Can anyone provide or give an expression in the sense of distribution theory for the functions $|x|^{s} , \log|x| $? I mean I would like to evaluate the Fourier transform $ \int_{-\infty}^{\infty}f(x)\exp(-iux) $ of these transforms in case it is possible.

  • 1
    $\begingroup$ Do you want the Fourier transform of $|x|^s$ or $|x|^s\log x$? (the first is in the body, the second in the title) $\endgroup$ Commented Apr 28, 2012 at 10:17
  • 1
    $\begingroup$ i am looking for the fourier transform of all $ |x|^{s} 4 and $log|x| $ although by differntiation with respect to 'x' i supsect they are all related. $\endgroup$ Commented Apr 28, 2012 at 10:50
  • $\begingroup$ In higher dimension, you might want to look here math.stackexchange.com/questions/3723136/… $\endgroup$
    – LL 3.14
    Commented Apr 27, 2021 at 12:45
  • $\begingroup$ @JoseGarcia Hi Jose. I added an answer that handles evaluates the Fourier Transform of $|x|^\alpha$ for all real values of $\alpha$. This actually was a lot more work than I had suspected initially. Please let me know how I can improve my answer. I really want to give you the best answer I can. $\endgroup$
    – Mark Viola
    Commented May 21, 2021 at 17:49
  • $\begingroup$ Related to Calculate the Fourier transform of $\log |x| $. $\endgroup$
    – robjohn
    Commented Aug 7, 2021 at 18:22

5 Answers 5


Concerning functions in question are not integrable on the line, the Fourier transform has to be considered in the sense of distributions. Particularly for the logarithm, it is known that (Vladimirov, Equations of Mathematical Physics, $\S2.5$) $$ F\left[{\cal P}\frac1{|x|}\right]=-2\gamma-2\log|\xi|, $$ where $\gamma$ is the Euler constant and ${\cal P}\frac1{|x|}$ is a distribution defined by $$ ({\cal P}\frac1{|x|},\varphi)= \int_{|x|\le 1}\frac{\varphi(x)-\varphi(0)}{|x|}\,dx+ \int_{|x|> 1}\frac{\varphi(x)}{|x|}\,dx. $$ With inverse FT one can get from here the FT of $\log|x|$: $$ F[\log|x|](\xi)=-2\pi\gamma\delta(\xi)-\frac\pi{|\xi|}, $$ taking into account that FT is defined in this book as $$ F[f](\xi)=\int_{\mathbb R}f(x)e^{ix\xi}\,dx. $$



Let $\psi_\alpha$ be the function $\psi_\alpha(x)=|x|^\alpha$. For $-1<\alpha<0$ the Fourier Transform is given by

$$\mathscr{F}\{\psi_\alpha\}(k)=\int_{-\infty}^\infty \psi_\alpha(x)e^{ikx}\,dx\tag1$$

For $\alpha \in [0,\infty)$, $\psi_\alpha(x)$ is locally integrable and yields a tempered distribution $\left(\psi_\alpha\right)_{D_1}=\left(|x|^\alpha\right)_{D_1}$ such that for any $\phi \in \mathbb{S}$ (i.e., $\phi$ is a Schwarz Space function)

$$\begin{align} \langle \left(\psi_\alpha\right)_{D_1},\phi\rangle&= \int_{-\infty}^\infty |x|^\alpha \phi(x)\,dx\tag2 \end{align}$$

However, for $\alpha\le -1$, $\psi_\alpha$ is not locally integrable and is, therefore, not a tempered distribution. We can define, however, a distribution that permits our defining the Fourier Transform.

Let $n\in \mathbb{N}_+$. For $\alpha\in (-(n+1),-n)$, we define the distribution $\left(\psi_\alpha\right)_{D_2}=\left(\frac1{|x|^{|\alpha|}}\right)_{D_2}$ such that for any $\phi\in \mathbb{S}$

$$\langle \left(\psi_\alpha\right)_{D_2}, \phi\rangle = \int_{-\infty}^\infty\frac{\phi(x)-\sum_{m=0}^{n-1} \frac{\phi^{m}(0)}{m!}x^m}{|x|^{|\alpha|}}\,dx\tag3$$

Equipped with $(1)-(3)$, we proceed to determine the Fourier Transform of $\psi_\alpha(x)=|x|^\alpha$ for $\alpha\in (-1,0)$, $\left(\psi_\alpha\right)_{D_1}$ for $\alpha\ge 0$, and $\left(\psi_\alpha\right)_{D_2}$ for $\alpha\le -1$.

