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

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?



Answers 2


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

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

jcarne
jcarne
August 13, 2019 17:28 PM

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

Dr Zafar Ahmed DSc
Dr Zafar Ahmed DSc
August 13, 2019 17:36 PM

Related Questions


Updated April 22, 2018 12:20 PM

Updated August 14, 2018 06:20 AM

Updated May 16, 2019 16:20 PM