Results 1 to 6 of 6

Thread: Challenge

  1. #1
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641

    Challenge

    We have some good integrators here...well...let see this one done. Prove

    $\displaystyle \int_0^{\infty}\frac{x^m}{(a^n+x^n)^p}dx=\frac{(-1)^{p-1}\cdot\pi\cdot{a^{m+1-n\cdot{p}}}\Gamma\left[\frac{m+1}{n}\right]}{n\sin\left[\frac{(m+1)\pi}{n}\right](p-1)!\Gamma\left[\frac{m+1}{n-p+1}\right]}$

    Good luck...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    11,152
    Thanks
    731
    Awards
    1
    Quote Originally Posted by Mathstud28 View Post
    We have some good integrators here...well...let see this one done. Prove

    $\displaystyle \int_0^{\infty}\frac{x^m}{(a^n+x^n)^p}dx=\frac{(-1)^{p-1}\cdot\pi\cdot{a^{m+1-n\cdot{p}}}\Gamma\left[\frac{m+1}{n}\right]}{n\sin\left[\frac{(m+1)\pi}{n}\right](p-1)!\Gamma\left[\frac{m+1}{n-p+1}\right]}$

    Good luck...
    From the way your solution is written I am presuming m, n, and p are all positive integers?

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Dec 2007
    From
    Anchorage, AK
    Posts
    276

    First step

    Well, right away I see you can make the substitution $\displaystyle x=au$. Then we have
    $\displaystyle \int_0^{\infty}\frac{x^m}{(a^n+x^n)^p}\,dx=\int_0^ {\infty}\frac{a^mu^m}{(a^n+a^nu^n)^p}a\,du$
    $\displaystyle =\int_0^{\infty}\frac{a^{m+1}u^m}{a^{np}(1+u^n)^p} \,du$
    $\displaystyle =a^{m+1-np}\int_0^{\infty}\frac{u^m}{(1+u^n)^p}\,du$
    Which takes care of the term containing a. Proving that
    $\displaystyle \int_0^{\infty}\frac{u^m}{(1+u^n)^p}\,du=\frac{(-1)^{p-1}\pi\Gamma\left(\frac{m+1}{n}\right)}{n\sin\left( \frac{(m+1)\pi}{n}\right)(p-1)!\Gamma\left(\frac{m+1}{n-p+1}\right)}$
    will require a bit more thought. I'm thinking a beta function integral may be involved at some point.

    --Kevin C.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by topsquark View Post
    From the way your solution is written I am presuming m, n, and p are all positive integers?

    -Dan
    No integers just

    $\displaystyle 0<m+1<np$ Although you may be right due to the $\displaystyle (p-1)!$

    It should also be noted that I do not have the solution to this, that is why I am so curious.
    Last edited by Mathstud28; Jun 15th 2008 at 09:43 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Dec 2007
    From
    Anchorage, AK
    Posts
    276

    Next steps

    Next, I see trying to find
    $\displaystyle \int_0^{\infty}\frac{u^m}{(1+u^n)^p}\,du$
    by the substitution $\displaystyle x=1+u^n$. Thus gives $\displaystyle dx=nu^{n-1}$, and thus we see:
    $\displaystyle \int_0^{\infty}\frac{u^m}{(1+u^n)^p}\,du=\frac{1}{ n}\int_1^{\infty}\frac{(x-1)^{\frac{m-n+1}{n}}}{x^p}\,du$.

