Splitting a 2d integral into 1d integrals

Question

Can 2D integral of the form

$$ I=\int_{-\infty}^\infty dxdy~ e^{i x y} f(x)g(y) $$

be split into a finite sequence of 1D integrals? We assume that $f$ and $g$ are nice enough, square integrable, functions.

Answer 1

One can replace $e^{xy}$ by its series expansion:

$$\ \sum_{k=0}^{\infty}\frac{1}{k!}x^ky^k$$ and integrate term by term, giving

$$\ \tag{1} \sum_{k=0}^{\infty}\frac{1}{k!} \ m_k(f) m_k(g)$$

where

$$\ m_k(f)=\int_{-\infty}^{\infty}x^kf(x)dx$$

is called the $k$-th moment of $f.$

Relationship (1) is of a formal nature.

We have to be in conditions

• allowing integration term by term and

• warranting the convergence of the resulting series (1)

A sufficient condition for this to hold is that all the integral terms $\int ... \int...$ are uniformly bounded by a certain constant $M$. In this way the series will be dominated by $M\sum_{k=0}^{\infty}\frac{1}{k!}=Me$.

(1) constitutes an infinite sum. But it is possible to have it reduced if for example all moments of $f$ vanish for $k$ bigger than a certain $k_0$, or if odd moments of $f$ are zero, and the same for even moments of $g$ above a certain $k_0$. This would happen if for example, $f$ is even, and $g$ odd.