by Ruben Kruepper
Last Updated August 13, 2019 18:20 PM

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?

$\frac{\Gamma{(\alpha+\beta)}}{\Gamma{(\alpha)}\Gamma{(\beta)}}t^{x-1}(1-t)^{y-1}$ is the Beta density and the area under the curve is 1

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

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 $$\int_{0}^{1} \frac{dt}{\sqrt{t}} =2,~~ \int_{0}^{1} \frac{dt}{(1-t)^{1/3}}=3/4, ~ \int_{-1}^{1} \frac{dt}{\sqrt{1-t^2}}=\pi $$

Updated December 20, 2018 04:20 AM

Updated April 24, 2018 15:20 PM

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