Calculate the Fourier transform of $\log |x| $

How can one prove that the Fourier transform of $\log |x|$ is $$-\pi \mathrm{pf} \frac{1}{|\xi|} +C \delta,$$ where $\mathrm{pf}\frac{1}{|x|} = D(\mathrm{sign}(x)\log|x|)$ (in the sense of distributions) and how can I compute the constant $C$?

Solution 1:

I will start from the well-known expression (see here) for the Euler-Mescheroni constant $\gamma$:

$$\gamma=\int_0^1\frac{1-\cos t}{t}dt-\int_{1}^\infty\frac{\cos t}{t}dt$$ Now, if for $ x>0$ we consider $$F(x)=\int_0^x\frac{1-\cos t}{t}dt-\int_{x}^\infty\frac{\cos t}{t}dt$$ Then we conclude from $ F(1)=\gamma$ and $F'(x)=1/x$ that $$\eqalign{\gamma+\ln x&= \int_0^x\frac{1-\cos t}{t}dt-\int_{x}^\infty\frac{\cos t}{t}dt\cr &=\int_0^1\frac{1-\cos(xt)}{t}dt-\int_{1}^\infty\frac{\cos (xt)}{t}dt }$$ And since the right side of the above formula is even we conclude that $$ \gamma+\ln|x|=\int_0^\infty\frac{\mathbb{I}_{[0,1]}(t)-\cos(xt)}{t}dt\tag1 $$ For every nonzero $x$.

Let the regular distribution assosiated with the function $x\mapsto \gamma+\ln|x|$ be denoted by $T$. What is the action of $T$ on some test function $\phi$?

Indeed, if $\phi$ is a function from $\mathcal{S}$ then using $(1)$ we see that $$\eqalign{\langle T,\phi\rangle &=\int_{\mathbb{R}}(\gamma+\ln|x|)\phi(x)dx\cr &=\int_0^\infty\frac{2\mathbb{I}_{[0,1]}(t)\hat{\phi}(0)-\hat{\phi}(t)-\hat{\phi}(-t)}{2t}dt\cr &=\int_0^\infty\frac{2\mathbb{I}_{[0,1]}(t)\hat{\phi}(0)-\hat{\phi}(t)-\check{\hat{\phi}}(t)}{2t}dt\tag2 }$$ Indeed, since $\hat{\phi}(t)=\int_{\mathbb{R}}\phi(x)e^{-ixt}dx$ we see easily that $$\hat{\phi}(0)=\int_{\mathbb{R}}\phi(x)dx\quad\hbox{and}\quad \hat{\phi}(t)+\hat{\phi}(-t)=2\int_{\mathbb{R}}\phi(x)\cos(xt)dx$$ Thus, applying (2) to $\hat {\phi}$ and noting that $\hat{\hat{\phi}}=2\pi\check{\phi}$ we conclude that $$\eqalign{\langle T,\hat{\phi}\rangle &=2\pi\int_0^\infty\frac{2\mathbb{I}_{[0,1]}(t)\phi(0)-\phi(t)-\phi(-t)}{2t}dt\cr &=\pi\int_0^1\frac{2\phi(0)-\phi(t)-\phi(-t)}{t}dt-\pi \int_1^\infty\frac{\phi(t)+\phi(-t)}{t}dt\cr &=\pi\int_{0}^\infty (\ln t)(\phi'(t)-\phi'(-t))dt\qquad\hbox{(integration by parts)}\cr &=\pi\int_{\mathbb{R}}{\rm sign}(t)\ln|t|\phi'(t)dt\cr &=-\pi\langle{\rm pf}\frac{1}{|x|},\phi\rangle }$$ So, $\hat{T}=-\pi\,{\rm pf}\frac{1}{|x|}$. But $$\widehat{(\gamma+\ln|x|)}=2\pi\gamma\delta+\widehat{\ln|x|}$$ Thus $$\widehat{\ln|x|}=-\pi\,{\rm pf}\frac{1}{|x|}-2\pi\gamma\delta$$ Which is the desired conclusion.

