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Math Help - Challenge

  1. #1
    MHF Contributor Mathstud28's Avatar
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    Challenge

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

    \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...
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    We have some good integrators here...well...let see this one done. Prove

    \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
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  3. #3
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    First step

    Well, right away I see you can make the substitution x=au. Then we have
    \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
    =\int_0^{\infty}\frac{a^{m+1}u^m}{a^{np}(1+u^n)^p}  \,du
    =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
    \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.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    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

    0<m+1<np Although you may be right due to the (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; June 15th 2008 at 10:43 PM.
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  5. #5
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    Next steps

    Next, I see trying to find
    \int_0^{\infty}\frac{u^m}{(1+u^n)^p}\,du
    by the substitution x=1+u^n. Thus gives dx=nu^{n-1}, and thus we see:
    \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 u=\frac{1}{x}. This gives:
    \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).
    =\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:
    \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
    =\frac{1}{n}B(\frac{np-m-n-1}{n}+1,\frac{m-n+1}{n}+1)
    =\frac{1}{n}B(\frac{np-m-1}{n},\frac{m+1}{n})
    Where B(x,y) is the Beta function.

    Now, B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}
    so
    \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})}
    =\frac{1}{n}\frac{\Gamma(\frac{np-m-1}{n})\Gamma(\frac{m+1}{n})}{\Gamma(p)}
    =\frac{1}{n(p-1)!}\Gamma(\frac{np-m-1}{n})\Gamma(\frac{m+1}{n})

    A little more to follow.

    --Kevin C.
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  6. #6
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    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:
    \Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin{\pi{z}}}.
    So:
    \Gamma(z)=\frac{\pi}{\sin(\pi{z})\Gamma(1-z)}
    and so:
    \Gamma(\frac{np-m-1}{n})=\frac{\pi}{\sin(\pi\frac{np-m-1}{n})\Gamma(1-\frac{np-m-1}{n})}
    =\frac{\pi}{\sin\left(\pi{p}-\pi\frac{m+1}{n}\right)\Gamma(\frac{n+m+1-np}{n})}
    =\frac{(-1)^p\pi}{\sin\left(-\frac{(m+1)\pi}{n}\right)\Gamma(\frac{n+m+1-np}{n})}
    =\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:
    \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.
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