# Fourier transform using principal value

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.

• Hint: Isn't $1/|x|^{n-\alpha}$ regarded as a frequency versus $|x|$? – al-Hwarizmi Jul 12 '13 at 11:36

$$f(x) = e^{-\pi \delta |x|^2},$$ 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.
• And the inevitability of the result follows from nice general properties of Fourier transform: $1/|x|^{n-\alpha}$ is a tempered distribution, rotation-invariant, and homogeneous under dilations. Fourier transform preserves rotation-invariance, and converts homogeneity of degree $n-\alpha$ to $\alpha$. Then the constant can be determined by the computation at the end of @RayYang's answer. – paul garrett Jul 12 '13 at 14:51
• @RayYang How did you identify your expression with the fourier transform of $1/|x|^{n-\alpha}$ ? – smiley06 Jul 13 '13 at 11:53