1. ## beta function convergence

my problem is to find out for what values of m and n the beta function

( defined by: [integral from 0 to 1] t^(m-1) * (1-t)^(n-1)dt)

is convergent... I have worked on it for several hours and am stuck

obviously the problems are at the endpoints, so i broke it up into two integrals, an integral from 0 to 1/2 and an integral from 1/2 to 1.

Now my only inclination is to use the comparison theorem for different combinations of possible values for m and n...I have shown divergence or convergence in several cases but can't narrow it down to only being convergent for m,n > 0, can anyone help?

2. Originally Posted by minivan15
my problem is to find out for what values of m and n the beta function

( defined by: $\displaystyle \int_0^1 t^{m-1}(1-t)^{n-1}dt$ )

is convergent... I have worked on it for several hours and am stuck

obviously the problems are at the endpoints, so i broke it up into two integrals, an integral from 0 to 1/2 and an integral from 1/2 to 1.

Now my only inclination is to use the comparison theorem for different combinations of possible values for m and n...I have shown divergence or convergence in several cases but can't narrow it down to only being convergent for m,n > 0, can anyone help?
The only fact to use is that $\displaystyle \int_0^1 t^\alpha dt$ is convergent if, and only if $\displaystyle \alpha>-1$.

Indeed, as you said, the problem lies at endpoints, and we can split the integral in two parts.
In the first part, for $\displaystyle 0\leq t\leq \frac{1}{2}$, we have $\displaystyle 0\leq t^{m-1}(1-t)^{n-1} \leq t^{m-1}$, hence it converges if $\displaystyle m-1>-1$ (i.e. if $\displaystyle m>0$). On the other hand, we have, for the same $\displaystyle 0\leq t \leq \frac{1}{2}$, $\displaystyle 0<t^{m-1}\left(\frac{1}{2}\right)^{n-1}\leq t^{m-1}(1-t)^{n-1}$ if $\displaystyle n-1>0$ and $\displaystyle 0<t^{m-1}\leq t^{m-1}(1-t)^{n-1}$ if $\displaystyle n-1\leq 0$, and in both cases this shows that the integral diverges if $\displaystyle m-1\leq -1$. Hence this part converges iff $\displaystyle m>0$.
You can do the same on the interval $\displaystyle \left[\frac{1}{2},1\right]$ by letting $\displaystyle u=1-t$. This part converges iff $\displaystyle n>0$.

Finally, the integral converges iff $\displaystyle m>0$ and $\displaystyle n>0$.

3. ## Re: beta function convergence

Originally Posted by Laurent
The only fact to use is that $\displaystyle \int_0^1 t^\alpha dt$ is convergent if, and only if $\displaystyle \alpha>-1$.

Indeed, as you said, the problem lies at endpoints, and we can split the integral in two parts.
In the first part, for $\displaystyle 0\leq t\leq \frac{1}{2}$, we have $\displaystyle 0\leq t^{m-1}(1-t)^{n-1} \leq t^{m-1}$, hence it converges if $\displaystyle m-1>-1$ (i.e. if $\displaystyle m>0$). On the other hand, we have, for the same $\displaystyle 0\leq t \leq \frac{1}{2}$, $\displaystyle 0<t^{m-1}\left(\frac{1}{2}\right)^{n-1}\leq t^{m-1}(1-t)^{n-1}$ if $\displaystyle n-1>0$ and $\displaystyle 0<t^{m-1}\leq t^{m-1}(1-t)^{n-1}$ if $\displaystyle n-1\leq 0$, and in both cases this shows that the integral diverges if $\displaystyle m-1\leq -1$. Hence this part converges iff $\displaystyle m>0$.
You can do the same on the interval $\displaystyle \left[\frac{1}{2},1\right]$ by letting $\displaystyle u=1-t$. This part converges iff $\displaystyle n>0$.

Finally, the integral converges iff $\displaystyle m>0$ and $\displaystyle n>0$.
I agree with all the reasoning but I do not see why the inequality 0\leq t^{m-1}(1-t)^{n-1} \leq t^{m-1} should always hold if m,n > 0. What if m = n = 1/2, for instance?

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# prove integral of beta function converges

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