Fourier transform of the distribution $\left[\frac{1}{|x|^d}\right]$ on $\mathbb{R}^d$ [duplicate]

Define $$\left[\frac{1}{|x|^d}\right]$$ on $$\mathbb{R}^d$$ as $$\left[\frac{1}{|x|^d}\right](\varphi) = \int_{|x|\le 1}\frac{\varphi(x)-\varphi(0)}{|x|^d} dx + \int_{|x| > 1}\frac{\varphi(x)}{|x|^d} dx.$$

It is easy to show that $$\left[\frac{1}{|x|^d}\right]$$ is a tempered distribution. It also agrees with $$1/|x|^d$$ away from the origin.

According to my textbook, the Fourier transform of the distribution $$\left[\frac{1}{|x|^d}\right]$$ equals $$c_1 \log |\xi| + c_2$$, with $$c_1 \neq 0$$.

However, I don't see how to prove this claim. Below is how far I've got, with $$F$$ standing for $$\left[\frac{1}{|x|^d}\right]$$.

\begin{align} \hat{F}(\varphi) &= F(\hat{\varphi}) = \int_{|x|\le 1}\frac{\hat{\varphi}(x)-\hat{\varphi}(0)}{|x|^d} dx + \int_{|x| > 1}\frac{\hat{\varphi}(x)}{|x|^d} dx \\&= \int_{\mathbb{R}^d} \left(\int_{|x|\le 1}\frac{\varphi(\xi)(e^{-2\pi i \xi \cdot x}-1)}{|x|^d} dx + \int_{|x| > 1}\frac{\varphi(\xi)e^{-2\pi i \xi \cdot x}}{|x|^d} dx\right) d\xi \\&= \int_{\mathbb{R}^d} \varphi(\xi) \left(\int_{|x|\le 1}\frac{(e^{-2\pi i \xi \cdot x}-1)}{|x|^d} dx + \int_{|x| > 1}\frac{e^{-2\pi i \xi \cdot x}}{|x|^d} dx\right) d\xi. \end{align}

So it is left to prove that $$\int_{|x|\le 1}\frac{(e^{-2\pi i \xi \cdot x}-1)}{|x|^d} dx + \int_{|x| > 1}\frac{e^{-2\pi i \xi \cdot x}}{|x|^d} dx = c_1 \log|\xi|+c_2.$$

• For $d=1$, see Equation $(18)$ of THIS ANSWER. Follow an analogous procedure to arrive at the expected result. Jan 3, 2023 at 22:30
• Thank you both, MarkViola and LL3.14. Jan 4, 2023 at 21:15

I have figured out a proof.

(As pointed out in a few comments, the original question was already answered in another post. The answer there is actually a lot more elegant.)

First let us define $$\Phi(\xi) = \int_{|x|\le 1}\frac{e^{-2\pi i \xi \cdot x}-1}{|x|^d} dx + \int_{|x| > 1}\frac{e^{-2\pi i \xi \cdot x}}{|x|^d} dx$$

The first term of $$\Phi(\xi)$$ converges because $$e^{-2\pi i \xi \cdot x} - 1 = O(|x|)$$.

The second term also clearly converges if we integrate using polar coordinates.

Due to rotational symmetry, $$\Phi(\xi)$$ depends only on $$|\xi|$$. We simply take the value of $$\Phi(\xi)$$ at $$|\xi| = 1$$ to be $$c_2$$.

Now we can take the derivative of $$\Phi(\xi)$$ with respective to $$\xi_i$$, multiply it by $$\xi_i$$, and finally sum all such terms over $$i=1,2,\cdots,d$$. We get $$\xi \cdot \nabla \Phi(\xi) = \int_{\mathbb{R}^d} \frac{(-2 \pi i \xi \cdot x) e^{-2 \pi i \xi \cdot x}}{|x|^d} dx.$$

For $$d \ge 2$$, this integral does not depend on $$\xi$$, as long as $$\xi \neq 0$$. (Again this can be seen through a change of integration variable $$x \rightarrow x/a$$.) It convergence can be seen by setting $$|\xi| = 1$$ and doing the integral in spherical coordinates. So we have $$\xi \cdot \nabla \Phi(\xi)$$ equal to a constant, or with slight abuse of notation, $$r \Phi'(|\xi|) = c_1$$. This gives us $$\Phi(|\xi|) = c_1 \log |\xi| + c_2$$.

