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

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 $\int_0^1 t^\alpha dt$ is convergent if, and only if $\alpha>-1$.

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

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

3. ## Re: beta function convergence

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

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

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