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Math Help - beta function convergence

  1. #1
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    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?
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  2. #2
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    Quote Originally Posted by minivan15 View Post
    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<t^{m-1}\left(\frac{1}{2}\right)^{n-1}\leq t^{m-1}(1-t)^{n-1} if n-1>0 and 0<t^{m-1}\leq t^{m-1}(1-t)^{n-1} 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.
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  3. #3
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    Re: beta function convergence

    Quote Originally Posted by Laurent View Post
    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<t^{m-1}\left(\frac{1}{2}\right)^{n-1}\leq t^{m-1}(1-t)^{n-1} if n-1>0 and 0<t^{m-1}\leq t^{m-1}(1-t)^{n-1} 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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