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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?

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    $\begingroup$ Can you show $\int_0^1t^{x-1}~\mathrm dt$ converges for all $x>0$? $\endgroup$ – Simply Beautiful Art Aug 13 at 17:26
  • $\begingroup$ @SimplyBeautifulArt I'm afraid I don't know how... $\endgroup$ – Ruben Kruepper Aug 13 at 17:33
  • $\begingroup$ You don't know how to integrate $t^{x-1}$? $\endgroup$ – Simply Beautiful Art Aug 13 at 17:38
  • $\begingroup$ @SimplyBeautifulArt Oh god sorry of course I do its $\frac{1}{x}t^x$ $\endgroup$ – Ruben Kruepper Aug 13 at 17:39
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    $\begingroup$ Can you show $\int_0^1(1-t)^{y-1}~\mathrm dt$ converges for all $y>0$? Can you show that $(1-t)^{y-1}$ is bounded for $t\in[0,0.5]$ and $t^{x-1}$ is bounded for $t\in[0.5,1]$? $\endgroup$ – Simply Beautiful Art Aug 13 at 17:44
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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 $(1-t)^{y-1}$.

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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.

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  • $\begingroup$ I think I see what you're trying to do, is there a way to formalize that result? I fear examples will be enough for the prof. ;) $\endgroup$ – Ruben Kruepper Aug 13 at 17:42
  • $\begingroup$ #@Ruben Kruepper I have added some more stuff, you may see it now. $\endgroup$ – Dr Zafar Ahmed DSc Aug 13 at 18:27
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$\frac{\Gamma{(\alpha+\beta)}}{\Gamma{(\alpha)}\Gamma{(\beta)}}t^{x-1}(1-t)^{y-1}$ is the Beta density and the area under the curve is 1

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

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