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 .

$$I(x)=\lim_{T\rightarrow \infty}\frac{1}{2\pi i}\int_{2-iT}^{2+iT}\frac{x^{s}}{s}\left(\sum_{k=1}^{\infty}\frac{\zeta(ks)-1}{k} \right )^{n}ds$$

$\zeta(s)$ : is the Riemann zeta function.

$n \geqslant 2 $

any insights are more than welcome .

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    $\begingroup$ I am not sure what you mean by "attacking the integral", but you can certainly write the $n$-th power in the integrand as a Dirichlet series $F(s)=\sum c_m m^{-s}$, and then Perron's formula yields that $I(x)$ equals $\sum_{m\leq x}c_m$. More precisely, for $x\in\mathbb{Z}$, the term $m=x$ gets weight $1/2$. $\endgroup$ – GH from MO Mar 17 '12 at 19:34
  • $\begingroup$ well, that IS the problem - more or less - the coefficients in the Dirichlet series $ c_{m} $ ..what form do they assume ? $\endgroup$ – mohammad-83 Mar 17 '12 at 19:41
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    $\begingroup$ Well, obviously, $c_m=\sum_{m=m_1^{k_1}\dots m_n^{k_n}}(k_1\dots k_n)^{-1}$, where the sum is over all decompositions $m=m_1^{k_1}\dots m_n^{k_n}$ with $m_j\geq 2$ and $k_j\geq 1$. $\endgroup$ – GH from MO Mar 17 '12 at 19:46
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    $\begingroup$ Do you have a simple formula/approximation for $n=1$, using the Laurent expansion you mention? If yes, please include it in your post, and we might get some ideas from it. At any rate, taking the $n$-th power of a Laurent expansion yields a Laurent expansion in the obvious way, but it is unclear to me how a Laurent expansion helps here (hence my question). $\endgroup$ – GH from MO Mar 17 '12 at 20:28
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    $\begingroup$ Perhaps it is tricky (although not impossible!) to use the $c_m$'s to approximate the $I(x)$, but my question is: can you use the Laurent series to approximate it better? You see, your last integral $\int_{2-i\infty}^{2+i\infty}\frac{x^{s}}{s}(s-1)^{n}ds$ diverges. In fact, if it were convergent, then shifting the contour to $(\sigma)$ with any $\sigma>0$ would imply that $I(x)=\beta_{-1}x+O(x^\sigma)$, which is false! $\endgroup$ – GH from MO Mar 17 '12 at 23:09

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.

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

$$I(x)=\sum _{k\leq \log_2 x}J_k(x)$$ where


$$J_k(x)=\sum_{n\leq \log_2x}\frac{\pi_k(x^{1/n})}{n}=\frac{1}{2\pi i}\int_{\sigma-i\infty}^{\sigma+i\infty}\log\prod_{p_1,p_2,...,p_k}\left(1-\frac{1}{(p_1p_2\cdots p_k)^s}\right)^{-1}\frac{x^sds}{s}$$ 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 $x$.

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.

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.


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 $$I(x)=\sum_{n\leq x}\frac{\log\prod_{i}n_i(n)}{\log n}.$$

  • $\begingroup$ thanks for the deep insights . i have just came across an interesting paper on the Dirichlet series : $$\kappa(s)=\prod_{n=2}^{\infty}\frac{1}{1-n^{-s}}=\exp\left(\sum_{k=1}^{\infty}\frac{\zeta(ks)-1}{k} \right )=\sum_{n=1}^{\infty}\frac{a_{n}}{n^{s}}$$ where $a_{n}$ is Oppenheim's unordered multiplicative partition function . barrel.ih.otaru-uc.ac.jp/bitstream/10252/2469/1/… in the paper, the author proposes an asymptotic formula for $\sum_{n\leq x}a_{n}$ $\endgroup$ – mohammad-83 Mar 20 '12 at 17:18
  • $\begingroup$ Interesting. I would've bet a tenner it cant be continued past the imaginary axis. $\endgroup$ – Kevin Smith Mar 20 '12 at 17:34
  • $\begingroup$ Any way, would you enlighten us on what you were trying to count? $\endgroup$ – Kevin Smith Mar 20 '12 at 17:36
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    $\begingroup$ in the paper the author shows that the function $$\kappa(s)=\exp\left(\sum_{k=1}^{\infty}\frac{\zeta(ks)-1}{k} \right )$$ has simple poles at $1,1/2,1/3 ... $with residues $\frac{1}{n^{s}}$ $n=1,2,3,...$ excuse my bluntness; but why can't we compute the mellin integral : $$\frac{1}{2\pi i }\int_{\sigma-i\infty}^{\sigma+i\infty} \kappa(s)\frac{x^{s}}{s} ds$$ using Cauchy residue theorem !?!? $\endgroup$ – mohammad-83 Mar 20 '12 at 17:46
  • $\begingroup$ as for analytic continuation, consider the following : $$\ln\kappa(s)=\sum_{n=1}^{\infty}\frac{\zeta(ns)-1}{n}$$ for $\Re(s)>1$ $$\frac{\zeta(ns)-1}{n}=\frac{s}{2\Gamma\left(1+\frac{ns}{2} \right )}\left(\int_{0}^{\infty}\frac{\left(x\pi \right )^{ns/2}}{x}\left(\frac{\theta(x)-1}{2}\right)-\frac{e^{x}}{x}x^{ns/2} dx\right)$$ $\theta(x)$ is the jacobi theta function $$\ln\kappa(s)=s\int_{0}^{\infty} \frac{1}{2x}\left(E_{s/2}((\pi x)^{s/2})-1\right)\omega(x)dx -s\int_{0}^{\infty}\frac{e^{x}}{2x}\left(E_{s/2}(( x)^{s/2})-1\right)dx$$ where $E(s)$ is the mittag-leffler function . $\endgroup$ – mohammad-83 Mar 20 '12 at 18:11

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