Solution 2:

For this answer, we will use the Fourier Transform indicated in the question, $$ \hat{f}(\xi)=\int_{-\infty}^\infty f(x)\,e^{-ix\xi}\,\mathrm{d}x\tag{FT} $$ for which the inverse transform is $$ f(x)=\frac1{2\pi}\int_{-\infty}^\infty\hat{f}(\xi)\,e^{ix\xi}\,\mathrm{d}\xi\tag{IFT} $$ Computing the Fourier Transform

One standard way to compute the Fourier Transform of this kind of function is to multiply by $e^{-\epsilon x^2}$ and let $\epsilon\to0$. $$ \begin{align} &\lim_{\epsilon\to0}\int_{-\infty}^\infty e^{-\epsilon x^2}\log\!|x|\,e^{-ix\xi}\,\mathrm{d}x\\ &=2\lim_{\epsilon\to0}\int_0^\infty e^{-\epsilon x^2}\log(x)\,\cos(x|\xi|)\,\mathrm{d}x\tag{1a}\\ &=\frac2{|\xi|}\lim_{\epsilon\to0}\int_0^\infty e^{-\epsilon x^2}\log(x)\,\mathrm{d}\sin(x|\xi|)\tag{1b}\\ &=-\frac2{|\xi|}\lim_{\epsilon\to0}\int_0^\infty e^{-\epsilon x^2}\frac{\sin(x|\xi|)}x\,\mathrm{d}x +\frac2{|\xi|}\lim_{\epsilon\to0}\int_0^\infty2\epsilon xe^{-\epsilon x^2}\log(x)\sin(x|\xi|)\,\mathrm{d}x\tag{1c}\\ &=-\frac\pi{|\xi|}+\frac2{|\xi|}\lim_{\epsilon\to0}\int_0^\infty2xe^{-x^2}(\log(x)-\log(\epsilon)/2)\sin(x|\xi|/\sqrt\epsilon)\,\mathrm{d}x\tag{1d}\\ &=-\frac\pi{|\xi|}+\frac2{|\xi|}\lim_{\epsilon\to0}\int_0^\infty2xe^{-x^2}\log(x)\sin(x|\xi|/\sqrt\epsilon)\,\mathrm{d}x\\ &\phantom{{}=-\frac\pi{|\xi|}}+\lim_{\epsilon\to0}\frac{\sqrt\epsilon\log(\epsilon)}{|\xi|^2}\int_0^\infty\left(2-4x^2\right)e^{-x^2}\cos(x|\xi|/\sqrt\epsilon)\,\mathrm{d}x\tag{1e}\\ &=-\frac\pi{|\xi|}\tag{1f} \end{align} $$ Explanation:
$\text{(1a)}$: apply symmetry
$\text{(1b)}$: prepare to integrate by parts
$\text{(1c)}$: integrate by parts
$\text{(1d)}$: $\int_0^\infty\frac{\sin(x)}{x}\,\mathrm{d}x=\frac\pi2$ and substitute $x\mapsto x/\sqrt\epsilon$
$\text{(1e)}$: distribute the integral over $\log(x)-\log(\epsilon)/2$
$\phantom{\text{(1e):}}$ and integrate the $\log(\epsilon)/2$ piece by parts
$\text{(1f)}$: the first integral vanishes by Riemann-Lebesgue
$\phantom{\text{(1f):}}$ the second by Riemann-Lebesgue or $\lim\limits_{\epsilon\to0}\sqrt\epsilon\log(\epsilon)=0$

Expanding the Set of Test Functions

$\text{(1f)}$ gives the Fourier Transform of $\log(|x|)$ when the test function vanishes at the origin. That is, $$ \int_{-\infty}^\infty\hat{\varphi}(x)\log(|x|)\,\mathrm{d}x=-\pi\int_{-\infty}^\infty\frac{\varphi(\xi)}{|\xi|}\,\mathrm{d}\xi\tag2 $$ Since $-\frac\pi{|\xi|}$ is not integrable near $0$, the right side of $(2)$ does not converge if $\varphi(0)\ne0$.

