Let us consider the Fourier transform for a continious function $f(x)$ $\lim\limits_{ x \to \pm \infty} f(x)=0$: $$ \mathcal{F}\{f(x) \}=F(u)=\int_{-\infty}^{+\infty} f(x) e^{-2\pi i ux} dx $$ We have
\begin{gather*} F(u)=\int_{-\infty}^{+\infty} f(x) e^{-2\pi i ux} dx=\int_{-\infty}^{+\infty} f(x) \sum_{n=0}^{\infty} \frac{(-2\pi i ux)^n}{n!} dx=\\=\sum_{n=0}^{\infty} \frac{(-2\pi i u)^n}{n!} \int_{-\infty}^{+\infty} f(x) x^n dx= \sum_{n=0}^{\infty} \frac{(-2\pi i u)^n}{n!} \mu_n, \end{gather*} where $$ \mu_n=\int_{-\infty}^{+\infty} f(x) x^n dx. $$
I want now to find the $\mu_n$ by using the inverse Fourier transform.
I know that $\mu_n$ can be found directly using the formula
$$
\mathcal{F}\left\{\int f(x) x^n \right\}=\frac{\mathcal{F}\{ f(x) x^n \}}{2 \pi x}
$$
but this way is not suitable, because then we have to find $\mathcal{F}\{ f(x) x^n \}$ for any $n$ separately.
Therefore I try to find a function $X(u)$ such that
$$
\mu_n=\mathcal{F}^{-1}\{ F(u) X(u) \}(n)
$$
I tried to expand $ X (u) $ in a series to find its coefficients but without success.
Question: Is is possible to find the $X(u)$ in a simple uniform way?
P.S. We can consider $ f $ not only as a continuous function but as smooth and more.
That $f$ is continuous isn't enough for the Fourier transform integral to converge. If $x^n f\in L^1$ for $n\le k$ then $F(u)=\int_{-\infty}^\infty e^{-2i\pi u x}f(x)dx$ is $C^k$ and $ \int_{-\infty}^\infty (-2i\pi x)^n f(x)dx= F^{(n)}(0)$ for $n\le k$.
If this holds for all $k$ then $F$ is smooth.
The condition for the Taylor series to converge and $F$ being equal to its Taylor series (analytic) is much more restrictive, a sufficient condition is that $f$ has exponential decay.