Help with a mellin-type integral - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T11:54:57Z http://mathoverflow.net/feeds/question/91485 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/91485/help-with-a-mellin-type-integral Help with a mellin-type integral mohammad-83 2012-03-17T19:13:21Z 2012-03-20T16:41:49Z <p>greetings . i've been trying to do this integral for many days now, with no clue on how to attack it . the integral is a mellin inverse of some kind, and appears in analytic number theory . </p> <p>$$I(x)=\lim_{T\rightarrow \infty}\frac{1}{2\pi i}\int_{2-iT}^{2+iT}\frac{x^{s}}{s}\left(\sum_{k=1}^{\infty}\frac{\zeta(ks)-1}{k} \right )^{n}ds$$</p> <p>$\zeta(s)$ : is the Riemann zeta function. </p> <p>$n \geqslant 2$</p> <p>any insights are more than welcome . </p> http://mathoverflow.net/questions/91485/help-with-a-mellin-type-integral/91724#91724 Answer by Kevin Smith for Help with a mellin-type integral Kevin Smith 2012-03-20T14:59:10Z 2012-03-20T14:59:10Z <p>Firstly I am wondering what, exactly, are you trying to count? The Dirichlet coefficients have been given explicitly by GH, so am I to assume this is where you started and now you are looking for another way to estimate the partial sums? Secondly, you cannot integrate the Laurent series termwise so that isn't going to help. My answer to your question merely explains why explicit formulae'' are not going to help you in the case $n=1$, which naturally impacts on the higher order cases, and offers an approach to simplifying your problem another way.</p> <p>Formally speaking, your function $I(x)$ has the same relationship to $\lfloor x \rfloor-1$ as Riemann's prime counting function $J(x)$ has to the ordinary prime counting function $\pi (x)$ - they are related by a particular type of Mobius transform, which amounts to taking logarithms of the infinite products comprised of the terms from the original Dirichlet series. You have $$I(x)=\sum_{n\leq \log_2x}\frac{\lfloor x^{1/n} \rfloor-1}{n}=\frac{1}{2\pi i}\int_{\sigma-i\infty}^{\sigma+i\infty}\log\prod_{n\geq 2}\left(1-\frac{1}{n^s}\right)^{-1}\frac{x^sds}{s}$$ and Riemann had $$J(x)=\sum_{n\leq \log_2x}\frac{\pi(x^{1/n})}{n}=\frac{1}{2\pi i}\int_{\sigma-i\infty}^{\sigma+i\infty}\log\prod_{p}\left(1-\frac{1}{p^s}\right)^{-1}\frac{x^sds}{s}$$ $$=\frac{1}{2\pi i}\int_{\sigma-i\infty}^{\sigma+i\infty}\log\zeta(s)\frac{x^sds}{s},$$ where Riemann's explicit formula'' is derived by evaluating the integral in terms of the residues at the poles of the integrand, which is where the zeros and pole of $\zeta(s)$ come into play. Now, if you are looking for an explicit formula of the type you mention in your comments above then, by the fundamental theorem of arithmetic, you can write your sum as </p> <p>$$I(x)=\sum _{k\leq \log_2 x}J_k(x)$$ where</p> <p>(1) </p> <p>$$J_k(x)=\sum_{n\leq \log_2x}\frac{\pi_k(x^{1/n})}{n}=\frac{1}{2\pi i}\int_{\sigma-i\infty}^{\sigma+i\infty}\log\prod_{p_1,p_2,...,p_k}\left(1-\frac{1}{(p_1p_2\cdots p_k)^s}\right)^{-1}\frac{x^sds}{s}$$ and the product runs over all products of $k$ not necessarily distinct primes. Here, $\pi_k(x)$ counts the number of such products whose value is less than $x$.</p> <p>Of course, now you have an even bigger problem: in order to use your explicit formula, you need to know something about the zeros and poles of a whole load of new functions (the integrands in (1)), and establish that Cauchy's theorem can be applied to evaluate each of their inverse Mellin transforms. I hope at this point you can see that invoking the zeros of $\zeta(s)$ and related objects, i.e. involving the primes, is apparently unhelpful here.</p> <p>I think it is a much better idea to follow GH's advice above and work directly with the arithmetic structure that your Dirichlet series encodes. However, I must do this in a second answer because the software is struggling with my latex. </p> http://mathoverflow.net/questions/91485/help-with-a-mellin-type-integral/91733#91733 Answer by Kevin Smith for Help with a mellin-type integral Kevin Smith 2012-03-20T16:21:22Z 2012-03-20T16:41:49Z <p>Continued from my earlier post: I would therefore suggest approaching it in the counting domain'' as follows: from the explicit definition of the coefficients given by GH above, you can easily see that your Dirichlet series in the case $n=1$ is $$\sum_{n=2}^{\infty}\left(\sum_{n=n_i^{k_i}(n)}\log n_i(n)\right)\frac{1}{n^s\log n}=\sum_{n=2}^{\infty}\frac{\log\prod_{i}n_i(n)}{n^s\log n}.$$ From here you get $$I(x)=\sum_{n\leq x}\frac{\log\prod_{i}n_i(n)}{\log n}.$$</p>