Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am trying to do contour integration on $\int_c \zeta(s)\zeta(2s) \frac{x^s}{s} ds$ and $\int_c \frac{x^s}{s^ks}ds$ where c is the line segment joining c-iT c+iT

I understand the basic theory behind it, how to find residues and Cauchy's theorem. I am just having trouble with bounding these integrals when using Cauchy's theorem.

What I mean to say here is let's say I am doing contour integration for x>1 on a square say for $\int_c \frac{x^s}{s}ds$ I can bound the horizontal integral by $\frac{1}{2\pi T}\frac{y^c}{log(y)}$ can I bound the horizontal integral for $\int_c \zeta(s)\zeta(2s) \frac{x^s}{s^ks} ds$ by $\frac{1}{2\pi T^kT}\frac{y^c}{log(y)}$?

share|improve this question

1 Answer 1

First multiply out $\zeta(s)\zeta(2s)$ as a Dirichlet series $\sum_{n = 1}^{\infty} \frac{a_n}{n^s}$. (I'll let you compute the values of $a_n$.) Your integral is then equal to $$\sum_{n \geq 1} a_n \int_c \bigl(\frac{x}{n}\bigr)^s \frac{ds}{s}.$$ Now if we let $T \to \infty$, then $\int_c x^s\frac{dx}{s}$ is equal to $0$, $\pi i$, or $2\pi i$ depending on whether $0 < x < 1$, $x = 1$, or $x > 1$ (assuming that the real part of the vertical line $c$ is positive).

Thus your integral is equal to $2\pi i \sum_{n \leq x} a_n,$ with the convention that you weight the final summand by $1/2$ if $x$ is an integer.

If you really want $T$ to be a finite value, rather than infinite, you will have to compute more carefully.

The first few chapters of Edward's book Riemann's zeta function are a good reference for these sorts of integrals, including the precise issues related to convergence, interchanging the summation and integration, and so on.

share|improve this answer
    
Thank you. I am not a fan of contour integrals and am not sure how to work with them and I am terrible at bounding them –  Jess Mar 17 '11 at 18:43
    
How did you compute the integral for the cases with $0 < x < 1$ or $x >1$? I tried using a rectangle with a "bump" around the origin to avoid a singularity, but got stuck. –  modnar Jun 2 at 23:58
    
@modnar: Dear modnar, Did you look at the reference I suggested (i.e. the book by Edwards)? My memory is that he explains this. If you can't find the explanation there, or in some other analytic number theory book (it is standard, so should be in any such book that discusses Mellin transforms) then you could ask this as a separate question, rather than in comments. Regards, –  Matt E Jun 3 at 3:16
    
I haven't taken a look at it, but found another reference. Thanks for the suggestion! –  modnar Jun 4 at 6:30

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.