CASE $1$: ($-1<\alpha <0)$

For $-1<\alpha<0$, $\psi_\alpha(x)=|x|^\alpha$ and its Fourier Transform can be computed directly from $(1)$ as

$$\begin{align} \mathscr{F}\{\psi_\alpha\}(k)&=\int_{-\infty}^\infty |x|^\alpha e^{ikx}\,dx\\\\ &=2\text{Re}\left(\int_0^\infty x^\alpha e^{i|k|x}\,dx\right)\\\\ &=\frac{2}{|k|^{1-|\alpha|}}\text{Re}\left(\int_0^\infty x^\alpha e^{ix}\,dx\right)\\\\ &=\frac{2\sin(\pi |\alpha|/2)\Gamma(1-|\alpha|)}{|k|^{1-|\alpha|}} \end{align}$$


We evaluated the integral $\int_0^\infty x^\alpha e^{ix}\,dx$ applying Cauchy's Integral Theorem to deform the integration from the real axis to the the imaginary axis. Proceeding, we find that

$$\begin{align} \int_0^\infty x^{\alpha}e^{ix}\,dx=e^{i\pi(\alpha+1)/2}\Gamma(1-|\alpha|) \end{align}$$

Therefore, the Fourier transform of $\psi_\alpha(x)=|x|^\alpha$, $\alpha\in (-1,0)$ is

$$\bbox[5px,border:2px solid #C0A000]{\mathscr{F}\{\psi_\alpha \}(k)=\frac{2\sin(\pi |\alpha|/2)\Gamma(1-|\alpha|)}{|k|^{1-|\alpha|}}}\tag4$$

CASE $2$: ($\alpha \in \mathbb{R}_{+}\setminus \mathbb{N}_{+})$

Let $n\in \mathbb{N}_{+}$. For $\alpha\in (n,n+1)$ and $\phi\in \mathbb{S}$ we have

$$\begin{align} \langle \mathscr{F}\{\left(\psi\right)_{D_1}\},\phi\rangle &=\langle \left(\psi\right)_{D_1},\mathscr{F}\{\phi\}\rangle \\\\ &=2\text{Re}\left(\int_0^\infty x^\alpha \int_{-\infty}^\infty \phi(k)e^{ikx}\,dk\,dx\right)\tag5 \end{align}$$

Integrating by parts $n+1$ times the inner integral in $(5)$ reveals

$$\begin{align} \langle \mathscr{F}\{\left(\psi\right)_{D_1}\},\phi\rangle &=2\text{Re}\left(e^{i(n+1)\pi/2}\int_0^\infty x^{\alpha-n-1} \int_{-\infty}^\infty \phi^{(n+1)}(k)e^{ikx}\,dk\,dx\right)\\\\ &=-2\text{Im}\left(e^{in\pi/2} \int_{-\infty}^\infty \phi^{(n+1)}(k)\int_0^\infty x^{\alpha-n-1}e^{ikx}\,dx\,dk\right)\tag6 \end{align}$$


To justify the interchange of integrals that led to $(6)$, we used the fact that for $\beta\in (0,1)$, there exists a number $C_\beta >0$ such that for any $L>0$, $\left|\int_0^L \frac{e^{it}}{t^\beta}\,dt\right|<C_\beta$. Then, we applied Fubini's Theorem and finished by appealing to the Dominated Convergence Theorem.

We can evaluate the inner integral in $(6)$ by enforcing the substitution $x\mapsto x/k$ and then applying Cauchy's Integral Theorem to deform the integration from the real axis to the the imaginary axis. Proceeding, we find that

$$\begin{align} \int_0^\infty x^{\alpha-n-1}e^{ikx}\,dx=e^{i\text{sgn}(k)\pi(\alpha-n)/2}\frac{\Gamma(\alpha - n)}{|k|^{\alpha -n}}\tag7 \end{align}$$

