Calculating Fourier Transform of $1/|t|^n$

Question

I have found the Fourier Transform of $x(t)=|t|^{n}$ and i can't calculate the Fourier Transform of $x(t)=|t|^{-n}$. Any suggestions?

Answer 1

The Fourier transform is defined for tempered distributions. The function $x(t)=|t|^{-n}$ is not a distribution since it is not locally integrable. Therefore, it does not have a Fourier transform.

Loosely speaking: if the transform of $|t|^{-n}$ existed, it would be a constant function, because the Fourier transform of a radially symmetric homogeneous distribution of degree $d$ is a radially symmetric distribution of degree $-d-n$. With $d=-n$ this gives $0$. However, the constant function is the Fourier transform of the Dirac $\delta$. This leaves $|t|^{-n}$ out in the cold.

Related threads: ($n$-dimensional) Inverse Fourier transform of $\frac{1}{\| \mathbf{\omega} \|^{2\alpha}}$, What's the Fourier transform of these functions?.

Answer 2

The hardest part is actually properly define the function as a tempered distribution before actually calculating the Fourier transform. The issue is that you defined unambiguously a distribution on $\mathbb R^*$, but you need to extend it to $\mathbb R$ as a tempered distribution. For complex $n$ that are not negative integers, you can do this by meromorphic continuation described in more detail in homogeneous distribution, which gives the result of line 313 of Fourier transform. In particular, for negative even integer, you can use functional derivatives, using the fact that $|x|^{-n} = x^{-n}$.

For negative odd integers, your function is ill defined. You can already anticipate the problem as formally extending the result of the table gives you infinite values. Indeed, according to the same wikipedia article the only homogeneous distributions of negative odd integer degree are odd, so there is no canonical way to extend your distribution. You could always find some ad hoc surrogates, but you'll have to make some arbitrary choices. For example, you could use a method similar to finite parts by subtracting the lower powers of the Taylor series of the test function at $0$ to get a convergent integral.

Hope this helps.