Basically, it seems you need a multi-dimensional justification of Parseval's formula for this application. In my

paper on the related 1D case, you will see (my Th 3.1) that I rely on a Theorem in Titschmarsh. He comments at the end of his theorem proof that having [$](f(x),g(x)) \in L^1 \equiv L^1(R)[$] and one of them bounded is sufficient, but his general conditions are weaker than that. Here [$]R = (-\infty,\infty)[$]. In my paper, I show how to get/use weaker conditions by introducing generalized Fourier transforms.

Presumably something similar holds in n-dimensions.

In your application, [$]f(x)[$] is a payoff function and [$]g(x)[$] is the probability density with [$]x \in R^n[$], using log-prices. For n-dimensional GBM, for example, [$]g(x)[$] will be an n-dimensional normal density, so bounded and in [$]L^1(R^n)[$]. Other densities may not be bounded, but (being probability densities) we know they are [$]L^1[$].

Arranging the payoff to be bounded should not be too hard: usually it is some put-call parity type trick, removing the + (max) condition from the payoff, valuing that, and then focusing on the payoff that's left. In my paper, I show that Titschmarsh's theorem can be applied as long as [$]f(x)[$] is (i) Fourier integrable in a strip and (ii) bounded. This avoids needing an [$]L^1[$] payoff.* For example, a put option payoff [$](K - e^x)^+[$] is bounded, not in [$]L^1[$], but has a generalized Fourier transform in a [$]z[$]-plane strip.

So, to summarize, my suggestion is to try to adapt that generalized Fourier transform approach.

----------------------------------------------------------------------------------------------

* But the strip has to overlap the analyticity strip for the characteristic function of [$]g(x)[$] at [$]-z[$]. This requirement is never a problem for a normal density, but for some others, it may require care.