The summatory totient function is 3 over pi squared times x squared plus O(x log x)

By For every positive integer $n$, ${\sum }_{d\mid n,d>0}\phi (d)=n$, one has ${\sum }_{d\mid n}\phi (d)=n={\mathrm{id}}_1(n)$, where ${\mathrm{id}}_1$ is from The power functions ${\mathrm{id}}_k$ and the divisor-power-sum functions ${\sigma }_k$. Applying Classical Möbius inversion over positive divisors gives $\phi (n)={\sum }_{d\mid n}\mu (d)\frac{n}{d}\mathrm{.}$ The same inversion applied to the identity ${\sum }_{d\mid n}\epsilon (d)=1$, with $\epsilon$ from The Dirichlet-convolution identity and the constant-one function, yields ${\sum }_{d\mid n}\mu (d)=\epsilon (n)=\{\begin{array}{ll}1,& n=1,\\ 0,& n>1.\end{array}$

Summing the divisor formula from step 1.1 over $n\le x$ and writing $n=dm$ gives ${\sum }_{n\le x}\phi (n)={\sum }_{d\le x}\mu (d){\sum }_{m\le x\mathrm{/}d}m\mathrm{.}$ Also, multiplying the second identity of step 1.1 by $1\mathrm{/}{n}^2$ and summing over $n\le x$ yields the finite identity $1={\sum }_{d\le x}\frac{\mu (d)}{d^2}{\sum }_{m\le x\mathrm{/}d}\frac{1}{m^2}\mathrm{.}$

For a positive integer $Y$, let $S(Y):={\sum }_{m\le Y}1\mathrm{/}{m}^2$. For every $m\ge Y+1$ one has $\frac{1}{m^2}\le \frac{1}{m(m-1)}=\frac{1}{m-1}-\frac{1}{m},$ so ${\sum }_{m>Y}\frac{1}{m^2}\le {\sum }_{m>Y}(\frac{1}{m-1}-\frac{1}{m})=\frac{1}{Y}\mathrm{.}$ Since The Basel sum is pi squared over six by a residue computation gives ${\sum }_{m=1}^{\mathrm{\infty }}1\mathrm{/}{m}^2={\pi }^2\mathrm{/}6$, it follows that $S(Y)={\pi }^2\mathrm{/}6+O(1\mathrm{/}Y)$. Apply this in the second formula of step 2.1 with $Y=y_d$. Because $1\mathrm{/}{y}_d\le 2d\mathrm{/}x$ for every $d\le x$, one gets ${\sum }_{d\le x}\frac{1}{d^2{y}_d}=O\text{\hspace{0.17em}⁣}\left(\frac{1}{x}{\sum }_{d\le x}\frac{1}{d}\right)\mathrm{.}$ Together with $\mathrm{\mid }\mu (d)\mathrm{\mid }\le 1$ from The number-theoretic Möbius function $\mu (n)$ from prime factorisation and The harmonic sum is log x plus gamma plus O(1/x), this yields ${\sum }_{d\le x}\frac{\mu (d)}{d^2}=\frac{6}{{\pi }^2}+O\text{\hspace{0.17em}⁣}\left(\frac{\log x}{x}\right)\mathrm{.}$

For each $d\le x$, ${\sum }_{m\le x\mathrm{/}d}m=\frac{y_d(y_d+1)}{2}=\frac{y_d^2}{2}+O(y_d),y_d=\frac{x}{d}+O(1)\mathrm{.}$ Therefore step 2.1 becomes ${\sum }_{n\le x}\phi (n)=\frac{x^2}{2}{\sum }_{d\le x}\frac{\mu (d)}{d^2}+O\text{\hspace{0.17em}⁣}\left(x{\sum }_{d\le x}\frac{1}{d}\right)\mathrm{.}$ Using step 3.1 and The harmonic sum is log x plus gamma plus O(1/x) now yields ${\sum }_{n\le x}\phi (n)=\frac{3}{{\pi }^2}{x}^2+O(x\log x)\mathrm{.}$