For $$d=1$$, the integral gives us $$-2 + \lim_{r\rightarrow\infty} 2 \sin 2\pi\xi r.$$ The divergent term disappears when it acts on $$\varphi(\xi)$$, because $$\lim_{r\rightarrow\infty} \int_\mathbb{R} \sin (2\pi\xi r) \varphi(\xi) d\xi = \lim_{r\rightarrow\infty} \frac{1}{2\pi r} \int_\mathbb{R} \cos (2\pi\xi r) \varphi'(\xi) d\xi = 0,$$ where we integrated by parts in the last step.

• this computation is quite close to this: math.stackexchange.com/questions/3723136/… Jan 4, 2023 at 6:07
• Thanks for pointing it out! Jan 4, 2023 at 21:15
• Your quick remark showing $c_1\neq 0$ for $d\ge2$ has disappeared in the new edit (While I'm here, just to name-drop: $e^{ix\xi} \to 0$ in $\mathcal D'$ as $\xi\to\infty$ is the Riemann-Lebesgue lemma.) Jan 5, 2023 at 2:29

I learned this via the following proof, which assumes you first understood the analogous question for $$1/|x|^{d-\lambda}$$, $$\lambda>0$$. I don't think it is better than your solution but thought it worthwhile to include a different proof which also computes explicit constants (they match with LL3.14's answer here). I'm using the $$L^2$$-isometric Fourier transform $$\mathcal Ff(\xi) = \int_{\mathbb R^d} f(x) e^{-2\pi i x\cdot \xi} dx$$.

Fix a test function $$\phi$$ and define for $$\lambda>0$$ $$I(\lambda) := \int_{\mathbb R^d}|x|^{-d+\lambda}\phi(x)dx.$$ We will evaluate $$\lambda I(\lambda)$$ to first order, then differentiate and send $$\lambda\to 0$$.

