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Math Help - Applying Mellin Transformation

  1. #1
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    Applying Mellin Transformation

    Which Dirichlet Series is obtained as the Mellin transformation of
    \frac{1}{e^t+ 1}= \sum_{n=0}^\infty (-1)^n (e^{-t})^{n+1}
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by EinStone View Post
    Which Dirichlet Series is obtained as the Mellin transformation of
    \frac{1}{e^t+ 1}= \sum_{n=0}^\infty (-1)^n (e^{-t})^{n+1}
     \left\{\mathcal{M}f\right\}(s) = \int_0^\infty \frac{x^{s-1}}{e^x+1}dx = \sum_{n=0}^\infty (-1)^n\int_0^\infty x^{s-1}e^{-x(n+1)} dx

    Now let  t=x(n+1) which means  dt=(n+1)dx .

    We then get  \left\{\mathcal{M}f\right\}(s) = \sum_{n=0}^\infty \frac{(-1)^n}{(n+1)^s}\int_0^\infty t^{s-1}e^{-t} dt

    Therefore  \left\{\mathcal{M}f\right\}(s) = \Gamma(s)\sum_{n=0}^\infty \frac{(-1)^n}{(n+1)^s} = \Gamma(s)\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^s}

    This Dirichlet series can then be written in terms of  \zeta(s) .

    Note: We require  x>0 for the original sum to converge. It also converges absolutely for  x>0 and that's why we can move the integral sign into the summation.
    Last edited by chiph588@; April 3rd 2010 at 09:55 AM.
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    Ok thanks, what can you say about \frac{\left\{\mathcal{M}f\right\}(s)}{\zeta(s)} ? Deduce that

    \zeta(s) = \frac{1}{\Gamma(s)(1-2^{1-s})} \int_0^s \frac{t^{s-1}}{e^t+1}
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by EinStone View Post
    Ok thanks, what can you say about \frac{\left\{\mathcal{M}f\right\}(s)}{\zeta(s)} ? Deduce that

    \zeta(s) = \frac{1}{\Gamma(s)(1-2^{1-s})} \int_0^s \frac{t^{s-1}}{e^t+1}
    Just use the equation  \zeta(s)(1-{2^{1-s}})= \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^s} .
    Last edited by chiph588@; April 3rd 2010 at 11:36 AM.
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  5. #5
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by chiph588@ View Post
    Just use the equation  \zeta(s)(1-{2^{1-s}})= \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^s} .
    Here's a proof of this.
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  6. #6
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    Ok good, can you still evaluate <br />
\frac{\left\{\mathcal{M}f\right\}(s)}{\zeta(s)}<br />
?
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  7. #7
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by EinStone View Post
    Ok good, can you still evaluate <br />
\frac{\left\{\mathcal{M}f\right\}(s)}{\zeta(s)}<br />
?
    What do you mean?
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    OK I will ask from beginning. I learned that the Mellin transformation transforms Power series in Dirichlet Series. If
    and
    then .

    Now in my case F(s) = \sum_{n=0}^\infty (-1)^n (e^{-s})^{n+1}

    Then we should have f(s) = \frac{1}{\Gamma(s)} \int_0^\infty \sum_{n=0}^\infty ((-1)^n (e^{-e^{-t}})^{n+1}) * t^{s-1}dt. How do I get your result from here?
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  9. #9
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by EinStone View Post
    OK I will ask from beginning. I learned that the Mellin transformation transforms Power series in Dirichlet Series. If
    and
    then .

    Now in my case F(s) = \sum_{n=0}^\infty (-1)^n (e^{-s})^{n+1}

    Then we should have f(s) = \frac{1}{\Gamma(s)} \int_0^\infty \sum_{n=0}^\infty ((-1)^n (e^{-e^{-t}})^{n+1}) * t^{s-1}dt. How do I get your result from here?
    The definition of the Mellin transformation is  \left\{\mathcal{M}f\right\}(s) = \int_0^\infty x^{s-1}f(x)dx

    We have  f(x) = \sum_{n=0}^\infty (-1)^n e^{-x(n+1)} , so  \left\{\mathcal{M}f\right\}(s) = \int_0^\infty x^{s-1}\sum_{n=0}^\infty (-1)^n e^{-x(n+1)}dx .

    Now simplify a bit to get  \left\{\mathcal{M}f\right\}(s) = \sum_{n=0}^\infty (-1)^n \int_0^\infty x^{s-1}e^{-x(n+1)}dx .

    Now let  t=x(n+1) which means  dt=(n+1)dx .

    We then get  \left\{\mathcal{M}f\right\}(s) = \sum_{n=0}^\infty \frac{(-1)^n}{(n+1)^s}\int_0^\infty t^{s-1}e^{-t} dt

    Therefore  \left\{\mathcal{M}f\right\}(s) = \Gamma(s)\sum_{n=0}^\infty \frac{(-1)^n}{(n+1)^s} = \Gamma(s)\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^s}


    So now, let  g(s) = \sum_{n=0}^\infty \frac{a_n}{n^s} and  G(z) = \sum_{n=1}^\infty a_n z^n , where  a_n=(-1)^{n+1} .
    Note  G(z) = \sum_{n=0}^\infty a_{n+1} z^{n+1} , thus  G(e^{-t}) = \sum_{n=0}^\infty a_{n+1} (e^{-t})^{n+1} .

    So as you can see (hopefully), we do get your identity for this case, namely  g(s) = \frac{1}{\Gamma(s)}\int_0^\infty G(e^{-t})t^{s-1}dt .
    Last edited by chiph588@; April 3rd 2010 at 05:55 PM.
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  10. #10
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    ah I found my mistake. My Power series had coefficient e^{-t} and not just t. Thanks for the post .
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