# residues over simple poles

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

1. ### temurman of no wordsRegistered 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.

3. ### camilusthe villain with x-ray glassesRegistered Senior Member

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895
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

5. ### temurman of no wordsRegistered Senior Member

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1,330
How did you get the last equality?

7. ### TachBannedBanned

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5,265
From here
Depending on normalization, there may or may not be an extra 1/2 in front of the expression.

8. ### TachBannedBanned

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5,265
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. ### temurman of no wordsRegistered Senior Member

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