t^{x-1}$ is defined for all $x$, and furthermore is bounded on $[c,1)$. Now since $t^{x-1}(1-t)^{y-1}\leq t^{x-1}$ when $y-1\geq 0$ we have

$\int_{c}^{1}t^{x-1}(1-t)^{y-1}dt\leq\int_{c}^{1}t^{x-1}dt\hspace{20pt}$ for all $x$,$\hspace{5pt}y-1\geq0$,

which is again finite. Thus putting the two integrals together we see that $\beta(x,y)$ converges when $x,y\geq1$, however we need to show this occurs for $x,y>0$.

What I would like to know is why $c=0.5$? What is the significance of this specific value? Perhaps the answer to this lies in the answer supplied in the previous question as to why $x,y>0$ are sufficient for convergence of $\beta(x,y)$. That is apparently there exists a constant $m_{1}$ such that $t^{x-1}(1-t)^{y-1}\leq m_{1}(1-t)^{y-1}$ when $x> 0$, and for all $t\in(0,0.5]$. I am assuming something similar can be done with the other part of the integral. That is there exists a constant $m_{2}$ such that $t^{x-1}(1-t)^{y-1}\leq m_{2}t^{x-1}$ when $y> 0$, and for all $t\in[c,1)$.

Clearly the above method will then work with these new functions $m_{1}(1-t)^{y-1}$ and $m_{2}t^{x-1}$ as the bounding functions. Any help would be greatly appreciated.

Answer 1

The value $c=0.5$ has no meaning whatsoever, and the first step of a proof could be to note that the function $u:t\mapsto t^{x-1}(1-t)^{y-1}$ to be integrated is bounded on $(0.1,0.9)$ hence its integral on $(0,1)$ converges if and only if those on $(0,0.1)$ and $(0.9,1)$ do.

For the $(0,0.1)$ part, note that there exists some positive finite constants $C$ and $C'$ such that $Cu_x\leqslant u\leqslant C'u_x$ on $(0,0.1)$, where $u_x:t\mapsto t^{x-1}$, hence the integral of $u$ on $(0,0.1)$ converges if and only if the integral of $u_x$ on $(0,0.1)$ does. Likewise, the integral of $u$ on $(0.9,1)$ converges if and only if the integral of $v_y$ on $(0.9,1)$ does, where $v_y:t\mapsto(1-t)^{y-1}$. But $v_y(t)=u_y(1-t)$ hence we are left with the condition that the integrals of $u_x$ and $u_y$ on $(0,0.1)$ both converge.

Can you finish? As you see, it all boils down to know the values of $z$ such that the integral of $u_z$ converges at $0^+$. But we know a primitive of $u_z$, don't we?