Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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).

share|improve this question
    
If I may ask, how do you prove this? –  quid Apr 8 '13 at 20:10
1  
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
    
Thank you for the explanation! –  quid Apr 8 '13 at 23:28
add comment

1 Answer 1

up vote 5 down vote accepted

I am not sure if this result is explicitly mentioned in the literature, but it certainly is classical.

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.)

From a cursory glance of Chowla's proof, the negative powers of a logarithm stem from the prime number theorem applied to the summatory function of the Möbius function, so it is likely that this bound could be improved with more modern estimates for this.

For what it's worth, I answered a question closely related to this here.

share|improve this answer
    
Do you mean $R(t)$ rather than $E(t)$ in the integral? –  Barry Cipra Apr 8 '13 at 21:03
    
Yep, thanks. All fixed now. –  Peter Humphries Apr 8 '13 at 21:09
    
Marvellous. I knew the partial summation but not the estimate. I think it is sufficient for my purposes. Thank you. –  Kevin Smith Apr 8 '13 at 21:12
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.