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.



Quantitative Perron formula with weights
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.


