# On the oscillation of the summatory totient about its average

Let $$R(x)=\sum_{n\leq x}\phi(n)-\frac{3x^2}{\pi^2}.$$ Montgomery has shown that $R(x)=\Omega_{\pm}(x\sqrt{\log\log x})$, which is the best known lower bound. It seems interesting therefore that $$\int_0^{\infty}\frac{R(x)dx}{x^2}=0,$$ because it tells us that the oscillations (which continue indefinitely) are particularly regular.

I cannot find any references for this integral, so I am wondering if it is known. I would particularly like to find other work of this nature as I cannot prove anything about the rate of convergence of the improper integral (other than $o(1)$ as $X\rightarrow\infty$ where $X$ is the upper limit of integration).

• If I may ask, how do you prove this? – user9072 Apr 8 '13 at 20:10
• It is quite lengthy, but the essence is that the integral over a finite interval may be written in terms of a uniformly convergent (for $X>1$) sum over the zeros of $\zeta(s)$. The necessary estimates to justify the limit of the contour are available. The uniform convergence and zero free region enables you to arrive at a contradiction supposing the limit as $X\rightarrow\infty$ is not $0$. More can be probably be said- it appears that the Mellin transform converges on the line $\sigma=1$. – Kevin Smith Apr 8 '13 at 20:34
• The Mellin transform of $R(x)$, that is. – Kevin Smith Apr 8 '13 at 20:39

Let $$R(x) = \sum_{n \leq x}{\varphi(n)} - \frac{3x^2}{\pi^2}, \qquad H(x) = \sum_{n \leq x}{\frac{\varphi(n)}{n}} - \frac{6x}{\pi^2}.$$
Then by partial summation, $$\int^{x}_{0}{\frac{R(t)}{t^2} \: dt} = H(x) - \frac{R(x)}{x}.$$ A classical result of Chowla states that $$H(x) - \frac{R(x)}{x} = O\left((\log x)^{-4}\right).$$ See Lemma 13 of S. Chowla, "Contributions to the analytic theory of numbers", Mathematische Zeitschrift 35:1 (1932), 279-299. (If you have access to Springer Link then it is available here.)
• Do you mean $R(t)$ rather than $E(t)$ in the integral? – Barry Cipra Apr 8 '13 at 21:03