    Next, try the substitution $\displaystyle u=\frac{1}{x}$. This gives:
    $\displaystyle \frac{1}{n}\int_1^{\infty}\frac{(x-1)^{\frac{m-n+1}{n}}}{x^p}\,du=\frac{1}{n}\int_1^0u^p\left(\fr ac{1}{u}-1\right)^{\frac{m-n+1}{n}}\left(-\frac{du}{u^2}\right)$.
    $\displaystyle =\frac{1}{n}\int_0^1u^{\frac{np-m-n-1}{n}}(1-u)^{\frac{m-n+1}{n}}\,du$.
    This is the Eulerian integral of the first kind, and so we have:
    $\displaystyle \int_0^{\infty}\frac{u^m}{(1+u^n)^p}\,du=\frac{1}{ n}\int_0^1u^{\frac{np-m-n-1}{n}}(1-u)^{\frac{m-n+1}{n}}\,du$
    $\displaystyle =\frac{1}{n}B(\frac{np-m-n-1}{n}+1,\frac{m-n+1}{n}+1)$
    $\displaystyle =\frac{1}{n}B(\frac{np-m-1}{n},\frac{m+1}{n})$
    Where $\displaystyle B(x,y)$ is the Beta function.

    Now, $\displaystyle B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$
    so
    $\displaystyle \int_0^{\infty}\frac{u^m}{(1+u^n)^p}\,du=\frac{1}{ n}\frac{\Gamma(\frac{np-m-1}{n})\Gamma(\frac{m+1}{n})}{\Gamma(\frac{np-m-1}{n}+\frac{m+1}{n})}$
    $\displaystyle =\frac{1}{n}\frac{\Gamma(\frac{np-m-1}{n})\Gamma(\frac{m+1}{n})}{\Gamma(p)}$
    $\displaystyle =\frac{1}{n(p-1)!}\Gamma(\frac{np-m-1}{n})\Gamma(\frac{m+1}{n})$

    A little more to follow.

    --Kevin C.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Dec 2007
    From
    Anchorage, AK
    Posts
    276

    I'm not getting the same result

    I'm getting a different result on the last steps:

    The gamma function has a relationship known as Euler's reflection formula:
    $\displaystyle \Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin{\pi{z}}}$.
    So:
    $\displaystyle \Gamma(z)=\frac{\pi}{\sin(\pi{z})\Gamma(1-z)}$
    and so:
    $\displaystyle \Gamma(\frac{np-m-1}{n})=\frac{\pi}{\sin(\pi\frac{np-m-1}{n})\Gamma(1-\frac{np-m-1}{n})}$
    $\displaystyle =\frac{\pi}{\sin\left(\pi{p}-\pi\frac{m+1}{n}\right)\Gamma(\frac{n+m+1-np}{n})}$
    $\displaystyle =\frac{(-1)^p\pi}{\sin\left(-\frac{(m+1)\pi}{n}\right)\Gamma(\frac{n+m+1-np}{n})}$
    $\displaystyle =\frac{(-1)^{p-1}\pi}{\sin\left(\frac{(m+1)\pi}{n}\right)\Gamma(\ frac{n+m+1-np}{n})}$

    This combines with the earlier results to give:
    $\displaystyle \int_0^{\infty}\frac{x^m}{(a^n+x^n)^p}\,dx=\frac{(-1)^{p-1}\pi{a}^{m+1-np}\Gamma\left(\frac{m+1}{n}\right)}{n\sin\left(\f rac{(m+1)\pi}{n}\right)(p-1)!\Gamma\left({\color{red}\frac{n+m+1-np}{n}}\right)}$.

    I'm not sure why I obtained a different value for the arguement of that gamma.

    --Kevin C.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Challenge. (1)
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Mar 6th 2010, 03:37 PM
  2. Challenge #1
    Posted in the Math Challenge Problems Forum
    Replies: 4
    Last Post: Oct 19th 2009, 08:23 PM
  3. a challenge
    Posted in the Algebra Forum
    Replies: 0
    Last Post: Mar 15th 2009, 08:02 PM
  4. Challenge of the Week
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: Feb 7th 2009, 01:36 AM
  5. Anyone up to a matrix challenge???
    Posted in the Math Challenge Problems Forum
    Replies: 0
    Last Post: Jan 4th 2009, 05:14 PM

Search Tags


/mathhelpforum @mathhelpforum