Contour shift of the holomorphic part of the smoothed Perron integral to $\sigma_1$: $O\bigl(M(X^{\sigma_1}/\varepsilon+X^{\sigma_0}/(\varepsilon T))\bigr)$

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 $\nu$) with the following property. Let $H:\mathbb C\to\mathbb C$, let $P$ be a finite set of points, let $U$ be an open set containing the closed rectangle $R=[\sigma_1,\sigma_0]\times[-T,T]$, where $3/4\le\sigma_1<1<\sigma_0\le2$ and $T>3$, and let $M\ge0$. Assume $H$ is holomorphic on $U\setminus P$, $|H(s)|\le M$ on $U\setminus P$, and every $p\in P$ has $\operatorname{Re}p<\sigma_0$. Then for $X>3$ and $0<\varepsilon<1$,

Since $H$ is bounded near each point of $P$, these are removable singularities and $H$ extends holomorphically to $U$; Cauchy's theorem on the rectangle then moves the integral to the left side $\operatorname{Re}s=\sigma_1$ and the two horizontal sides, and the bound $|\mathcal M\widetilde{1_\varepsilon}(s)|\ll1/(\varepsilon|s|^2)$ (valid for $1/4\le\operatorname{Re}s\le2$) gives $\ll MX^{\sigma_1}/\varepsilon$ for the left side and $\ll MX^{\sigma_0}/(\varepsilon T^2)$ for the horizontal sides. In the contour method, $H=G-\sum_{p\in P}r(p)/(s-p)$ is the generating function with its polar parts removed, $M\asymp\log^2(qT)$ is the bound supplied by the zero-free region, and $X^{\sigma_1}$ with $\sigma_1=1-c/\log(qT)$ produces the saving $\exp(-c\sqrt{\log X})$.

Formalization Note. Icc σ₁ σ₀ ×ℂ Icc (-T) T is the closed rectangle; the integral is written as $\frac1{2\pi i}\cdot i\int_{-T}^{T}(\cdots)(\sigma_0+it)\,dt$. The set $P$ may meet the rectangle, including its boundary; the value of $H$ at points of $P$ is irrelevant.