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

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

2 Answers 2

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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.

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    $\begingroup$ this computation is quite close to this: math.stackexchange.com/questions/3723136/… $\endgroup$ Jan 4, 2023 at 6:07
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    $\begingroup$ Thanks for pointing it out! $\endgroup$ Jan 4, 2023 at 21:15
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    $\begingroup$ 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.) $\endgroup$ Jan 5, 2023 at 2:29
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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 $$ \mathcal F \left[\frac1{|x|^d}\right] = -A_d\log |x| - C_dA_d = \frac{2\pi^{d/2}}{\Gamma(d/2)} \left(-\log |x| +\frac{\psi^{(0)}(d/2)-\gamma}2 - \log\pi \right).$$
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    $\begingroup$ 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 $$I(\lambda)=\int_{\mathbb R^d}\frac{\phi(x)-\phi(0)}{|x|^{d-\lambda}}dx+\int_{|x|>1}\frac{\phi(0)}{|x|^{d-\lambda}}dx+\phi(0)\int_{|x|<1}\frac{1}{|x|^{d-\lambda}}dx?$$ $\endgroup$
    – Mark Viola
    Jan 4, 2023 at 15:30
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    $\begingroup$ +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 ... $\endgroup$
    – LL 3.14
    Jan 4, 2023 at 21:05
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    $\begingroup$ @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 $\endgroup$ Jan 4, 2023 at 23:34
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    $\begingroup$ I suspect that the first term in $I(\lambda)$ should be integrated over $|x| \le 1$ rather than $\mathbb{R}^d$. $\endgroup$ Jan 4, 2023 at 23:53
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    $\begingroup$ @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 $\endgroup$ Jan 5, 2023 at 1:48

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