2
$\begingroup$

For $\kappa >1$ and $t,X\geq 1$ $$\sum _{n\leq X}a_n=\frac {1}{2\pi i}\int _{\kappa \pm iT}\frac {\mathcal F(s)X^sds}{s}+\mathcal O\left (x^\kappa \sum _{=1}^\infty \frac {1}{n^\kappa (1+T|\log (X/n)|)}\right )$$ where $a_n\ll 1$ and $$\mathcal F(s)=\sum _{n=1}^\infty \frac {a_n}{n^s}.$$ This is a quantitative version of Perron's formula - the above is taken from Page 132 of Tenenbaum's Introduction to Probabilistic Number Theory.

On page 134, display (11), we have a qualitative version of Perron wit the first Cesaro weight $$\sum _{n\leq X}a_n(X-n)=\frac {1}{2\pi i}\int _{\kappa \pm i\infty }\frac {\mathcal F(s)X^{s+1}ds}{s(s+1)}.$$

Is there anywhere a quantative version of this weighted version? I can prove it, but surely it must already be somewhere.

$\endgroup$
3
  • $\begingroup$ It’s not in Chapter 5 of Montgomery-Vaughan? $\endgroup$ Commented Dec 3, 2019 at 14:43
  • $\begingroup$ only the qualitative result :( $\endgroup$
    – tomos
    Commented Dec 3, 2019 at 16:06
  • $\begingroup$ Hmmm. I don't know where else you could hope to find it. I would just prove it from scratch (and in an analytic number theory paper, I would be very terse, since it's essentially classical). $\endgroup$ Commented Dec 3, 2019 at 16:33

1 Answer 1

3
$\begingroup$

If $|a_n|\ll 1$ and $c>1$, then

$\displaystyle\sum_{n\leq x}(x-n)a_n = \frac{1}{2\pi i}\int_{c-iT}^{c+iT}\mathcal{F}(s)\frac{x^{s+1}}{s(s+1)}ds+O\Big(\frac{x^{c+1}(\log x)^2}{T^2}\Big)$.

A detailed proof can be found in Murty's "Problems in Analytic Number Theory", solution to Problem 4.1.8.

$\endgroup$
7
  • $\begingroup$ $|F(s)|\le \sum_n |a_n| n^{-c}$ on $\Re(s)=c$ thus it is completely obvious $\endgroup$
    – reuns
    Commented Dec 21, 2019 at 22:31
  • $\begingroup$ @reuns I agree, but a reference was requested. $\endgroup$
    – 2734364041
    Commented Dec 22, 2019 at 6:31
  • $\begingroup$ only x^c/T? i was actually hoping over T^2... i'll have a look in your reference $\endgroup$
    – tomos
    Commented Jan 6, 2020 at 13:08
  • $\begingroup$ @tomos you are correct; I was being inefficient. $\endgroup$
    – 2734364041
    Commented Jan 6, 2020 at 13:09
  • $\begingroup$ no worries :) efficiency once appeared on the street where i live, then left. the street stayed the same though. $\endgroup$
    – tomos
    Commented Jan 6, 2020 at 13:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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