Fourier representation of a smooth Selberg weight
= Fourier representation of a smooth Selberg weight
If $f$ is smooth and supported on $[-1,1]$ and $g(t)=\int_{\mathbb R}e^xf(x)e(-tx)\,dx$, <Fourier inversion theorem>[Fourier inversion] gives
$$
f\left(\frac{\log d}{\log D}\right)
=\int_{\mathbb R}g(t)d^{-(1-2\pi it)/\log D}\,dt.
$$
Expanding 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.