Using $(7)$ in $(6)$, we obtain

$$\begin{align} \langle \mathscr{F}\{\left(\psi\right)_{D_1}\},\phi\rangle &=-2\Gamma(\alpha-n)\text{Im}\left(e^{in\pi/2} \int_{-\infty}^\infty \frac{\phi^{(n+1)}(k)}{|k|^{\alpha-n}}e^{i\text{sgn}(k)\pi(\alpha-n)/2}\,dk\right)\\\\ &=-2\sin(\pi \alpha/2)\Gamma(\alpha-n)\int_{-\infty}^\infty \frac{\phi^{(n+1)}(k)}{|k|^{\alpha-n}}\left(\text{sgn}(k)\right)^{n+1}\,dk\tag8 \end{align}$$

Integrating by parts $n+1$ times the integral on the right-hand side of $(8)$ results in the following

$$\begin{align} \langle \mathscr{F}\{\left(\psi\right)_{D_1}\},\phi\rangle&=-2\sin(\pi \alpha/2)\Gamma(\alpha+1)\int_{-\infty}^\infty \frac{\phi(k)-\sum_{m=0}^{n}\frac{\phi^{(m)}(0)}{m!}k^m}{|k|^{\alpha+1}}\,dk\tag9 \end{align}$$

from which we infer that the Fourier Transform of $\psi=|x|^\alpha$ for $\alpha \in \mathbb{R}_{>0}\setminus \mathbb{N}_{>0}$ is given by

$$\begin{align} \bbox[5px,border:2px solid #C0A000] {\mathscr{F}\{\left(\psi\right)_{D_1}\}(k)=-2\sin(\pi \alpha/2)\Gamma(\alpha+1)\left(\frac1{|k|^{\alpha+1}}\right)_{D_2}}\\\\ \tag{10} \end{align}$$

In the Appendix, we provide a development to show that for $\alpha=n$, the Fourier Transform of $\left(|x|^n\right)_{D_1}$ is given by

$$\begin{align} \bbox[5px,border:2px solid #C0A000] {\mathscr{F}\{\left(\psi\right)_{D_1}\}(k)=-2\sin(\pi n/2)\Gamma(n+1)\text{PV}\left(\frac1{|k|^{n+1}}\right)_{D_2}+2\pi \cos(n\pi/2)\delta^{(n)}(k)}\\\\ \tag{11} \end{align}$$

where $\text{PV}$ denotes the Cauchy Principal value, which applies in $(11)$ for odd values of $n$, and is given by

$$\begin{align} \text{PV}\left(\frac1{|k|^{n+1}}\right)_{D_2}&=\lim_{\delta\to 0^+}\left(\int_{|x|\ge \delta} \frac{\phi(k)-\sum_{m=0}^{n}\frac{\phi^{(m)}(0)}{m!}k^m }{k^{n+1}}\,dk\right)\\\\ &=\lim_{\delta\to 0^+}\left(\int_{|x|\ge \delta} \frac{\phi(k)-\sum_{m=0}^{n-1}\frac{\phi^{(m)}(0)}{m!}k^m }{k^{n+1}}\,dk\right)\tag{12} \end{align}$$

CASE $3$: ($\alpha<-1, \alpha \notin \mathbb{Z}_{<0})$

We appeal to the result in $(10)$ and the Fourier Inversion Theorem to immediately arrive at the Fourier Transform of $\left(\psi_\alpha\right)_{D_2}=\left(\frac1{|x|^{|\alpha|}}\right)_{D_2}$ for $\alpha\in (-(n+1),-n)$, $n>1$

$$\begin{align} \bbox[5px,border:2px solid #C0A000] {\mathscr{F}\{\left(\psi\right)_{D_2}\}(k)=2\sin(\pi |\alpha|/2)\Gamma(1-|\alpha|)\left(|k|^{|\alpha|-1}\right)_{D_1}}\tag{13} \end{align}$$

We can extend the result in $(13)$ to include $\alpha=-n$ for even values of $n$. Thus, for $\alpha=-n$, $n$ even we have

$$\begin{align} \bbox[5px,border:2px solid #C0A000] {\mathscr{F}\{\left(\psi_n\right)_{D_2}\}(k)=\frac{\pi \cos(n\pi/2)}{\Gamma(|n|)}\left(|k|^{|n|-1}\right)_{D_1}}\tag{14} \end{align}$$