1. We can write $$I(\lambda)=\int_{|x|<1}\frac{\phi(x)-\phi(0)}{|x|^{d-\lambda}}dx+\int_{|x|>1}\frac{\phi(x)}{|x|^{d-\lambda}}dx+\phi(0)\int_{|x|<1}\frac{1}{|x|^{d-\lambda}}dx.$$ The last term is clearly $$A_d \phi(0)/\lambda$$, with $$A_d = \frac{2\pi^{d/2}}{\Gamma(d/2)}\neq 0$$ the $$(d-1)$$-dimensional surface area of the ball. So we have \begin{align}\lambda I(\lambda)&=A_d \phi(0)+\lambda\left(\int_{|x|<1}\frac{\phi(x)-\phi(0)}{|x|^{d-\lambda}}dx+\int_{|x|>1}\frac{\phi(x)}{|x|^{d-\lambda}}dx\right)\\ &=A_d \phi(0)+\lambda\left(\int_{|x|<1}\frac{\phi(x)-\phi(0)}{|x|^{d}}dx+\int_{|x|>1}\frac{\phi(x)}{|x|^{d}}dx\right) + O(\lambda^2) .\end{align} So $$\lim_{\lambda\downarrow0}(\lambda I(\lambda))' = \int_{|x|<1}\frac{\phi(x)-\phi(0)}{|x|^{d}}dx+\int_{|x|>1}\frac{\phi(x)}{|x|^{d}}dx = \left[\frac1{|x|^d}\right](\phi).$$
2. At the same time, it should be already known that $$I(\lambda) = \int_{\mathbb R^d} |x|^{-d+\lambda} \phi(x) dx = C_{d,\lambda} \int_{\mathbb R^d} |x|^{-\lambda} \widehat \phi(x)dx,$$ where $$C_{d,\lambda}:=\frac{\Gamma(\lambda/2)}{\Gamma((d-\lambda)/2)}\pi^{\frac d2 + \lambda} = \frac2\lambda\frac{\Gamma(1+\lambda/2)}{\Gamma((d-\lambda)/2)}\pi^{\frac d2 + \lambda}.$$ Now, $$1/C_{d,\lambda}$$ is analytic near $$\lambda=0$$, and it can be computed with some elbow grease (or computer) that we have the expansion $$\frac1{C_{d,\lambda}} = \underbrace{\frac{\Gamma(d/2)}{2\pi^{d/2}}}_{=1/A_d}\lambda + \underbrace{\frac{\Gamma(d/2)}{4\pi^{d/2}}(-\psi^{(0)}(d/2) + \gamma + 2\log\pi)}_{=:C_d/A_d}\lambda^2 + O(\lambda^3)$$ ($$\psi^{(0)}=\Gamma'/\Gamma$$ is the digamma function and $$\gamma$$ is the Euler-Mascheroni constant) which means that $$\int_{\mathbb R^d} |x|^{-\lambda} \widehat \phi(x)dx = \left(\frac{1}{A_d}\lambda + \frac{C_d}{A_d} \lambda^2 + O(\lambda^3)\right) I(\lambda).$$ Differentiating in $$\lambda$$ (note $$\frac{d}{d\lambda} a^{\lambda} = (\log a)a^\lambda$$) we get $$-\int_{\mathbb R^d} (\log |x|) |x|^{-\lambda} \widehat\phi(x)dx = \frac{1}{A_d}(\lambda I(\lambda))' + \frac{C_d}{A_d}(\lambda^2 I(\lambda))' + O(\lambda)$$ We used $$I(\lambda) = c_1 + \frac{c_2}\lambda + O(\lambda) = O(1/\lambda)$$ and $$I'(\lambda)=O(1/\lambda^2)$$ from step 1. In addition, step 1 also gives $$(\lambda^2 I(\lambda))' = \lambda I(\lambda) + \lambda (\lambda I(\lambda))' = A_d \phi(0) + o(1).$$ Putting these facts together, we can send $$\lambda\to 0$$ to get $$-\int_{\mathbb R^d} \log|x| \widehat \phi (x) dx =\frac{1}{A_d}\left[\frac1{|x|^d}\right](\phi) + C_d \phi(0).$$ In other words: $$-\mathcal F\log|x| = \frac1{A_d}\left[\frac1{|x|^d}\right] + C_d \delta_0.$$ Finally, taking the Fourier transform gives $$\mathcal F \left[\frac1{|x|^d}\right] = -A_d\log |x| - C_dA_d = \frac{2\pi^{d/2}}{\Gamma(d/2)} \left(-\log |x| +\frac{\psi^{(0)}(d/2)-\gamma}2 - \log\pi \right).$$
• You wrote $$I(\lambda)=\int_{\mathbb R^d}\frac{\phi(x)-\phi(0)}{|x|^{d-\lambda}}dx+\int_{|x|>1}\frac{\phi(x)}{|x|^{d-\lambda}}dx+\phi(0)\int_{|x|<1}\frac{1}{|x|^{d-\lambda}}dx$$ But if $I(\lambda)$ is defined as $I(\lambda) = \int_{\mathbb R^d}|x|^{-d+\lambda}\phi(x)dx$, then should we not have $$I(\lambda)=\int_{\mathbb R^d}\frac{\phi(x)-\phi(0)}{|x|^{d-\lambda}}dx+\int_{|x|>1}\frac{\phi(0)}{|x|^{d-\lambda}}dx+\phi(0)\int_{|x|<1}\frac{1}{|x|^{d-\lambda}}dx?$$ Jan 4, 2023 at 15:30
• +1 to Mark Viola. But your constant at the end seems correct as it agrees with the one I found here math.stackexchange.com/questions/3723136/…. The constant $\tilde{\tilde{C}}_d$ is also computed there ... Jan 4, 2023 at 21:05
• @LL3.14 in adapting my handwritten notes which was only for d=2 I forgot to cross check with your answer which I commented under OP’s own answer. Thanks for pointing this out! I will make these changes when I get to a computer Jan 4, 2023 at 23:34
• I suspect that the first term in $I(\lambda)$ should be integrated over $|x| \le 1$ rather than $\mathbb{R}^d$. Jan 4, 2023 at 23:53
• @LL3.14 I have fixed the proof; since I was one step away from computing $\tilde{\tilde C_d}$ I went ahead and did that and also found it matches with yours. Thanks again Jan 5, 2023 at 1:48