I am looking for references and results on integrals with product of two infinite sums:
$$S_1=\int_0^{\infty} \sum_{n=0}^{\infty} f(nx) \sum_{k=0}^{\infty} \overline{ g(kx)} dx$$
Above integral is well defined with complex functions $f$ and $g$ in $C^1$, of rapid decrease at infinity and verifying:
$$\int_0^{\infty} f(x) dx= \int_0^{\infty} g(x) dx=0$$ and $$f(0)=g(0)=0$$
In case $f=g$ is there any way to express following sum without any sum (may be imposing some condition on $f$?) :
$$S_2=\int_0^{\infty} \sum_{n=0}^{\infty} f(nx) \sum_{k=0}^{\infty} \overline{ f(kx)} dx$$
For example it is not hard to prove (with $f$ as defined above)(using Poisson Summation formula - note that we cannot intervene integral and sum!):
$$\int_0^{\infty} \sum_{n=0}^{\infty} f(nx) dx = \int_0^{\infty} \frac{1}{x}\int_x^{\infty} f(y) dy dx$$
But with double product it is not so simple to get rid of the sum! and may be not possible in general case ?
For info and as demanded in comment, here is the demo of above formula
we can split the integral in two:
$$\int_0^{\infty}\sum_{n=1}^{\infty} f(nx) dx= \int_0^{\epsilon}\sum_{n=1}^{\infty} f(nx) dx + \int_{\epsilon}^{\infty}\sum_{n=1}^{\infty} f(nx) dx$$
and chose $\epsilon$ small to have first integral as small as we want, on the second one we can interchange sum and integral and if we note $F(x)$ the primitive of $f(x)$ such that $F(0)=0$ we have (as $\lim_{x \to 0} F(x)=0$ due to the condition imposed on $f(x)$: $\int_0^{\infty} f(x) dx=0$):
$$\int_{\epsilon}^{\infty}\sum_{n=1}^{\infty} f(nx) dx =\sum_{n=1}^{\infty} \int_{\epsilon}^{\infty} f(nx) dx= -\sum_{n=1}^{\infty} \frac{1}{n} F(n \epsilon)$$
Now we can apply the Poisson summation formula ($\lim_{x \to 0}\frac{F(x)}{x}=0$) to obtain (we note $\mathcal{F}$ the Fourier tansform):
$$ \sum_{n=1}^{\infty} \frac{1}{n \epsilon} F(n \epsilon) = \frac{1}{\epsilon} \sum_{n=1}^{\infty} \mathcal{F}(\frac{F(|x|)}{|x|})(\frac{n}{\epsilon})+\frac{1}{2 \epsilon} \mathcal{F}(\frac{F(|x|)}{|x|})(0)$$
And we see that, as the terms $\mathcal{F}(\frac{F(|x|)}{|x|})(\frac{n}{\epsilon})$ can be as small as we want for $\epsilon$ small, only one term remains and taking the limit with $\epsilon \to 0$:
$$\int_{0}^{\infty}\sum_{n=1}^{\infty} f(nx) dx = -\frac{1}{2} \mathcal{F}(\frac{F(|x|)}{|x|})(0) $$
So finally, we find the expected result:
$$\int\limits_{0}^{\infty} \sum\limits_{n =1}^{\infty} f(nx) = \int_{0}^{\infty} \frac{1}{x} \int_{x}^{\infty} f(y) dy dx $$