SPECIAL CASE $4$: ($\alpha=-n)$, $n$ odd

Finally, we redefine the distribution in $(1)$ for the case in which $\alpha =-n$, $n$ odd and for $\phi\in \mathbb{S}$ to write

$$\begin{align} \langle \mathscr{F}\{\left(\psi_{n}\right)_{D_2}\},\phi\rangle &=\langle \left(\psi_{n}\right)_{D_2},\mathscr{F}\{\phi\}\rangle \\\\ &=2\text{Re}\int_0^\infty \int_{-\infty}^\infty \phi(k) \frac{ e^{ikx}-\sum_{m=0}^{|n|-2}\frac{(ikx)^m}{m!}-\frac{(ikx)^{(|n|-1} \xi_{[0,1]}(x)}{(|n|-1)!}}{x^{|n|}}\,dk\,dx\\\\ &=\frac2{(|n|-1)!}\text{Re} \int_0^\infty \frac1x \int_{-\infty}^\infty \phi(k)(ik)^{|n|-1}(e^{ikx}-\xi_{[0,1]}(x))\,dk\,dx\\\\ &+\frac{2\sin(|n|\pi/2)H_{|n|-1}}{(|n|-1)!}k^{|n|-1}\\\\ &=\frac{2\sin(|n|\pi/2)}{(|n|-1)!}\int_{-\infty}^\infty k^{|n|-1}\phi(k)\int_0^\infty \frac{\cos(|k|x)-1\xi_{[0,1]}(x)}{x}\,dx\,dk\\\\ &+\frac{2\sin(|n|\pi/2)H_{|n|-1}}{(|n|-1)!}k^{|n|-1}\tag{15} \end{align}$$

In the question posted HERE, I showed using both real analysis and complex analysis that the inner integral on the right-hand side of $(15)$ is given by

$$ \int_0^\infty \frac{\cos(|k|x)-1\xi_{[0,1]}(x)}{x}\,dx=-\gamma-\log(|k|)\tag{16}$$

Using $(16)$ in $(15)$ reveals

$$\begin{align} \langle \mathscr{F}\{\left(\psi_n\right)_{D_2}\},\phi\rangle &=\frac{2\sin(|n|\pi/2)}{(|n|-1)!}\int_{-\infty}^\infty k^{|n|-1}(-\gamma-\log(|k|)+H_{|n|-1})\phi(k)\,dk\tag{17} \end{align}$$

from which we deduce that in distribution that the Fourier transform of $\left(\psi_n(x)\right)_{D_2}=\left(\frac1{|x|^{|n|}}\right)_{D_2}$, $n<0$, $n$ odd is

$$\begin{align} \bbox[5px,border:2px solid #C0A000] {\mathscr{F}\{\left(\psi_{n}\right)_{D_2}\}(k)=-2\frac{\sin(|n|\pi/2)}{\Gamma(|n|)}\,|k|^{|n|-1}(\gamma+\log(|k|)-H_{|n|-1})}\tag{18} \end{align}$$

APPENDIX: Direct Development for the Case $(\alpha\in \mathbb{N}_0)$

Let $n\in \mathbb{N}_{\>0}$. For $\alpha=n$, $\psi(x)=|x|^n$, and $\phi\in \mathbb{S}$ we have

$$\begin{align} \langle \mathscr{F}\{\psi\},\phi\rangle &=\langle \psi,\mathscr{F}\{\phi\}\rangle \\\\ &=2\text{Re}\left(\int_0^\infty x^n \int_{-\infty}^\infty \phi(k)e^{ikx}\,dk\,dx\right)\tag{A1} \end{align}$$

Integrating by parts $n$ times the inner integral in $(A1)$ reveals

$$\begin{align} \langle \mathscr{F}\{\psi\},\phi\rangle &=2\text{Re}\left(e^{in\pi/2}\int_0^\infty \int_{-\infty}^\infty \phi^{(n)}(k)e^{ikx}\,dk\,dx\right)\\\\ &=\cos(n\pi/2)\int_{-\infty}^\infty \int_{-\infty}^\infty \phi^{(n)}(k)e^{ikx}\,dk\,dx\\\\&-2\sin(n\pi/2)\int_0^\infty \int_{-\infty}^\infty \phi^{(n)}(k)\sin(kx)\,dk\,dx\tag{A2} \end{align}$$