However, as mentioned in Exercise 13 (Distributional interpretation of $1/|x|$) of Terry Tao's blog "245C, Notes 3: Distributions", we can evaluate principal-value tests against $\frac1{|x|}$ by computing $$ \int_{-\infty}^\infty\hat{\varphi}(x)L_r(x)\,\mathrm{d}x=-\pi\int_{|\xi|\gt r}\frac{\varphi(\xi)}{|\xi|}\,\mathrm{d}\xi-\pi\int_{|\xi|\le r}\frac{\varphi(\xi)-\varphi(0)}{|\xi|}\,\mathrm{d}\xi\tag3 $$ If $\varphi(0)=0$, $(3)$ agrees with $(2)$, but $(3)$ converges even if $\varphi(0)\ne0$.

For any $\varphi$ so that $\int_{-\infty}^\infty\hat{\varphi}(x)\,\mathrm{d}x=0$, $\varphi(0)=0$, so subtracting $(2)$ from $(3)$ gives $$ \int_{-\infty}^\infty\hat{\varphi}(x)(L_r(x)-\log(|x|))\,\mathrm{d}x=0\tag4 $$ That is, for any $\hat\varphi$ that is orthogonal to $1$, $\hat\varphi$ is orthogonal to $L_r(x)-\log(|x|)$. Therefore, there is a constant, $\lambda_r$, so that, in the sense of distributions, $$ \lambda_r=L_r(x)-\log(|x|)\tag5 $$ Thus, $$ \begin{align} \frac{2\pi}r\varphi(0) &=\partial_r\int_{-\infty}^\infty\hat{\varphi}(x)L_r(x)\,\mathrm{d}x\tag{6a}\\ &=\partial_r\int_{-\infty}^\infty\hat{\varphi}(x)(L_r(x)-\log(|x|))\,\mathrm{d}x\tag{6b}\\ &=\partial_r\lambda_r\int_{-\infty}^\infty\hat{\varphi}(x)\,\mathrm{d}x\tag{6c}\\[6pt] &=\partial_r\lambda_r2\pi\varphi(0)\tag{6d} \end{align} $$ Explanation:
$\text{(6a)}$: take the derivative of $(3)$
$\text{(6b)}$: $\int_{-\infty}^\infty\hat{\varphi}(x)\log(|x|)\,\mathrm{d}x$ is constant in $r$
$\text{(6c)}$: apply $(5)$
$\text{(6d)}$: apply $\text{(IFT)}$

Therefore, for some constant $K$, $$ \lambda_r=K+\log(r)\tag7 $$

Computing $\boldsymbol{K}$

Use $\varphi(\xi)=e^{-\xi^2/2}$ and $\hat\varphi(x)=\sqrt{2\pi}\,e^{-x^2/2}$ in $(3)$: $$ \begin{align} \sqrt{2\pi}\int_{-\infty}^\infty e^{-x^2/2}L_r(x)\,\mathrm{d}x &=-\pi\int_{|\xi|\gt r}\frac{e^{-\xi^2/2}}{|\xi|}\,\mathrm{d}\xi-\pi\int_{|\xi|\le r}\frac{e^{-\xi^2/2}-1}{|\xi|}\,\mathrm{d}\xi\tag{8a}\\ &=\pi\log\left(r^2/2\right)+\pi\gamma\tag{8b} \end{align} $$ and in the left hand side of $(2)$: $$ \sqrt{2\pi}\int_{-\infty}^\infty e^{-x^2/2}\log(|x|)\,\mathrm{d}x=\pi(-\gamma-\log(2))\tag9 $$ Subtracting $(9)$ from $(8)$ and applying $(5)$ gives $$ \begin{align} \pi\log\left(r^2\right)+2\pi\gamma &=\sqrt{2\pi}\int_{-\infty}^\infty e^{-x^2/2}\,\overbrace{(L_r(x)-\log(|x|))}^{\lambda_r}\,\mathrm{d}x\tag{10a}\\ &=2\pi\lambda_r\tag{10b} \end{align} $$ from which we get $$ \lambda_r=\gamma+\log(r)\tag{11} $$ That is, $K=\gamma$, the Euler-Mascheroni constant.

