# residues over simple poles

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

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

Messages:
895
my book on prime numbers has a line where it skims over a residue computation, and Im in dire need of clarification. It's rather simple, and I may very well be the one mistaken, but Im getting a extra factor of 1/2 in the residue whereas in the book it does not appear and apparently isn't a typo either.

We have $\psi (z) = \sum_{n \in \mathbb{N}} e^{-n^2\pi z} = {1 \over 2\pi i}\int_{c-i\infty}^{c+i\infty} {\xi (s) \over s(s-1)}z^{{-1 \over 2}s}ds$ for c>1 and R(z)>0.

next, we move the line of integration left to R(s)=1/2 passing the simple pole at s=1 with residue 1/sqrt(z):

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

and for some reason Im getting that the last term should be ${1 \over 2\sqrt{z}}$The extra 1/2 is coming from the fact that $\xi (1) = 1/2$ when computing the residue of Res(f,1) when $f(s) = {\xi (s) \over s(s-1)}z^{{-1 \over 2}s}$

Last edited: Jun 1, 2011

to hide all adverts.
3. ### temurman of no wordsRegistered Senior Member

Messages:
1,330
Did you check $\xi(1)$ is really 1/2?

to hide all adverts.
5. ### TachBannedBanned

Messages:
5,265
Yes, it is 1/2. See here
Use the fact that $\Gamma(1/2)=\pi^{1/2}$

and

$(s-1)\zeta(s)->1$ when $s->1$

to hide all adverts.
7. ### temurman of no wordsRegistered Senior Member

Messages:
1,330
Yes, but $\xi$ tends to have different normalizations in different books.

8. ### TachBannedBanned

Messages:
5,265
So, this may be the problem, on the wiki page the normalization is 1/2.

camilus,
what does your book use? 1?

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

Messages:
895
not sure what you're asking. The book just states that passing the simple pole leaves a residue of 1/sqrt(z), just as posted above.

Basically Im computing the residue for $f(s) = {\xi (s) \over s(s-1)}z^{{-1 \over 2}s}$ at s=1 as such:

$Res(f,1) = \lim_{s \rightarrow 1} (s-1)*{\xi (s) \over s(s-1)}z^{{-1 \over 2}s} = {\xi (1) \over 1}z^{{-1 \over 2}}={1 \over 2\sqrt{z}$

the following link computes the same residue without the z^-s/2 factor. since there is no problems for letting s->1 in that factor so just multiply the link by z^-1/2...

http://www.wolframalpha.com/input/?i=residue of (xi function)/(z(z-1)) at z=1

10. ### TachBannedBanned

Messages:
5,265
Yes, we know this, this is not the issue.

Did you check to see if in your book $\xi(1)$ is really defined to be 1/2?

Last edited: Jun 1, 2011
11. ### camilusthe villain with x-ray glassesRegistered Senior Member

Messages:
895
of course it is, why wouldnt it be.

Xi is universally defined as $\xi (s) = {1 \over 2}s(s-1)\pi^{-s \over 2}\Gamma ({s \over 2}) \zeta (s)$

and $\pi^{-s \over 2}\Gamma ({s \over 2}) \zeta (s) = \int_0^\infty \psi (x)x^{{s \over 2} -1}dx = {1 \over s(s-1)} + \int_1^\infty \psi(x)(x^{s/2} + x^{1-s \over 2}){dx \over x}$ by the functional equation of the Jacobi theta function,

thus making

$\xi (s) = {1 \over 2}s(s-1)\pi^{-s \over 2}\Gamma ({s \over 2}) \zeta (s) = {1 \over 2}s(s-1) \left [{1 \over s(s-1)} + \int_1^\infty \psi(x)(x^{s/2} + x^{1-s \over 2}){dx \over x} \right ] = {1 \over 2} + {1 \over 2}s(s-1)\int_1^\infty \psi(x)(x^{s/2} + x^{1-s \over 2}){dx \over x}$

Now you can clearly see Xi(0) = Xi(1) = 1/2

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

Messages:
895
which is also implied by the functional equation of xi, Xi(z) = Xi(1-z) for z=0.

13. ### temurman of no wordsRegistered Senior Member

Messages:
1,330
Well, just to illustrate that there are some variations in the normalization of xi, Ivic's book defines it without the 1/2. You can see it makes sense because why should you include 1/2? Riemann's original definition had a 1/2, because he was using Gauss' "gamma" function $\Pi(\frac{s}2)$, which is related to the usual (Legendre's) gamma function by $\Pi(\frac{s}{2})=\Gamma(\frac{s}{2}+1)=\frac{1}{2}{\Gamma}(\frac{s}{2})$.

