contour integration $\zeta(s)\zeta(2s)$ and $x^s/s^(k+1)ds$

Question

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 $\frac{1}{2\pi T^kT}\frac{y^c}{log(y)}$?

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.