Logarithmically smoothed Perron formula (${\sum }_{n\le x}{a}_n\log (x/n)$)

If the Dirichlet series $F(s)={\sum }_n{a}_n{n}^{-s}$ is absolutely convergent at $\sigma >0$, thenThe scalar kernel is $(2\pi {)}^{-1}{\int }_{\mathbb{R}}{e}^{v(\sigma +it)}(\sigma +it{)}^{-2}dt=v_{+}$. This follows by Fourier inversion applied to $u{e}^{-\sigma u}{1}_{u\ge 0}$, whose Fourier transform is $(\sigma +it{)}^{-2}$. Absolute convergence of the series and of $\int |\sigma +it{|}^{-2}dt$ justifies interchange. The factor $ds=idt$ is essential.

Suppose $|a_n|\le (\log (2n){)}^k$ and $A(x)={\sum }_{n\le x}{a}_n\log (x/n)=O(x(\log x{)}^{\beta })$, with $0\le \beta \le k$. For $0<h\le 1$, direct subtraction givesThe last term has absolute value $O((x{h}^2+h)(\log x{)}^k)$. Choose $h=(\log x{)}^{(\beta -k)/2}$ to obtain the displayed bound, with an additional $O((\log x{)}^k)$ term harmless as $x\to \infty$ for fixed $k$. This elementary finite-difference argument converts a logarithmically smoothed Perron formula bound to cancellation in its original partial sum.