Fourier transform of $| t |^n$ I understand that the Fourier transform of $|t|^n$ where $n$ is a positive integer is related to derivatives of the delta function -- see e.g. these posts: What's the Fourier transform of these functions?, Fourier Transform of f(t)=|t| and This link.

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*For $n=1$ : $ \mathcal F[|t|](\lambda) = -{2 \over \lambda^2 } $

*For $n=2$ : $\mathcal F[t^2](\lambda) = -2\pi \delta''(\lambda)$

*For $n=3$ : $\mathcal F[|t|^3](\lambda) = {12 \over \lambda^4}$

*For $n=4$ : $\mathcal F[t^4](\lambda) = 2\pi \delta^{(4)}(\lambda)$
where $\mathcal F[|t|^n](\lambda) := \int e^{-i \lambda t} |t|^n dt $.  For general $n$ Mathematica gives FourierTransform[Abs[t]^n, t, f, FourierParameters -> {1, -1}] --> -2 Abs[f]^(-1 - n) Gamma[1 + n] Sin[(n \[Pi])/2] which only works for $n$ odd...
I am guessing the general formula is

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*For $n$ odd: $\mathcal F[|t|^n](\lambda) = ({-1 / \lambda^2} )^{(n+1)/2} \ 2n!$

*For $n$ even: $\mathcal F[|t|^n](\lambda) = (-1)^{n/2} \ 2\pi \ \delta^{(n)}(\lambda)$
Is this correct? Can anyone maybe provide a reference?  There is so much published about Fourier transforms over the Internet that it's hard to find anything!
 A: We have $\mathcal F[t f(t)](\lambda) = i \frac{d}{d\lambda} [\mathcal F[ f(t)](\lambda)]$ (where signs and constants might differ depending upon which definition of the Fourier transform you use).
Also $\mathcal F(1)(\lambda) = 2\pi \delta(\lambda)$, and $\mathcal  F[\text{sign}(t)](\lambda) = -2\frac i\lambda$.  Now use $|t|^{2n} = t^{2n} \cdot 1$, and $|t|^{2n+1} = t^{2n+1} \cdot \text{sign}(t)$.
Iteration gives the formulas you conjectured.
A: This is a great, classic, iconic sort of question. The tempered distributions $|x|^{-s}$ and such, even for complex $s$ (under meromorphic continuation!), have understandable Fourier transforms: away from poles, the Fourier transform of $|x|^{-s}$ on $\mathbb R^n$ is a constant multiple of $|x|^{-(n-s)}$, and the constant can be determined by sample evaluation.
Some comments: unsurprisingly, one must be careful about "the distribution" $|x|^{-s}$, since it's not always locally integrable, so some sort of regularization is necessary. E.g., on $\mathbb R^1$, "$1/x$" really has to be the Cauchy principal value integral.
Also, e.g. on $\mathbb R^1$, the meromorphic family of distributions $s\to |x|^{-s}$ has a pole at $s=1$, with residue (a constant multiple of) $\delta$.
Some of my course notes discuss such things from various viewpoints. See the notes on http://www.math.umn.edu/~garrett/m/real/, perhaps especially the "Iconic examples of distributions" a bit down the page.
I should say that I learned about these things from Stein-and-Weiss, "Fourier analysis on Euclidean spaces", and various volumes of Gelfand-et-al's "Generalized functions", and later finally saw L. Schwartz and A. Grothendieck's discussions of such things.
EDIT: prompted by a comment-question... Yes, on $\mathbb R^1$, for even $n$, $|x|^n=x^n$, and we know/compute that its Fourier transform is a constant multiple of $\delta^{(n)}$. For odd $n$, the Fourier transform of $|x|^n$ is not just supported at $0$ (which would make it a derivative of $\delta$), but is (for colorful language) a multiple of the Hadamard "finite part" (in French "partie finie") of $|\xi|^{-(n+1)}$. For $n=1,3,5,\ldots$ this is not locally $L^1$ at $0$, so direct integration against it cannot be the way to evaluate it. (Riesz proved that it is the analytic/meromorphic continuation of the tempered distribution $|x|^s$, which does also give one way to compute it.)
Again, a basic discussion of parity and homogeneity shows that there is no distribution supported at $0$ that could be this tempered distribution.
It is interesting to compare $\hbox{sgn}(x)\cdot |x|^s$ with even integers $s$. For example, with $s=0$, the Fourier transform of the sign function is a scalar multiple of the Cauchy Principal value integral against $1/x$. That is, since $1/x$ is not locally integrable at $0$, it cannot quite be "integrate against $1/x$", but it still as close to that as possible.
The odd-integer case is already illuminated by $|x|^1$, whose Fourier transform is a scalar multiple of (a regularization of) $1/|x|^2$. Again, this is not locally $L^1$ at $0$, so cannot be the literal integral-against functional. Hadamard and Riesz give descriptions of this distribution.
(Again, several existence and uniqueness assertions are carefully proven in the notes mentioned above.)
