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Thread: Applying Mellin Transformation

  1. #1
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    Applying Mellin Transformation

    Which Dirichlet Series is obtained as the Mellin transformation of
    $\displaystyle \frac{1}{e^t+ 1}= \sum_{n=0}^\infty (-1)^n (e^{-t})^{n+1}$
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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
    $\displaystyle \frac{1}{e^t+ 1}= \sum_{n=0}^\infty (-1)^n (e^{-t})^{n+1}$
    $\displaystyle \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 $\displaystyle t=x(n+1) $ which means $\displaystyle dt=(n+1)dx $.

    We then get $\displaystyle \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 $\displaystyle \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 $\displaystyle \zeta(s) $.

    Note: We require $\displaystyle x>0 $ for the original sum to converge. It also converges absolutely for $\displaystyle x>0 $ and that's why we can move the integral sign into the summation.
    Last edited by chiph588@; Apr 3rd 2010 at 08:55 AM.
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    Ok thanks, what can you say about $\displaystyle \frac{\left\{\mathcal{M}f\right\}(s)}{\zeta(s)}$ ? Deduce that

    $\displaystyle \zeta(s) = \frac{1}{\Gamma(s)(1-2^{1-s})} \int_0^s \frac{t^{s-1}}{e^t+1}$
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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 $\displaystyle \frac{\left\{\mathcal{M}f\right\}(s)}{\zeta(s)}$ ? Deduce that

    $\displaystyle \zeta(s) = \frac{1}{\Gamma(s)(1-2^{1-s})} \int_0^s \frac{t^{s-1}}{e^t+1}$
    Just use the equation $\displaystyle \zeta(s)(1-{2^{1-s}})= \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^s} $.
    Last edited by chiph588@; Apr 3rd 2010 at 10:36 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 $\displaystyle \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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    Ok good, can you still evaluate $\displaystyle
    \frac{\left\{\mathcal{M}f\right\}(s)}{\zeta(s)}
    $ ?
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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 $\displaystyle
    \frac{\left\{\mathcal{M}f\right\}(s)}{\zeta(s)}
    $ ?
    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 $\displaystyle F(s) = \sum_{n=0}^\infty (-1)^n (e^{-s})^{n+1}$

    Then we should have $\displaystyle f(s) = \frac{1}{\Gamma(s)} \int_0^\infty \sum_{n=0}^\infty ((-1)^n (e^{-e^{-t}})^{n+1}) * t^{s-1}dt$. 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 $\displaystyle F(s) = \sum_{n=0}^\infty (-1)^n (e^{-s})^{n+1}$

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

    We have $\displaystyle f(x) = \sum_{n=0}^\infty (-1)^n e^{-x(n+1)} $, so $\displaystyle \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 $\displaystyle \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 $\displaystyle t=x(n+1) $ which means $\displaystyle dt=(n+1)dx $.

    We then get $\displaystyle \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 $\displaystyle \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 $\displaystyle g(s) = \sum_{n=0}^\infty \frac{a_n}{n^s} $ and $\displaystyle G(z) = \sum_{n=1}^\infty a_n z^n $, where $\displaystyle a_n=(-1)^{n+1} $.
    Note $\displaystyle G(z) = \sum_{n=0}^\infty a_{n+1} z^{n+1} $, thus $\displaystyle 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 $\displaystyle g(s) = \frac{1}{\Gamma(s)}\int_0^\infty G(e^{-t})t^{s-1}dt $.
    Last edited by chiph588@; Apr 3rd 2010 at 04:55 PM.
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  10. #10
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    ah I found my mistake. My Power series had coefficient $\displaystyle e^{-t}$ and not just $\displaystyle t$. Thanks for the post .
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