User anon - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T21:37:02Z http://mathoverflow.net/feeds/user/24305 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/99095/limit-of-an-inverse-mellin-transform Limit of an inverse Mellin transform anon 2012-06-08T08:00:01Z 2012-06-08T08:00:01Z <p>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</p> <p>$$I(\beta) := \int_{a-i\infty}^{a+i\infty} \frac{\log(1-\frac s{\beta})}{s^2} x^s ds.$$</p> <p>[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?</p> http://mathoverflow.net/questions/99095/limit-of-an-inverse-mellin-transform Comment by anon anon 2012-06-09T09:21:05Z 2012-06-09T09:21:05Z It's all clear now, thanks a lot! Edwards' suggestion was misleading. http://mathoverflow.net/questions/99095/limit-of-an-inverse-mellin-transform Comment by anon anon 2012-06-08T19:16:14Z 2012-06-08T19:16:14Z 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. http://mathoverflow.net/questions/99095/limit-of-an-inverse-mellin-transform Comment by anon anon 2012-06-08T15:50:23Z 2012-06-08T15:50:23Z 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?