Conclusion

Equations $(5)$ and $(11)$ say that $L_r(x)-\log(x)=\gamma+\log(r)$; therefore, $(3)$ can be written as $$ \int_{-\infty}^\infty\hat\varphi(x)\overbrace{(\log(|x|)+\gamma+\log(r))}^{L_r(x)}\,\mathrm{d}x=-\pi\int_{-\infty}^\infty\frac{\varphi(\xi)-\varphi(0)[|\xi|\lt r]}{|\xi|}\,\mathrm{d}\xi\tag{12} $$ $\text{(FT)}$ says that $\hat\delta=1$ and $\text{(IFT)}$ says that $\hat1=2\pi\delta$. Setting $C_r=-2\pi(\gamma+\log(r))$ and $\varphi_r(\xi)=\varphi(\xi)-\varphi(0)[|\xi|\lt r]$, where $[\dots]$ are Iverson brackets, we have $$ \int_{-\infty}^\infty\hat\varphi(x)\log(|x|)\,\mathrm{d}x=\int_{-\infty}^\infty\left(\left(-\frac\pi{|\xi|}\right)\varphi_r(\xi)+C_r\delta(\xi)\varphi(\xi)\right)\mathrm{d}\xi\tag{13} $$ Thus, depending on the $r$ we choose in $(3)$, we get a different $C_r$ in $(13)$. If we choose $r=1$, we get $C_r=-2\pi\gamma$, which is the $C$ obtained by Omran Kouba. If we choose $r=e^{-\gamma}$, then we get $C_r=0$, which removes the need for a delta function. However, no choice of $r$ removes the need for using $\varphi_r$ to permit dividing by $|\xi|$ while retaining integrability.

Solution 3:

Yes, such computations are standard, but/and can be done in several ways. One approach is to first observe that ${d\over dx}(\log|x|)$ is the principal-value integral $u$ against $1/x$ (not $1/|x|$). This principal value integral is not a literal integrate-against functional, since $1/x$ is not locally integrable (nor is $1/|x|$). Even though it's not a literal integral, one still shows directly that $x\cdot u = 1$, where on the left multiplication by the smooth function (of moderate growth...) on tempered distributions is as usual. (We simply cannot "divide" in a pointwise sense.)

Fourier transform has an easily-verified effect on positive-homogeneity, and parity: the FT of $|x|^{-s}$ is a constant multiple of $|x|^{1-s}$, literally so for $0<\Re(s)<1$, and then by meromorphic continuation. Thus, the Fourier transform of $u$ is a constant multiple of $\mathrm{sgn}\,x$. By integrating against $xe^{-\pi x^2}$, for example (or almost any other odd Schwartz function) one finds that the constant is $-i\pi$ (maybe!).

Thus, letting $F$ be Fourier transform, $$ -i\pi \mathrm{sgn}\,x \;=\; F u \;=\; F{d\over dx}\log|\cdot| \;=\; -2\pi ix \cdot F\log|\cdot| $$ Thus, $2x\cdot F\log|\cdot|= \mathrm{sgn}\,x$. Again, we cannot quite divide pointwise. However, the kernel of the multiplication-by-$x$ operator on tempered distributions consists of distributions supported at $\{0\}$, which (essentially by the theory of Taylor-Maclaurin series) is just finite linear combinations of Dirac $\delta$ and its derivatives. Further, the only such linear combination annihilated by mult'n by $x$ are just multiples of $\delta$ itself. Thus, the relation $2x\cdot F\log|\cdot|=\mathrm{sgn}\,x$ determines that Fourier transform up to multiples of $\delta$.

