Consider $f(x)$ a function from $\mathbb{R^+}$ to $\mathbb{C}$ such that $f(x) \sim_0 x$ and $\int_{0}^{\infty} f(x) dx=\int_{0}^{\infty} x^2 f(x) dx=0$
Can we demonstrate that following integral is real?
$$I=\int_{0}^{1} \int_{0}^{\infty} \sum_{n=1}^{\infty} n^2 x^2f(nx) e^{2 i \pi n y} \sum_{k=1}^{\infty} \overline{f(kx) e^{2 i \pi k y}} dx dy$$
If we could interchange integrals we would then integrate by parts on $y$ and obtain the result:
$$ I=\overline{I}=\int_{0}^{\infty} \int_{0}^{1} \sum_{n=1}^{\infty} f(nx) e^{2 i \pi n y} \sum_{k=1}^{\infty} \overline{k^2 x^2 f(kx) e^{2 i \pi k y}} dx dy$$
But this interchange seems not valid as:
$$\int_{0}^{1} \sum_{n=1}^{\infty} f(nx) e^{2 i \pi n y} \sum_{k=1}^{\infty} \overline{k^2 x^2 f(kx) e^{2 i \pi k y}} dy = \sum_{n=1}^{\infty} n^2 x^2 f(nx) \overline{ f(nx) }$$
Above function has a limit of the form $\frac{C}{x}$ near zero, so the integral on $x$ will not converge (Poisson summation formula provides this limit).
Note: $I$ is well-defined as $\sum_{n=1}^{\infty} f(nx) e^{2 i \pi n y} $ and $\sum_{n=1}^{\infty} n^2 x^2 f(nx) e^{2 i \pi n y}$ do not have any singularity for $x$ near zero (this can be shown using Poisson summation formula, even for $y=0$ there is no singularity as we have $\int_{0}^{\infty} f(x) dx=\int_{0}^{\infty} x^2 f(x) dx=0$).