My major is Physics so I am very naive about the Fourier transform, and its main use for me is to simplify analysis of physical problems with translational symmetry, turning some PDEs into ODEs.
I am computing a quantity (a holographic correlation function) $f(\vec{x})$ and I am able to obtain its Fourier transform $\tilde{f}(\vec{p}) = \frac{1}{\sqrt{2\pi}}\int d^4\vec{x} f(\vec{x}) e^{-i\vec{p}\cdot\vec{x}}$ in a close form as a formal series, with each term having an explicit but complicated expression, so no hope to sum up the series. What I want is to extract physical quantities from the first few powers in $\vec{x}$ of $f(\vec{x})$, for example it can look like \begin{align} \frac{1}{(x_1^2+x_2^2)^{2\Delta}}(3x_1^2-x_2^2+o(\vec{x}^2)) \end{align} My question is how do I obtain the data of power series in $\vec{x}$ of $f(\vec{x})$ from the Fourier transform $\tilde{f}(\vec{p})$. There is no way to get a simple closed form for $f(\vec{p})$, but I may be able to do a $\frac{1}{\vec{p}^2}$ expansion term by term in the formal series expression, since "short distance corresponds to large momentum". Even if I get the $\frac{1}{\vec{p}^2}$ expansion, it seems complicated to Fourier transform the powers, for example my Mathematica is not able to compute the Fourier transform of $\frac{1}{(x_1^2+x_2^2)^{2\Delta}}(3x_1^2-x_2^2)$.
In one word, I need to understand how the short distance behavior (power series in $\vec{x}$) of $f(\vec{x})$ is encoded in the Fourier transform $\tilde{f}(\vec{p})$, and what kind of large momentum data I need to extract it.
The power series of $f$ in terms of it's Fourier transform $\hat f$
$$ f(x) = \int d^4 p \, e^{ip\cdot x} \hat{f}(p) $$
(note that I changed the normalization of the Fourier transform) can be simply obtained, formally, by expanding the exponential
$$ f(x) \, {"="} \sum_{k=0} \frac{(i x)^k}{k!} \mu_k $$
where the (wannabe) moments are
$$\mu_k =\int d^4 p \, p^k \hat{f}(p) \ . $$ Clearly for the moments to exist $\hat f$ should decay sufficiently fast at infinity. This however does not guarantee that the power series of $f$ converge. A necessary condition is that $\hat f$ decays exponentially (being compactly supported, for example, would do as well, and implies even better regularity property of $f$).