Fourier transform using principal value

Question

Can anyone help me compute the Fourier transform of $ 1/|x|^{n-\alpha} $ in $\mathbb{R}^n $ where $ 0 < \alpha < n $ ? Somehow it becomes the principal value of $ 1/|x|^\alpha $ which I can't figure out.

Answer 1

This is not all that simple a calculation. A standard technique (see, e.g., Stein's Singular Integrals, III.3) is to remember that the Fourier transform of the Gaussian is another Gaussian, and to use that as an auxiliary function. That is to say, if

$$f(x) = e^{-\pi \delta |x|^2},$$ then $$\hat{f}(\xi) = \delta^{-\frac{n}{2}} e^{-\frac{\pi |\xi|^2}{\delta}}.$$

This is useful because if one invokes a change of variable in the definition of the Gamma function one retrieves the following identity: $$ \int_0^\infty e^{-\pi \delta |x|^2} \delta^{\beta -1} d\delta = \frac{\Gamma(\beta)}{(\pi|x|^2)^\beta}$$ Taking the Fourier transform of the above expression with respect to $x$ yields $$\int_0^\infty \delta^{\beta - \frac{n}{2} -1} e^{- \pi |\xi|^2 / \delta} d\delta = \int_0^\infty s^{-1 + \frac{n}{2} -\beta} e^{-\pi s |\xi|^2} ds = \frac{\Gamma(\frac{n}{2} -\beta)}{(\pi |\xi|^2)^{\frac{n}{2} -\beta}} $$ Setting $\beta = \frac{n-\alpha}{2}$, we get the desired result.