From the Fourier transform inversion theorem, the first integral on the right-hand side of $(A2)$ is $2\pi \phi^{(n)}(0)$. For the second integral on the right-hand side of $(A2)$, we write

$$\begin{align} \int_0^\infty \int_{-\infty}^\infty \phi^{(n)}(k)\sin(kx)\,dk\,dx&=\lim_{L\to \infty}\int_0^L \int_{-\infty}^\infty \phi^{(n)}(k)\sin(kx)\,dk\,dx\\\\ &= \lim_{L\to\infty }\int_{-\infty}^\infty \phi^{(n)}(k)\frac{1-\cos(kL)}{k}\,dk\\\\ &=\lim_{\delta\to 0^+}\int_{|k|\ge \delta}\frac{\phi^{(n)}(k)}{k}\,dk\tag{A3} \end{align}$$

In arriving at $(A3)$ we made use of the fact that $\int_{|k|\le \delta} \phi^{(n)}(k)\frac{1-\cos(kL)}{k}\,dk=O(\delta)$ uniformly and applied the Riemann Lebesgue Lemma.

Now, integrating by parts $n$ times the integral on the right-hand side of $(A3)$, we see that

$$\int_0^\infty \int_{-\infty}^\infty \phi^{(n)}(k)\sin(kx)\,dk\,dx=n!\lim_{\delta \to 0^+}\int_{|k|\ge \delta}\frac{\phi(k)-\sum_{m=0}^{n-1}\frac{\phi^{(m)}(0)k^{m}}{m!}}{k^{n+1}}\,dk \tag{A4}$$

Using $(A4)$ in $(A2)$ we obtain

$$\begin{align} \langle \mathscr{F}\{\psi\},\phi\rangle &=\cos(n\pi/2)2\pi \phi^{(n)}(0)\\\\&-2\sin(n\pi/2)n!\lim_{\delta \to 0^+}\int_{|k|\ge \delta}\frac{\phi(k)-\sum_{m=0}^{n-1}\frac{\phi^{(m)}(0)k^{m}}{m!}}{k^{n+1}}\,dk\tag{A5} \end{align}$$

from which we deduce that in distribution that the Fourier transform of $\psi(x)=|x|^n$, $n\in \mathbb{N}_0$ is

$$\bbox[5px,border:2px solid #C0A000]{\mathscr{F}\{\psi \}(k)=\cos(n\pi/2)2\pi \delta^{(n)}(k)-2\sin(n\pi /2)\Gamma(n+1)\left(\frac{1}{k^{n+1}}\right)_{D_2}}\\\\\tag{A6}$$

where the the distribution $\left(\frac{1}{k^{n+1}}\right)_{D_2}$ in $(A6)$ is given by $(12)$.

  • $\begingroup$ Just to let you know, editing a deleted answer will still bump the question to the main page. If you need to workshop your answer, please use the sandbox. $\endgroup$
    – Asaf Karagila
    Commented May 7, 2021 at 21:25
  • $\begingroup$ @AsafKaragila I am unfamiliar with the sandbox and how to use it. But I have finally completed this rather lengthy post. Your feedback would be appreciated. $\endgroup$
    – Mark Viola
    Commented May 8, 2021 at 2:25
  • $\begingroup$ It's easy, Mark. You take an answer that is 'free for use', you use it to make all the edits, and when you're done, you copy/paste the answer to the intended page, and restore the answer to its 'free for use' state. $\endgroup$
    – Asaf Karagila
    Commented May 8, 2021 at 6:20
  • $\begingroup$ Go to that link. Find an answer that says "free for anyone to use." Click edit. Write your answer. Workshop your answer. Edit some more. Let it sit for two days. Get back to it. Fix all the typos you find, and all the small mistakes that crept in before. Rinse and repeat as necessary. When you're done, click edit, copy the whole content, come to the intended question (e.g. this one), click the "answer" button in the bottom, paste your copied answer (you might want to manually type one line to avoid the CAPTCHA script). Submit your answer. Go back to the sandbox. Rollback the box you used. Done $\endgroup$
    – Asaf Karagila
    Commented May 8, 2021 at 6:27
  • $\begingroup$ If you define $|x|^{-n}$ as $$(|x|^{-n}, \phi) = \int_{\mathbb R} |x|^{-n} \left( \phi(x) - \xi_{[-1, 1]}(x) \sum_{m = 0}^{n - 1} {\phi^{(m)}(0)} \frac {x^m} {m!} \right) dx$$ for $n \in \mathbb N$, as the formula for $\lambda_1$ seems to imply, then it's not true that $\mathcal F[|x|^{-2}](k) = -\pi |k|$ and not true that $\mathcal F[|x|^{-3}](k) = -k^2 (\gamma + \ln |k|)$. $\endgroup$
    – Maxim
    Commented Jul 13, 2021 at 17:22

I thought that it might be instructive to present an approach to deriving the Fourier transform of $\log(|x|)$ that is different from the regularization approach 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.


