Fourier representation of a smooth Selberg weight

If $f$ is smooth and supported on $[-1,1]$ and $g(t)={\int }_{\mathbb{R}}{e}^{x}f(x)e(-tx)dx$, Fourier inversion givesExpanding the square of the resulting Möbius-weighted divisor sum turns its mean value into an Euler product controlled by zeta functions near their pole at one.

For $Y\ge X\ge 2$, a smooth Selberg divisor weight of level $D=X^{1/10}$ equals one on every prime in $(Y,Y+X]$. Its second moment can be expressed through an Euler product $H$ whose zeta-factor bound contributes $O(1/\log D)$ after integration against rapidly decreasing Fourier transforms. Consequently