# Quantitative Perron formula with weights

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.

• It’s not in Chapter 5 of Montgomery-Vaughan? Dec 3 '19 at 14:43
• only the qualitative result :( Dec 3 '19 at 16:06
• 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). Dec 3 '19 at 16:33

If $$|a_n|\ll 1$$ and $$c>1$$, then
$$\displaystyle\sum_{n\leq x}(x-n)a_n = \frac{1}{2\pi i}\int_{c-iT}^{c+iT}\mathcal{F}(s)\frac{x^{s+1}}{s(s+1)}ds+O\Big(\frac{x^{c+1}(\log x)^2}{T^2}\Big)$$.
• $|F(s)|\le \sum_n |a_n| n^{-c}$ on $\Re(s)=c$ thus it is completely obvious Dec 21 '19 at 22:31