Results 1 to 3 of 3

Thread: beta function convergence

  1. #1
    Junior Member
    Joined
    Sep 2008
    Posts
    55

    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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    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: $\displaystyle \int_0^1 t^{m-1}(1-t)^{n-1}dt$ )

    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 $\displaystyle n>0$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2012
    From
    Czech Republic
    Posts
    2

    Re: beta function convergence

    Quote Originally Posted by Laurent View Post
    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?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. The beta function integration
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Aug 31st 2010, 11:11 AM
  2. [SOLVED] beta function
    Posted in the Advanced Statistics Forum
    Replies: 15
    Last Post: Feb 16th 2010, 03:39 PM
  3. Beta function proof
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Jun 14th 2009, 03:03 AM
  4. Beta function?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Feb 10th 2009, 03:22 PM
  5. beta function
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Nov 21st 2008, 03:57 PM

Search tags for this page

Search Tags


/mathhelpforum @mathhelpforum