# 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?

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

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
August 13, 2019 17:36 PM

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