2
$\begingroup$

In Edwards' very nice book ``Riemann's zeta function'' the following integral comes up in section 1.14. Suppose $\beta = \sigma + i\tau$ with $\sigma > 0$. Suppose $x > 1$. Fix some real number $a > \sigma$ and let

$$I(\beta) := \int_{a-i\infty}^{a+i\infty} \frac{\log(1-\frac s{\beta})}{s^2} x^s ds.$$

[Here, the logarithm is taken as $\log(s-\beta)-\log(-\beta)$ for $\tau \ne 0$ using the branch of logarithm defined away from the negative real axis, which is real on the positive real axis.] It's easy to see that this integral is absolutely convergent (note that $|x^s| = x^a$). Now Edwards claims that $\lim_{\tau \to \infty} I(\beta) = 0$ because ``it is not difficult to show, using the Lebesgue bounded convergence theorem ... that the limit of this integral is the integral of the limit, namely zero''. I didn't find a good way to bound $\log(1-\frac s{\beta})$ for fixed $s$. Is there a solution that avoids long and messy calculations?

Improve this question
$\endgroup$
5
  • 3
    $\begingroup$ For $s$ fixed, and $\beta$ big, you can use the Taylor expansion for $\log{1+z}$ around $z=0$. $\endgroup$
    – Matt Young
    Commented Jun 8, 2012 at 14:41
  • $\begingroup$ Sure, but for dominated convergence I would have to bound the absolute value of the integrand by an (integrable) function that only depends on $s$ on a region $(a-i\infty,a+i\infty) \times (\sigma+i\tau_0,\sigma+i\infty)$, but the Taylor expansion isn't valid on all of this region. But maybe I misunderstand what Edwards or you are suggesting? $\endgroup$
    – anon
    Commented Jun 8, 2012 at 15:50
  • 3
    $\begingroup$ You just said that you didn't find a good way to bound $\log(1-s/\beta)$ for fixed $s$. Now if you want to vary $s$ then it's another story. What I would do is get an asymptotic bound on $I(\beta)$ and let $\beta \rightarrow \infty$. In the region $|s| < |\beta|/2$ use the Taylor expansion, and for $|s| > |\beta|/2$ bound the log with absolute values and use the fact that the integral of $\log{y}/y^2$ from $Y$ to $\infty$ is $O(\log{Y}/Y)$. So maybe you can get a bound like $|I(\beta)| \leq \log(1 + |\beta|)/|\beta|$. $\endgroup$
    – Matt Young
    Commented Jun 8, 2012 at 16:43
  • $\begingroup$ Thanks! That sounds good, I'll think about it tomorrow. And my question was about the limit of $I(\beta)$, not about the log. I didn't phrase it well, sorry. $\endgroup$
    – anon
    Commented Jun 8, 2012 at 19:16
  • $\begingroup$ It's all clear now, thanks a lot! Edwards' suggestion was misleading. $\endgroup$
    – anon
    Commented Jun 9, 2012 at 9:21

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .