we know that the Fourier integrals
$$ \int_{-\infty}^{\infty}dx x^{m}e^{ixu} =F(u)$$
exists only in the sense of distribution
however what would happen if we approximate the complex exponential by a Pade Approximant
$$ e^{ixu} = \frac{ a_{0}(u)+a_{1}(u)x+...........+a_{m}(u) x^{m}}{b_{0}(u)+b_{1}(u)x+...........+b_{n}(u) x^{n}}= \frac{P(x,u)}{Q(x,u)} $$
and then we use some method to integrate $$ \int_{-\infty}^{\infty}dx x^{m}\frac{P(x,u)}{Q(x,u)} $$ as a function of 'u'
The integral will diverge just as before, as an illustration I'll use the second-order Pade approximation around $x=0$, $$ e^{i x u} = Pe(x) + O(x^5) , $$ with $$Pe(x)= \frac{-\frac{1}{12} u^2 x^2+\frac{i u x}{2}+1}{-\frac{1}{12} u^2 x^2-\frac{i u x}{2}+1}. $$ The fourier transform diverges because $$ \lim_{x \to \infty} |Pe(x) e^{i x u} | = 1.$$ To calculate the fourier transform of $x^m$ in the sense of distributions you must first verfiy why $$\delta(u) = \frac{1}{\sqrt{2}} \int_{-\infty}^{\infty} e^{i x u} dx$$ which can be found in standard text books [1].
[1] Kammler, David (2000), A First Course in Fourier Analysis, Prentice Hall,