Contribution of a simple pole $r/(s-p)$ to the truncated smoothed Perron integral: $\mathcal M\widetilde{1_\varepsilon}(p)X^p$ up to $O(X^{1/4}/\varepsilon+X\log X/(\varepsilon T))$

Throughout, $\nu$ is a fixed smoothing kernel: a $C^1$ function on $\mathbb R$ supported in $[1/2,2]$, nonnegative on $(0,\infty)$, with $\int_0^\infty\nu(x)\,dx/x=1$; $\widetilde{1_\varepsilon}=$ Smooth1 ν ε is the smoothed indicator of $(0,1]$ obtained by Mellin convolution with the delta-spike $\nu(x^{1/\varepsilon})/\varepsilon$ (it equals $1$ on $(0,1-\varepsilon\log2]$, $0$ on $[1+2\varepsilon\log 2,\infty)$, and lies in $[0,1]$), and $\mathcal M\widetilde{1_\varepsilon}(s)=\int_0^\infty\widetilde{1_\varepsilon}(x)x^{s-1}dx$ is its Mellin transform (Mathlib's mellin).

Statement. There is a constant $C>0$ (depending only on $\nu$) such that for all $X>3$, $0<\varepsilon<1$, $T>3$ and every $p\in\mathbb C$ with $1/2\le\operatorname{Re}p\le1$, writing $\sigma_0=1+1/\log X$,

The integrand is meromorphic in $\operatorname{Re}s>0$ with a single simple pole at $s=p$, of residue $\mathcal M\widetilde{1_\varepsilon}(p)X^p$; shifting the line of integration to $\operatorname{Re}s=1/4$ (where $|\mathcal M\widetilde{1_\varepsilon}(s)|\ll1/(\varepsilon|s|^2)$ and $|X^s|=X^{1/4}$) picks up this residue, and the parts of the line $\sigma_0$ with $|t|>T$ contribute $O(X\log X/(\varepsilon T))$ because $|s-p|\ge\sigma_0-1=1/\log X$ there. In the contour method for $\sum_{n<N}a(n)$ the polar parts $r(p)/(s-p)$ of the generating function are treated by this lemma exactly, so that the remaining holomorphic part can be shifted into the zero-free region without meeting any pole.

Formalization Note. The integral is written as $\frac1{2\pi i}\cdot i\int_{-T}^{T}(\cdots)(\sigma_0+it)\,dt$ with $\sigma_0=1+(\log X)^{-1}$ as a real number cast to $\mathbb C$; $X^{s}$ is the complex power of the positive real $X$.