To determine the constant, let $g(x)=e^{-\pi x^2}$, for example, and for arbitrary Schwartz function $f$, use the standard trick $$ v(f) \;=\; v(f-f(0)\cdot g)+f(0)v(g) $$ and then evaluate $v(f-f(0)g)$ by using the literal integral definition, since $f-f(0)g$ vanishes at $0$, etc.

EDIT: per request of the questioner, I'll give the determination-of-constant idea (in principle standard, but... etc) in further detail, though I would have to think more to express it in terms of the Euler-Mascheroni constant, etc.

That is, let $u=\widehat \log|\cdot|$. Suppose we know that $x\cdot u=a\cdot \mathrm{sgn}\,x$, where I've written another constant $a$ to accommodate possible earlier boo-boos, and make it easier to track. Also note that for a test function $f$, if $f(0)=0$, then $f(x)/x$ is also a test function. Let $g$ be the Gaussian, as above. Then $f(x)-f(0)\cdot g(x)$ is of the form $x\cdot h(x)$ for a test function $h$. Thus, $$ u(f) \;=\; u(f-f(0)g)+f(0)u(g) \;=\; u(x\cdot {f-f(0)g\over x}) + \delta f \cdot u(g) \;=\; (x\cdot u)({f-f(0)g\over x}) + \delta f\cdot u(g) $$ $$ \;=\; a\int \mathrm{sgn}\,(x)\cdot {f(x)-f(0)g(x)\over x}\;dx + \delta f\cdot u(g) \;=\; a\int {f(x)-f(0)g(x)\over |x|}\;dx + \delta f\cdot u(g) $$ The integral can be further explicated in various ways, e.g., integrating by parts. The most-unknown part of the business is the constant $u(g)$, which appears (maybe part of) the coefficient of $\delta$.

Edit-Edit: in response to some further questions: to see the vanishing at $0$ of $f-f(0)g$: $$ f(0)-f(0)\cdot g(0) \;=\; f(0) - f(0)\cdot 1 \;=\; 0 $$ The fact that $$ u(f) \;=\; u(f-f(0)g+f(0)g)\;=\;u(f-f(0)\cdot g) + u(f(0)\cdot g) \;=\; u(f-f(0)\cdot g) + f(0)\cdot u(g) $$ is the linearity of $u$. As to evaluating the constant which involves the Euler-Mascheroni constant, I do not have an easy answer. But the literal integral can be manipulated in several ways, for example integrating by parts, to get something like your 'pf' functional.

Solution 4:

I thought that it might be instructive to present an approach to deriving the Fourier transform of $\log(|x|)$ that uses a regularization approach. This way forward is distinct from the methodology I used in THIS ANSWER.

The result herein includes a distributional interpretation of $\frac1{|x|}$. Finally, we show that the distributional interpretation of $\frac1{|x|}$ is non-unique and that it differs from other interpretations by a multiple of the Dirac Delta distribution. With that introduction, we now proceed.

PRELIMARIES

Let $\psi(x)=\log(|x|)$ and let $\Psi$ denote its Fourier Transform . Then, we write

$$\Psi(x)=\mathscr{F}\{\psi\}(x)\tag 1$$

where $(1)$ is interpreted as a Tempered Distribution. That is, for any $\phi \in \mathbb{S}$, we can write

$$\langle \mathscr{F}\{\psi\}, \phi\rangle =\langle \psi, \mathscr{F}\{\phi\}\rangle$$