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

$$\begin{align} \langle \mathscr{F}\{\psi\}, \phi\rangle &=\langle \psi, \mathscr{F}\{\phi\}\rangle\\\\ &=\int_{-\infty}^\infty \log(|x|)\int_{-\infty}^\infty \phi(k)e^{ikx}\,dk\,dx\\\\ &=2\int_0^\infty \log(x)\int_{-\infty}^\infty \phi(k)\cos(kx)\,dk\,dx\\\\ &=4\phi(0)\int_0^\infty \frac{\sin(x)}{x}\,\log(x)\,dx\\\\ &+2\int_0^\infty \log(x) \left(\int_{-\infty}^\infty (\phi(k)-\phi(0)\xi_{[-1,1]}(k))\cos(kx)\,dk\right)\,dx\\\\ &=-2\pi \gamma \phi(0)\tag2\\\\ &+2\lim_{L\to \infty } \int_{-\infty}^\infty \frac{\phi(k)-\phi(0)\xi_{[-1,1]}(k)}k\left(\log(L)\sin(kL)-\int_0^L \frac{\sin(kx)}{x}\,dx\right)\,dk\\\\ &=-2\pi \gamma \phi(0)-\pi\int_{-\infty}^\infty \frac{\phi(k)-\phi(0)\xi_{[-1,1]}(k)}{|k|}\,dk\tag3 \end{align}$$


In arriving at $(2)$, we used the result I posted in THIS ANSWER and THIS ONE, and we used Fubini's theorem to justify interchanging integral $\int_0^L \,dx$ with the integral $\int_{-\infty}^\infty \,dk$.

In arriving at $(3)$, we used integration by parts to show that the limit of the term involving $\log(L)$ is $0$ and we appealed the the Dominated Convergence Theorem to justify interchanging the limit with the integration over $k$ for the second term on the right-hand side of $(3)$.

From $(3)$, we deduce the Fourier Transform of $\psi$ in distribution

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

where the distribution $\left(\frac1{|x|}\right)_1$ is defined by its action on and $\phi\in \mathbb{S}$ as

$$\int_{-\infty}^\infty \left(\frac1{|x|}\right)_1\phi(x)\,dx=\int_{-\infty}^\infty \frac{\phi(x)-\phi(0)\xi_{[-1,1]}(x)}{|x|}\,dx$$


It was arbitrary to split the integration in $(2)$ into intervals $|x|\le 1$ 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 $\phi\in \mathbb{S}$,

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

  • $\begingroup$ I don't disagree with the content of your answer, but there is significant overlap with this answer on MO, this answer on MSE and this answer on MSE, maybe adding a link to the best one would suffice $\endgroup$ Commented Apr 23, 2021 at 5:14
  • $\begingroup$ @CalvinKhor This methodology herein is completely different from the methodology used in both of the first two references you cited. And the problem herein (i.e. Fourier Transform of $\log(|x|)$) is different from that in the third reference (Fourier transform of $1/x$). And I thought it would be instructive to present a solution that is different from the solutions that both others and I have posted elsewhere. $\endgroup$
    – Mark Viola
    Commented Apr 23, 2021 at 13:49
  • $\begingroup$ This is a considerably non-standard notion of "principal value integral"... ?!? $\endgroup$ Commented May 5, 2021 at 22:24
  • $\begingroup$ @paulgarrett Hi Paul. The term "principal value" might be a poor choice. The distributional interpretation of $\frac1{|x|}$ is discussed in Exercise 13 of THIS NOTE written by Terence Tao. Perhaps I should drop the $\text{PV}$ and just write $\frac1{|x|}$ with the explanation that it is to be interpreted in distribution as described herein. I've edited accordingly, but feel free to suggest another notation. $\endgroup$
    – Mark Viola
    Commented May 6, 2021 at 0:46
  • $\begingroup$ @paulgarrett I haven't received your reply regarding your thoughts on a better notation for the regularization of the distribution $\frac1{|x|}$. In the meantime, I have posted a detailed solution that discusses the Fourier Transform of $|x|^\alpha$ for all real values of $\alpha$. Any feedback you have would be most appreciated. $\endgroup$
    – Mark Viola
    Commented May 8, 2021 at 2:28

