Timeline for Limit of an inverse Mellin transform
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 9, 2012 at 9:21 | comment | added | anon | It's all clear now, thanks a lot! Edwards' suggestion was misleading. | |
| Jun 8, 2012 at 19:16 | comment | added | anon | 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. | |
| Jun 8, 2012 at 16:43 | comment | added | Matt Young | 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 $|I(\beta)| \leq \log(1 + |\beta|)/|\beta|$. | |
| Jun 8, 2012 at 15:50 | comment | added | anon | 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? | |
| Jun 8, 2012 at 14:41 | comment | added | Matt Young | For $s$ fixed, and $\beta$ big, you can use the Taylor expansion for $\log{1+z}$ around $z=0$. | |
| Jun 8, 2012 at 8:00 | history | asked | anon | CC BY-SA 3.0 |