Now, let $\psi_\epsilon(k) =e^{-\varepsilon|k|}\log(|k|)$. Therefore, $\psi(k)=\lim_{\varepsilon\to 0^+}\psi_\varepsilon(k)$ and we see that

$$\begin{align} \lim_{\varepsilon\to 0^+}\langle \mathscr{F}\{\psi_\varepsilon\}, \phi\rangle&=\lim_{\varepsilon\to 0^+}\langle \psi_\varepsilon, \mathscr{F}\{\phi\}\rangle \\\\ &=\langle \psi,\mathscr{F}\{\phi\}\rangle\\\\ &=\langle \mathscr{F}\{\psi\}, \phi\rangle \end{align}$$

Next, we evaluate the Fourier transform of $\psi_\varepsilon$.

EVALUATING THE FOURIER TRANSFORM OF $\displaystyle \psi_\varepsilon$

Denote by $\Psi_\epsilon$, the Fourier transform of $\psi_\varepsilon$. Then, we have

$$\begin{align} \Psi_\varepsilon(x)&=\mathscr{F}\{\psi_\epsilon\}(x)\\\\ &=\int_{-\infty}^\infty e^{-\varepsilon|k|}\log(|k|) e^{ikx}\,dk\\\\ &=2\text{Re}\left(\int_0^\infty e^{-(\varepsilon -ix)k}\log(k) \,dk\right)\\\\ &=-\frac{2\varepsilon}{\varepsilon^2+x^2}\gamma -\frac{\varepsilon}{\varepsilon^2+x^2}\log(\varepsilon^2+x^2)-\frac{2x}{\varepsilon^2+x^2}\arctan(x/\varepsilon)\\\\ &=\psi^{(1)}_\varepsilon(x)+\psi^{(2)}_\varepsilon(x)+\psi^{(3)}_\varepsilon(x)\tag2 \end{align}$$

where

$$\begin{align} \psi^{(1)}_\varepsilon(x)&=-\frac{2\varepsilon}{\varepsilon^2+x^2}\gamma\\\\ \psi^{(2)}_\varepsilon(x)&=-\frac{\varepsilon}{\varepsilon^2+x^2}\log(\varepsilon^2+x^2)\\\\ \psi^{(3)}_\varepsilon(x)&=-\frac{2x}{\varepsilon^2+x^2}\arctan(x/\varepsilon) \end{align}$$

Next, we will find the distributional limits of $\psi^{(1)}_\varepsilon$, $\psi^{(2)}_\varepsilon$, and $\psi^{(3)}_\varepsilon$.

DISTRIBUTIONAL LIMITS OF $\displaystyle \psi^{(1)}_\varepsilon$, $\displaystyle \psi^{(2)}_\varepsilon$, and $\displaystyle \psi^{(3)}_\varepsilon$

Again, let $\phi\in \mathbb{S}$. Then,

$$\begin{align} \lim_{\varepsilon\to 0^+}\langle \psi^{(1)}_\varepsilon,\phi \rangle &=\lim_{\varepsilon\to 0^+}\int_{-\infty}^\infty \psi^{(1)}_\varepsilon(x)\phi(x)\,dx\\\\ &=\lim_{\varepsilon\to 0^+}\int_{-\infty}^\infty \left(-\frac{2\varepsilon}{\varepsilon^2+x^2}\gamma \right)\phi(x)\,dx\\\\ &=-2\gamma\lim_{\varepsilon\to 0^+}\int_{-\infty}^\infty \frac{\phi(\varepsilon x)}{x^2+1}\,dx\\\\ &=-2\pi \gamma \phi(0)\tag3 \end{align}$$

$$\begin{align} \langle \psi^{(2)}_\varepsilon,\phi \rangle &=\int_{-\infty}^\infty \left(-\frac{\varepsilon}{\varepsilon^2+x^2}\log(\varepsilon^2+x^2) \right)\phi(x)\,dx\\\\ &=-2\log(\varepsilon)\int_{-\infty}^\infty \frac{\phi(\varepsilon x)}{x^2+1}\,dx-\int_{-\infty}^\infty \frac{\log(1+x^2)}{1+x^2}\phi(\varepsilon x)\,dx\\\\ &= -2\pi \log(\varepsilon)\phi(0)-2\pi \log(2) \phi(0)+o(\varepsilon) \end{align}\tag4$$

