Fourier transform of $|x|^b$. So, if $x\in \mathbb{R}$, then the fourier coefficients of $|x|$ decay like $k^{-2}$. I had a homework problem that asked us to show that the fourier coefficients of $|x|^b$ decay like $k^{-1-b}$.  
My question is about doing this in $\mathbb{R}^n$. My friend and I were talking about it, and I think that if $x\in\mathbb{R}^n$, the higher dimensional fourier series should decay like $k^{-2n}$ and then if $0<a<n$ the fourier coefficients of $|x|^{a}$ should be like $k^{-n^2-a}$. 
Does this seem right? Is there a way to show this? 
 A: This question does broach an important idea! Yes, the tempered distribution given by the meromorphic (distribution-valued) continuation of $|x|^{-s}$ on $\mathbb R^n$ is a (determinable... apply it to a Gaussian and use properties of the Gamma function) constant multiple of $|x|^{-(n-s)}$. This is provable in the range where $|x|^{-s}$ is locally integrable essentially by the definition, and then ... by whatever semi-magical invocations one prefers... "by analytic continuation" for all values of $s\in\mathbb C$. 
(My own choice of "semi-magic" here would use the Snake Lemma to prove existence and uniqueness of an extension to the whole space of Schwartz functions of the distribution $|x|^{-s}$ on the subspace of Schwartz functions vanishing to infinite order at $0$...)
A: I'm new here, so my apologies if I am doing something wrong since this isn't a full answer, but I can't leave a comment. 
I've been thinking about something similar. First, let $k\in\mathbb{Z}^n$ and assume that $|k_1|=||k||_{\infty}$ (the other cases are similar.) Then $|k_1|\simeq ||k||_{1}$. So, integrate by parts $n$ times in the $x_1$ variable. This is why this isn't a proper answer: I don't know of a closed form for this $n$ fold derivative. But, it should be something like $|x|^{-a}$ (or maybe something slightly different, by the exponent should be something that is integrable locally.) Now, you essentially want to compute the fourier coefficient of $|x|^{-a}$. Now you integrate by parts $k$ more times until $k+1+a >n$ but $k+a<n$. Now you want to do a"fractional" integration by parts to get the rest of the decay. (You can see an example of that here: https://mathoverflow.net/questions/129830/integration-by-parts-for-the-fractional-laplacian). 
Again, I know there are some gaps above, but this is the right idea and I don't have any reputation so I can't just leave this in the comments. 
edit: change $\infty$ to $1$.
edit: I changed the wrong one.
