residues over simple poles

Discussion in 'Physics & Math' started by camilus, Jun 1, 2011.

  1. temur man of no words Registered Senior Member

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    1,330
    This is not true, you cannot write the same thing in two different ways and then make the 2 mysteriously disappear.

    Yeah, I am worried because now we already have two mistakes (typos?) on one page.
     
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  3. camilus the villain with x-ray glasses Registered Senior Member

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    Im sure of what Im saying, perhaps I wasnt clear enough. The first derived integral after the Mellin inversion and substitutions z->n*n*pi*z and s->s/2 was:

    \(\psi (z) = {1 \over 2 \pi i} \int_{c-i\infty}^{c+i\infty}\pi^{{-s \over 2}}\Gamma ({s \over 2}) \zeta (s) z^{{-s \over 2}} \frac{ds}{2}\)

    substituting \(\pi^{-s/2}\Gamma (s/2) \zeta (s)=\frac{2\xi(s)}{s(s-1)} \) we get

    \(\psi (z) = {1 \over 2 \pi i} \int_{c-i\infty}^{c+i\infty} \frac{2\xi(s)}{s(s-1)} z^{{-s \over 2}} \frac{ds}{2}= {1 \over 2 \pi i} \int_{c-i\infty}^{c+i\infty} \frac{\xi(s)}{s(s-1)} z^{{-s \over 2}} ds = {1 \over 2 \pi i} \int_{c-i\infty}^{c+i\infty}\pi^{{-s \over 2}}\Gamma ({s \over 2}) \zeta (s) z^{{-s \over 2}} ds\)

    now you see how the 2 cancelled. and you can see the last residue computation here:

    http://www.wolframalpha.com/input/?i=residue of pi^(-z/2)*gamma(z/2)*zeta(z) at z=1
     
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  5. temur man of no words Registered Senior Member

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    How did you get the last equality?
     
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  7. Tach Banned Banned

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    From here
    Depending on normalization, there may or may not be an extra 1/2 in front of the expression.
     
  8. Tach Banned Banned

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    You asked Wolfram to compute the residue for \(\pi^{{-s \over 2}}\Gamma ({s \over 2}) \zeta (s)\). Aren't you supposed to compute the residue for \(\pi^{{-s \over 2}}\Gamma ({s \over 2}) \zeta (s) z^{{-s \over 2}}\)?

    You should have stopped at:

    \(\psi (z) = {1 \over 2 \pi i} \int_{c-i\infty}^{c+i\infty} \frac{\xi(s)}{s(s-1)} z^{{-s \over 2}} ds\)

    because the residue is very easy to calculate (as you have actually already done at the beginning of the thread. The last equality only confuses things.
     
    Last edited: Jun 9, 2011
  9. temur man of no words Registered Senior Member

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    1,330
    What I meant is that the last inequality is inconsistent with

     

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