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I was recently working on arithmetic functions and using Perron's formula to obtain asymptotic estimates. One observation I made was that the Dirichlet series often can be written in terms of the Riemann zeta function.

More formally, let $f(n)$ be an arithmetic function and $F(s)=\sum^\infty_0 f(n)n^{-s}$ be its Dirichlet series. Is true that $F(s)=\frac{A(s)}{B(s)}$, where $A(s)$ or $B(s)$ are some factors of $\zeta(s)$, or even possibly $\zeta'(s)$ (the first derivative) as in the case of the von Mangoldt function?

It seems like this property is not essential to the rest of the analysis as I was just more concerned about where the poles but it did seem like an innate property as a result of the Euler product representation.

Also one question that I had to grapple with was the handling of an essential singularity. As I did not have to work out the details for any particular example, I was left wondering about the impact of this as opposed to having a pole of some finite order. Do we adjust the line of integrations to avoid the essential singularity or are we still able to carry on the analysis with no issues whatsoever.

Any insight provided will be helpful. Thanks!

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The function $F(s)$ will have the Euler product representation $$F(s)=\prod_{p}\left(1+f(p)p^{-s}+f(p^{2})p^{-2s}+\cdots\right).$$ In general, you will not be able to write $F(s)$ in terms of the zeta function, however often times we will pull out large zeta factors so that the left over series converges on a larger domain. This is achieved by writing $F(s)=H(s)\zeta(s)^{l}$ where $H(s)$ converges on a larger range, and we may use this the evaluate the average $\sum_{n\leq x}f(n)$. For example, suppose that $f(n)$ is fairly close to $1$ on the primes, such as the functions $\frac{\phi(n)}{n},$ $\frac{n}{\sigma(n)}$, $\frac{\sigma(n)}{n}$, etc. To quantify this closeness to $1$ , we define $h(n)=\left(\mu*f\right)(n)$, and require that $\sum_{n=1}^{\infty}\frac{|h(n)|}{n^{\alpha}}$ converges for some $\alpha<1$. In other words, if $H(s)=\sum_{n=1}^{\infty}h(n)n^{-s}$, then $F(s)=H(s)\zeta(s)$, and while $F(s)$ only converges absolutely for $\sigma>1$, the function $H(s)$ will converge absolutely for $\sigma>\alpha$. We may then show that $$\sum_{n\leq x}f(n)=H(1)x+O\left(x^{\alpha}\right)$$ by entirely elementary methods. Indeed, $$\sum_{n\leq x}f(n)=\sum_{n\leq x}\sum_{d|n}h(d)=\sum_{d\leq x}h(d)\left[\frac{x}{d}\right] $$

$$=x\sum_{d\leq x}\frac{h(d)}{d}+O\left(\sum_{d\leq x}h(d)\right)=x\sum_{d=1}^\infty\frac{h(d)}{d}+O\left(x^\alpha\right)$$ as

$$\sum_{d\leq x}h(d)+x\sum_{d>x}\frac{h(d)}{d}\leq \sum_{d\leq x}|h(d)| \frac{x^\alpha}{d^\alpha}+x\sum_{d>x}\frac{|h(d)|}{d^\alpha}\frac{1}{x^{1-\alpha}}=O\left(x^\alpha \sum_{d=1}^\infty \frac{|h(d)|}{d^\alpha}\right)=O\left(x^{\alpha}\right).$$

This approach may be similarly applied when taking out higher powers of $\zeta(s)$. The left over function will not necessarily be of the form $\zeta(s+1)$, as it is for $\phi(n)/n$, and so the constant coming from the Euler product $H(1)$ will not necessarily be expressible in terms of known constants.

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  • $\begingroup$ Thanks for your solution. Any insight to the second part of my question? $\endgroup$ – Haikal Yeo Dec 22 '13 at 13:48

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