Based on parity and homogeneity (which are treated well by Fourier transform), the (tempered distribution) Fourier transform of (the meromorphically continued family of tempered distributions) $1/|x|^s$ on $\mathbb R^n$ is a constant multiple of $1/|x|^{n-s}$. The constant can be determined by application to the Gaussian $e^{-\pi |x|^2}$, which is its own Fourier transform (in some normalization).

Yes, in $\mathbb R^1$, $\log |x|$ is the derivative with respect to $s$ of $|x|^s$, then evaluated at $s=0$. If we believe that this differentiation commutes with the other operations in play, then we obtain the Fourier transform of $\log |x|$.

For my tastes, the argument using meromorphic continuation of distributions with homogeneity and parity properties is more persuasive than truncation argument. Tastes vary.

EDIT: since the question got bumped-up anyway, perhaps I might as well include (what I think is) a useful alternative to other answers. Namely, to directly describe the Fourier transform of $\log|x|$ on $\mathbb R^1$, while dodging some of the complications of the more general $|x|^s\cdot \log|x|$, we can proceed as follows. First, the derivative of $\log|x|$ is $1/x$, pointwise. In fact, distributionally, it is a small exercise to show that derivative is the principal value $PV {1\over x}$.

It is fairly standard (based on homogeneity and parity) that the Fourier transform of $PV{1\over x}$ is $-\pi i\, \mathrm{sgn}(x)$.

Because of the way Fourier transform and derivative interact, we have $x\cdot \widehat{\log|x|}=-{1\over 2}\,\mathrm{sgn}(x)$.

If we imagine that we can divide by $x$, then $\widehat{\log|x|}=-{1\over 2}\,{1\over |x|}$. However, integration-against-$1/|x|$ does not extend to a homogeneous, even distribution on test functions. (Without the requirement of homogeneity, sure, Hahn-Banach gives lots of extensions...)

But it is true that $x\cdot \widehat{\log|x|}=-{1\over 2}\mathrm{sgn}(x)$. This implies that we can evaluate that Fourier transform by integration-against $f$ for Schwartz functions $f$ vanishing at $0$. In particular, and since we do already know that this Fourier transform exists, for general Schwartz functions $f$,

$$ \widehat{\log|x|}(f) \;=\; \widehat{\log|x|}(f-f(0)\cdot e^{-\pi x^2}) + \widehat{\log|x|} (e^{-\pi x^2}) $$ $$ \;=\; x\cdot \widehat{\log|x|}\Big({f-f(0)\cdot e^{-\pi x^2}\over x}\Big) + \widehat{\log|x|}(e^{-\pi x^2}) $$ $$ \;=\; -{1\over 2}\int_{\mathbb R} \mathrm{sgn}(x)\Big({f-f(0)\cdot e^{-\pi x^2}\over x}\Big)\;dx + \delta(f)\cdot\widehat{\log|x|}(e^{-\pi x^2}) $$ $$ \;=\; -{1\over 2}\int_{\mathbb R} {f-f(0)\cdot e^{-\pi x^2}\over |x|}\;dx + \delta(f)\cdot \widehat{\log|x|}(e^{-\pi x^2}) $$ The value of $\widehat{\log|x|}$ on the Gaussian can be evaluated directly, in some gruesome detail as indicated below.

Following through, perhaps the constants are correct in the claim that, for all Schwartz functions $f$, $$ \widehat{\log|x|}(f) \;=\; -{1\over 2}\int_{\mathbb R} {f(x)-f(0)\cdot e^{-\pi x^2}\over |x|}\;dx \;+\; \delta(f) \cdot \Big({-\log\pi\over 2} + {\Gamma'({1\over 2})\over 2\sqrt{\pi}}\Big) $$

(Edit: corrected some dropped constants, etc.)

