The average order of $\frac{\sigma_1(n)}{\sigma_0(n)}$

Question

I want to calculate the average order of $\frac{\sigma_1(n)}{\sigma_0(n)}.$ I know that for every $e\gt0$,$$f(x):=\sum_{1\le n\le x}\frac{\sigma_1(n)}{\sigma_0(n)}=o(x^{2-e})$$ I wonder if it's true that $$f(x)\sim{\frac{x^2}2\frac{\log\log(x)}{\log(x)}}.$$

Answer 1

No, $$f(x)\sim c\frac{x^2}{\sqrt{\log x}}$$ for some constant $c$. See here.

Answer 2

I don't know what you $\sigma_k$ means. If $\sigma_k(n)$ is defined as $\sum_{d|n}d^k$. Perron's formula givs the asymptotic formula. Particularly the L-function, $L(s)$ say, associated to $f(x)$ is $$L(s)=\sum_{n\geq 1}\frac{\sigma_1(n)}{\sigma_0(n)n^s}=\sum_{d\geq 1}\frac{1}{d^{s-1}} \sum_{m\geq 1}\frac{1}{\tau(dm)m^s}=\prod_{p}\Big\{\sum_{\alpha\geq 0,\beta\geq 0}\frac{1}{(\alpha+\beta+1)p^{\alpha(s-1)+\beta s}}\Big\}.$$ The sum in the bracket is $1+\frac{1}{2p^s}+\frac{1}{2p^{s-1}}+\frac{1}{3p^{2s-1}}+\frac{1}{4p^{3s-1}}+\cdots.$ We can see $L(s)$ is convergent absolutely for $\Re s>2$. (we can see when $s=2$, $\alpha(s-1)+\beta s=1$ iff $\alpha=1,\beta=0$.) Thus the dominant for the asymptotic formula is given by the residue at $s=2$. Write $L(s)=\zeta^{1/2}(s-1)U(s),$ where $U(s)$ is convergent absolutely for $\Re s>2-\epsilon$, any $\epsilon>0$. The residue at $s=2$ is a little more complicated, by some computation involving in complex analysis.