Average order of Eulers totient function squared

Question

I was wondering if one has a nice asymptotic formula for the sum $$\sum_{n\le x} \phi(n)^2$$ and if so, how does one calculate it. I know that one has $\sum_{n\le x} \phi(n) = \frac{3}{\pi^2}x^2+O(x\log x)$, but I can't seem to find similar results for other powers of $\phi$.

Answer 1

We can get the first term of the asymptotics with the Wiener-Ikehara theorem. For this purpose we need to evaluate $$L(s) = \sum_{n\ge 1} \frac{\varphi(n)^2}{n^s}.$$

Recall that $$ n = \prod_p p^v \quad\text{implies}\quad \varphi(n) = \prod_{p|n} (p^v - p^{v-1})$$ which we may either quote or derive using the fact that $\varphi(n)$ is multiplicative and the fact that there are $p^{v-1}$ values not coprime to $p^v$ in the interval $[1, p^v].$

This implies that $L(s)$ has the following Euler product: $$\prod_p \left(1 + \frac{(p-1)^2}{p^s} + \frac{(p^2-p)^2}{p^{2s}} + \frac{(p^3-p^2)^2}{p^{3s}} + \frac{(p^4-p^3)^2}{p^{4s}} + \cdots\right).$$

Expanding the squares we get $$\prod_p \left(1 + \sum_{q\ge 1} \frac{p^{2q}}{p^{qs}} - 2 \sum_{q\ge 1} \frac{p^{2q-1}}{p^{qs}} + \sum_{q\ge 1} \frac{p^{2q-2}}{p^{qs}}\right).$$ This is $$\prod_p \left(1 + \left(1 - \frac{2}{p} + \frac{1}{p^2} \right) \sum_{q\ge 1} \frac{p^{2q}}{p^{qs}} \right) = \prod_p \left(1 + \left(1 - \frac{2}{p} + \frac{1}{p^2} \right) \frac{p^{2-s}}{1-p^{2-s}} \right) \\ = \prod_p \left(1 + \left(1 - \frac{2}{p} + \frac{1}{p^2} \right) \frac{p^{2-s}}{1-1/p^{s-2}} \right) \\ = \zeta(s-2) \prod_p \left(1 - 1/p^{s-2} + p^{2-s} - 2 p^{1-s} + p^{-s} \right) \\ = \zeta(s-2) \prod_p \left(1 - 2 \frac{1}{p^{s-1}} + \frac{1}{p^s} \right).$$

There is a simple pole at $s=3$ (from the zeta function which contributes one to the residue and the product, which is readily seen to converge there) and hence by the Wiener-Ikehara Theorem we have $$\sum_{n\le x} \varphi(n)^2 \sim \frac{x^3}{3} \prod_p \left(1 - 2 \frac{1}{p^{2}} + \frac{1}{p^3} \right).$$

Convergence of the product follows because we have $$0 < \frac{2}{p^2} - \frac{1}{p^3} < 1 $$ and $$\sum_p \left( \frac{2}{p^{2}} - \frac{1}{p^3} \right)$$ converges e.g. by comparison with $\zeta(2)$ and $\zeta(3).$

The numeric value of the constant is given by $$\prod_p \left(1 - 2 \frac{1}{p^{2}} + \frac{1}{p^3} \right) \approx 0.428249.$$

This material is not original and can be found at OEIS 127473.