两个结论的证明记录
esquigybcu
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个人记录
powerful number 计数
记 \le x 的 powerful number 有 P(x) 个，那么
P(x)\sim\frac{\zeta(3/2)}{\zeta(3)}x^{1/2}
证明：注意到每个 powerful number 都可以被唯一地写成 a^3b^2 的形式，其中 a 无平方因子。于是
\begin{aligned}
P(x)&=\sum_{\scriptstyle a^3b^2\le x\atop\scriptstyle a\text{ is squarefree}}1\\
&=\sum_{\scriptstyle a\le x^{1/3}\atop\scriptstyle a\text{ is squarefree}}\left\lfloor\sqrt{x/a^3}\right\rfloor\\
&=\sum_{\scriptstyle a\le x^{1/3}\atop\scriptstyle a\text{ is squarefree}}\left(\sqrt{x/a^3}+O(1)\right)\\
&=x^{1/2}{\color{red}\sum_{\scriptstyle a\le x^{1/3}\atop\scriptstyle a\text{ is squarefree}}\sqrt{1/a^3}}+O(x^{1/3})
\end{aligned}
记红色的和为 S，注意到
\begin{aligned}
\sum_{a\text{ is squarefree}}a^{-3/2}&=\prod_p\left(1+p^{-3/2}\right)\\
&=\prod_p\left(\frac{1-p^{-3}}{1-p^{-3/2}}\right)\\
&=\frac{\zeta(3/2)}{\zeta(3)}
\end{aligned}
而且
\begin{aligned}
\sum_{\scriptstyle a>x^{1/3}\atop\scriptstyle a\text{ is squarefree}}a^{-3/2}&\le\sum_{a>x^{1/3}}a^{-3/2}\\
&=O(x^{-1/6})
\end{aligned}
故 S=\frac{\zeta(3/2)}{\zeta(3)}+O(x^{-1/6})，故 P(x)=\frac{\zeta(3/2)}{\zeta(3)}x+O(x^{1/3})。
1/\varphi 的求和
\sum_{n\le x}\frac{1}{\varphi(n)}=\frac{\zeta(2)\zeta(3)}{\zeta(6)}\ln x+o(\ln x)
证明：注意到 \mathrm{id}/\varphi=\mathbf{1}*(\mu^2/\varphi)，故
\begin{aligned}
\sum_{n\le x}\frac{1}{\varphi(n)}&=\sum_{n\le x}\frac{1}{n}\sum_{\scriptstyle d\mid n\atop\scriptstyle d\text{ is squarefree}}\frac{1}{\varphi(d)}\\
&=\sum_{\scriptstyle d\le x\atop\scriptstyle d\text{ is squarefree}}\frac{1}{\varphi(d)}\sum_{m\le x/d}\frac{1}{dm}\\
&=\sum_{\scriptstyle d\le x\atop\scriptstyle d\text{ is squarefree}}\frac{1}{d\varphi(d)}(\ln (x/d)+\gamma+O(d/x))\\
&=\sum_{\scriptstyle d\le x\atop\scriptstyle d\text{ is squarefree}}\frac{\ln x-\ln d+\gamma}{d\varphi(d)}+O\left(\sum_{\scriptstyle d\le x\atop\scriptstyle d\text{ is squarefree}}\frac{1}{x\varphi(d)}\right)
\end{aligned}
后面懒得敲了，自己推吧（（