I'm asking something that may be trivial for those who are deeply into Analytic Number Theory, but unfortunately I'm still not into that set. The core $k(n)$ of an integer $n$ is the product of all its prime factors, $$ k(n)=\prod_{p\mid n}p\quad (p\text{ prime}); $$ in other words, it is the greatest squarefree factor of $n\,$. Let $N(x,y)$ be the number of those integers $n\leq x$ with $k(n)\leq y\,$. I'm trying (with scarce success) to show the steps which bring to the double Perron's formula for $N(x,y)\,$, i.e. $$ N(x,y)=-\frac{1}{4\pi^2}\int_{\alpha-i\infty}^{\alpha+i\infty}\int_{\beta-i\infty}^{\beta+i\infty}f(s,t)\frac{x^sy^t}{st}{d}s\,{d}t,\quad\alpha,\beta>0,\quad\alpha+\beta>1, $$ where $$ f(s,t)=\sum_{n\geq 1}\frac{1}{n^s}\frac{1}{k(n)^t}=\prod_p\left(1+\frac{1}{p^t\left(p^s-1\right)}\right). $$ Thanks for any help, suggestions or clear references.
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1$\begingroup$ Perron's formula in $s$ gives you the terms with $n\le x$, and Perron's formula in $t$ gives you the terms with $k(n)\le y$. Hence the RHS equals $N(x,y)$. Also you might mention that you are trying to read the nice recent work of Robert and Tenenbaum: dossier.univ-st-etienne.fr/rool6510/www/index.html . $\endgroup$– LuciaJan 21, 2014 at 4:37
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$\begingroup$ I apologize if to not have mentioned Robert and Tenenbaum's recent great work may be considered as unfair. I really didn't mean it. I am actually looking for someone who could be so kind to write down the calculation to get the second Perron's formula in $t$, once having the (trivial) first one. What you said Lucia, is exactly the same that is stated in the recent Robert and Tenenbaum's work and in the less recent Tenenbaum's work at MR0990518. I'm sorry but up till now I couldn't find any more detailed references about. Many very thanks indeed in advance to all who could help me. Francesco $\endgroup$– user35352Jan 22, 2014 at 17:13
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