$$\begin{align} \langle \psi^{(3)}_\varepsilon,\phi \rangle &=\int_{-\infty}^\infty \left(-\frac{2x}{\varepsilon^2+x^2}\arctan(x/\varepsilon)\right)\phi(x)\,dx\\\\ &-\int_{|x|\le 1}\frac{2x}{\varepsilon^2+x^2}\arctan(x/\varepsilon) \phi(x)\,dx-\int_{|x|\ge 1}\frac{2x}{\varepsilon^2+x^2}\arctan(x/\varepsilon) \phi(x)\,dx\\\\ &=-\phi(0)\int_{|x|\le 1}\frac{2x}{\varepsilon^2+x^2}\arctan(x/\varepsilon) \,dx\\\\ &-\int_{|x|\le 1}\frac{2x}{\varepsilon^2+x^2}\arctan(x/\varepsilon) (\phi(x)-\phi(0))\,dx-\int_{|x|\ge 1}\frac{2x}{\varepsilon^2+x^2}\arctan(x/\varepsilon) \phi(x)\,dx\\\\ &= \left(2\pi \log(\varepsilon) +2\pi \log(2)\right)\phi(0)+o(\varepsilon)\\\\ &-\pi \int_{|x|\le 1}\frac{\phi(x)-\phi(0)}{|x|}\,dx-\pi \int_{|x|\ge 1}\frac{\phi(x)}{|x|}\,dx\tag5 \end{align}$$

FINAL RESULTS

Substituting $(3)$, $(4)$, and $(5)$ into $(2)$, we find that

$$\begin{align} \lim_{\varepsilon\to 0^+}\langle \mathscr{F}\{\psi_\varepsilon\},\phi\rangle =-2\pi \gamma \phi(0)-\pi \int_{|x|\le 1}\frac{\phi(x)-\phi(0)}{|x|}\,dx-\pi\int_{|x|\ge 1}\frac{\phi(x)}{|x|}\,dx\\\\ \end{align}$$

from which we assert that in distribution

$$\bbox[5px,border:2px solid #C0A000] {\mathscr{F}\{\psi\}(x)=-2\pi \gamma \delta(x)-\pi \left(\frac1{|x|}\right)_1}$$

where we interpret $\left(\frac1{|x|}\right)_1$ to mean that for any $\phi\in \mathbb{S}$,

$$\int_{-\infty}^\infty \left(\frac1{|x|}\right)_1 \phi(x)\,dx=\int_{|x|\le 1}\frac{\phi(x)-\phi(0)}{|x|}\,dx+ \int_{|x|\ge 1}\frac{\phi(x)}{|x|}\,dx$$

NOTE:

It was arbitrary to split the integration in $(5)$ into inervals $|x|\le 1$ and $|x|\ge 1$. Had we chosen instead the intervals $|x|\le \nu$ and $|x|\ge \nu$ for any $\nu>0$, we would have obtained

$$\bbox[5px,border:2px solid #C0A000] {\mathscr{F}\{\psi\}(x)=-2\pi (\gamma+\log(\nu)) \delta(x)-\pi \left(\frac1{|x|}\right)_\nu}$$

where we interpret $\left(\frac1{|x|}\right)_\nu$ to mean that for any $\phi\in \mathbb{S}$,

$$\int_{-\infty}^\infty \left(\frac1{|x|}\right)_\nu \phi(x)\,dx=\int_{|x|\le \nu}\frac{\phi(x)-\phi(0)}{|x|}\,dx+ \int_{|x|\ge \nu}\frac{\phi(x)}{|x|}\,dx$$