Which book are you using?

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

Messages:
895
the original problem is out of William and Fern Ellison's Prime Numbers. I seen Ivic at the library today, maybe I should've grabbed it instead of Titchmarsh. Im really looking for a copy of Edwards, which I will borrow from my friend tomorrow.

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

Messages:
895
I think you forgot an s in your last equation. $\Gamma(\frac{s}{2}+1)=\frac{s}{2}{\Gamma}(\frac{s}{2}) = \frac{1}{2}s\Gamma({s \over 2})$.

if you define it using Gauss's gamma the 1/2 is still implicitly there by the above equation.

if you want to get rid of the 1/2 completely, you'd have to do a change of variable from s/2 to s, which would turn s into 2s, thus having zeta(2s) instead of zeta(s) in the equations. Thats the whole reason behind the s/2.

16. ### TachBannedBanned

Messages:
5,265
look at definition (1.36).

17. ### temurman of no wordsRegistered Senior Member

Messages:
1,330
Thanks, I forgot an s there.

In case it was not clear, the sentence with Gauss' gamma I wrote was trying to explain why we have 1/2 in some definitions of xi. It is because Riemann did not use Gamma function as we define today; he used Gauss' factorial function all the way, which gives a factor 1/2 when converted to the language of Gamma functions. Edwards also uses Gauss' factorial function consistently. Actually it has some convenience over gamma because $\Pi(n)=n!$ instead of $\Gamma(n+1)=n!$. If Riemann were to use the Gamma function as we define today, his xi would have been without the factor 1/2, just as Ivic defines in his book.

Last edited: Jun 2, 2011
18. ### camilusthe villain with x-ray glassesRegistered Senior Member

Messages:
895
interesting indeed, although I believe it is irrelevant for my question. Doing some more calculations on wolframalpha, Im 100% sure Im correct, just see the following links. It didnt let me put z^(-s/2), but since integration is w.r.t s, then any number can take the place of z. I used 23 in this case, so just replace 23 by z and you get exactly the computation Im inquiring about.

http://www.wolframalpha.com/input/?...))((23*pi)^(-z/2))(zeta function(z))/2 at z=1

http://www.wolframalpha.com/input/?i=residue of 23^(-z/2)*(xi function)/(z(z-1)) at z=1

both links are saying the same thing, that the residue should contain that 1/2 factor. one uses xi, and the other uses zeta and gamma.

My book defines xi with the one-half several times, therefore Im assuming maybe it is an error. The reason that we are integrating xi(s)/(s(s-1)) and not 2xi(s)/(s(s-1)) is because the 2 canceled with the 1/2 from the change of s->s/2. If we were integrating 2xi/.., then the residue would definitely be 1/sqrt(z), as in this link:

http://www.wolframalpha.com/input/?i=residue of (2*xi function)/(z(z-1))23^(-z/2) at z=1

19. ### TachBannedBanned

Messages:
5,265
If that is the case, your original answer is correct, it must contain 1/2.

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

Messages:
895
thanks a lot tach and temur, i was afraid i wouldnt get a response. Have any of you studied the theory of RZF rigorously? if so can you offer an ordered list of topics I should cover? and/or what book have you found to be the clearest?

in case you wanted to know, this question is related to the proof of Hardy's theorem. Its a small error I dont think it compromises the rest of the proof.

21. ### TachBannedBanned

Messages:
5,265
I used a book by Courant, it is good but very expensive. This book costs a fraction and it is excellent.

22. ### temurman of no wordsRegistered Senior Member

Messages:
1,330
If you are studying Hardy's theorem it is already beyond my "expertise". Last year I did some reading from Edwards, Ivic, and Patterson to figure out a quick way to prove the prime number theorem with error bounds. My impression was that Edwards and Ivic have a more direct approach, while Patterson has in mind general techniques that apply to other kinds of zeta functions too.

Last edited: Jun 3, 2011
23. ### camilusthe villain with x-ray glassesRegistered Senior Member

Messages:
895
whats the name of the courant book, i googled courant and courant zeta and courant riemann and couldnt find anything. The last link is mostly about quantum field theory, which i think I'll pass on for now.

Titchmarsh proves some very important theorems but the majority of the book is still beyond me. Edwards is the way to go, but I cant believe the FIU library doesnt carry it. Hopefully a friend there will lend me his copy today. They do have Ivic, which I'll check out when I go to school today, but hopefully it doesnt throw me off too bad with its different nomalizations.

tach how far did you get in your studies of RZF? Im really in need of someone with whom to discuss this with. The only input my professor had was that he 'thought' I was right.