  • $\begingroup$ Hi Paul. I hope that you are doing well. Where did the multiplicative factor $-1/2$ come from in the final line? It certainly does not follow from the previous line. You also are missing multiplication by $f(0)$ on the term $\widehat{\log|x|} (e^{-\pi x^2})$. Finally, I've posted a solution on this page that derives the Fourier Transform of $|x|^\alpha$ for all real $\alpha$. I'd appreciate a second pair of eyes on it if you're interested and have a moment. ;-). $\endgroup$
    – Mark Viola
    Commented Aug 6, 2021 at 17:20
  • $\begingroup$ @MarkViola, ah, thanks for observing typos! :) $\endgroup$ Commented Aug 6, 2021 at 20:24
  • $\begingroup$ You're welcome Paul. Are you using the FT with a $\frac1{2\pi}$ scaling factor? $\endgroup$
    – Mark Viola
    Commented Aug 6, 2021 at 20:38
  • $\begingroup$ @MarkViola, my intention, though I may of course fail, is to put the $2\pi$ in the exponent, and have no adjustment of measure... $\endgroup$ Commented Aug 6, 2021 at 20:39

I'm not a mathematician, but I will give what I think is a relatively easy approach. The main point is the origin of the $\delta$-function. I leave it to the mathematicians to fill in the details -- or point out that I am wrong.

To begin, I use the result of H. Bateman, Table of Integral Transforms, (A. Erdelyi editor), (McGraw-Hill, NY, 1954), vol. I, p. 76, sec. 2.5, formula 3, $${\cal F}_s[(\ln|t|)/t](x) = -\frac{\pi}{2}\gamma - \frac{\pi}{2}\ln x $$ with $x>0$ where the Fourier sine transform is defined as $${\cal F}_s[f(t)](x) = \int_{-\infty}^{\infty}\, dx \; f(t)\sin xt . $$ For what is below, define the Fourier cosine transform as $${\cal F}_c[f(t)](x) = \int_{-\infty}^{\infty}\, dx \; f(t)\cos xt $$ and the Fourier transform as $${\cal F}[f(t)](x) = \int_{-\infty}^{\infty}\, dx \; f(t)e^{-ixt} . $$

The Fourier sine transform of $(\ln|t|)/t$ is related to the integral for ${\cal F}_s[(\ln|t|)/t](1)$ in sec. 539 of G.M. Fikhtengol'ts, Course in differential and integral calculus, (in Russian), vol. II, p. 784, $$ {\cal F}_s[(\ln|t|)/t](1) = -\frac{\pi}{2}\gamma . $$ One then uses the techniques discussed in Fikhtengol'ts to obtain ${\cal F}_s[(\ln|t|)/t](1)$ and from there ${\cal F}_s[(\ln|t|)/t](x)$

Note the we can write ${\cal F}[\ln |t|](x)$ in terms of ${\cal F}_c[\ln |t|](x)$: $$ {\cal F}[\ln |t|](x) = \int_{-\infty}^{\infty}\, \ln |t| e^{-ixt} \\ = 2 \int_{0}^{\infty}\, \ln |t| \cos xt \\ = 2{\cal F}_c[\ln |t|](x) . $$ We can extend ${\cal F}_c[\ln |t|](x)$ to all real $x$ noting that the integral expression for ${\cal F}_c[\ln |t|](x)$ is manifestly an even function of $x$.

Now, as noted, $$ {\cal F}_s[(\ln |t|)/t](x) $$ is known for $x>0$. But, the integral expression for ${\cal F}_s[(\ln |t|)/t](x)$ is an odd function in $x$.
Hence, for all real $x$ $$ {\cal F}_s[(\ln |t|)/t](x) = -\frac{\pi}{2} \, sgn\, x ( \ln |x| +\gamma ). $$ We next note, $$ 2 \frac{d}{dx} {\cal F}_s[(\ln |t|)/t](x) = 2 {\cal F}_c[\ln |t|](x) ={\cal F}[\ln |t|](x). $$ But from the previous expression for ${\cal F}_s[(\ln |t|)/t](x)$, we have $$ 2 \frac{d}{dx} {\cal F}_s[(\ln |t|)/t](x) = -\pi\frac{d}{dx} \, sgn \, x \,\ln |x| -\pi \gamma \frac{d}{dx}\, sgn\, x \\ -\pi\frac{1}{|x|} -2\pi\gamma\delta(x) = {\cal F}[\ln |t|](x). $$

Is that correct?


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