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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.

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  • $\begingroup$ It’s not in Chapter 5 of Montgomery-Vaughan? $\endgroup$
    – Peter Humphries
    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). $\endgroup$
    – Peter Humphries
    Commented Dec 3, 2019 at 16:33

1 Answer 1

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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.

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  • $\begingroup$ $|F(s)|\le \sum_n |a_n| n^{-c}$ 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

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