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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 $C$?

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  • $\begingroup$ If I remember correctly this computation is explained very well in Folland's book on real analysis, 2nd edition, the chapter on distributions. He also makes an interesting remark on the fact that the term $C\delta$ is a "normalization" one, similar to the normalizations physicists make in QFT (where they essentially subtract infinities). If all else fails, check Tao's blog: terrytao.wordpress.com/2009/04/19/245c-notes-3-distributions/… (Here he computes the Fourier transform of $|x|^{-1}$ which is the derivative of $\log |x|$ ) $\endgroup$ – Giuseppe Negro Jun 29 '17 at 0:34
  • $\begingroup$ @GiuseppeNegro In Folland, I could find the remark on quantum physics, but I cannot find the computation. Can you give me some pointers? I think that the transform of $|x|^{-s}$ is given as exercise in Tao's blog. $\endgroup$ – Dal Jun 29 '17 at 7:39
  • $\begingroup$ I am sorry, I was remembering incorrectly, the remark only concerns the definition of $|x|^{-1}$ and not its Fourier transform. See however Exercise 19, where Folland computes $\widehat{|x|^{-1}}=-\pi i \mathrm{sign}$. This computation, together with the fact that $\widehat{\log |x|}=i\xi \widehat{|x|^{-1}}$, explain the first summand in your formula for the Fourier transform. The delta summand comes from dividing by $|\xi|$ but I cannot think off the top of my head of an explanation for that. $\endgroup$ – Giuseppe Negro Jun 29 '17 at 8:08
  • $\begingroup$ @GiuseppeNegro Exactly, I agree with that. And the other problem is calculating $C$ precisely. $\endgroup$ – Dal Jun 29 '17 at 8:28
  • $\begingroup$ $C=-2\pi \gamma$, where γ is the Euler-Mascheroni constant. $\endgroup$ – Cosmas Zachos Jun 29 '17 at 23:17
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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.

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  • $\begingroup$ The second equality in $(2)$ is not obvious. Can you expand on it? $\endgroup$ – md2perpe Jul 12 '17 at 15:09
  • $\begingroup$ Thank you for your answer. I too would like to know about the second equality in $(2)$ and also why we conclude from $ F(1)=\gamma$ and $F'(x)=1/x$ that $$\eqalign{\gamma+\ln x&= \int_0^x\frac{1-\cos t}{t}dt-\int_{x}^\infty\frac{\cos t}{t}dt\cr &=\int_0^1\frac{1-\cos(xt)}{t}dt-\int_{1}^\infty\frac{\cos (xt)}{t}dt }$$ $\endgroup$ – Dal Jul 12 '17 at 20:14
  • $\begingroup$ @Dal, $F(x)-F(1)=\int_1^x F'(t)dt$ that is $ F(x)-\gamma=\ln x$, then make the chang of variables $t\leftarrow xt$. For the second equality in $(2)$, I just multiplied both sides of $(1)$ by $\phi$ and integrated and then used the expretions of the obtained integrals in terms of the fourier transform as explained after $(2)$. $\endgroup$ – Omran Kouba Jul 12 '17 at 22:58
  • $\begingroup$ @md2perpe, please see my explaination above. Thank you. $\endgroup$ – Omran Kouba Jul 12 '17 at 23:00
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    $\begingroup$ @OmranKouba, you mean "$\phi$ is compactly supported". I think that $\mathscr D$ is dense in $\mathscr S$, but taking $\phi \in \mathscr D$ doesn't help. I can't see that the function $\frac{\mathbb{I}_{[0,1]}(t) - \cos(xt)}{t} \, \phi(x)$ belongs to $L^1((x,t) \in \mathbb R \times [0, \infty])$, and if not we can't apply the Fubini-Tonelli theorem. $\endgroup$ – md2perpe Jul 13 '17 at 12:07
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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.

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    $\begingroup$ Thank you. Your answer is very educational. But do you also know how to solve the problem as originally posed? $\endgroup$ – Dal Jul 9 '17 at 8:30
  • $\begingroup$ @Robjohn How the Fourier transform of $\log|x|$ where $x\in R^2$? Is it $1/|\xi|^2$? $\endgroup$ – Bob Apr 8 '20 at 13:18
  • $\begingroup$ @BobOakley: Using this normalization of the Fourier Transform, in $\mathbb{R}^n$, away from $0$, $$\mathscr{F}(\log(|x|))=-\frac{\pi^{(n+1)/2}}{|\xi|^n}\frac{\Gamma(n)}{\Gamma\!\left(\frac{n+1}2\right)}$$ $\endgroup$ – robjohn Apr 10 '20 at 18:18
  • $\begingroup$ @robjohn Yes, I forgot the absolute value. Perhaps you are thinking of the classical principal value. Here it is in the distributional sense $\psi=\text{PV}\left(\frac1{|x|}\right)$ is defined as $$\langle \psi,\phi \rangle =\int_{|x|\ge 1}\frac{\phi(x)}{|x|}\,dx+\int_{|x|\le 1}\frac{\phi(x)-\phi(0)}{|x|}\,dx$$ $\endgroup$ – Mark Viola Apr 18 at 3:42
  • $\begingroup$ Then would $\delta$ be applied to $\phi(x)-\phi(0)$ since $\delta$ is $0$ for $|x|\ge1$? $\endgroup$ – robjohn Apr 18 at 4:46
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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.

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  • $\begingroup$ The part of your answer with the calculation of the constant is not clear to me. Could you carry it out in detail? $\endgroup$ – Dal Jun 30 '17 at 20:49
  • $\begingroup$ To be honest, your additional details didn't clear the matter for me: 1. why does $f - f(0)g$ does vanish at $0$? 2. why $u(f) = u(f-f(0)g)+f(0)u(g)$? 3. How do those step help in finding the constant in the expression I wrote in the question? 4. How does one find out the value of the constant and the integral in the last line of your calculations? $\endgroup$ – Dal Jul 2 '17 at 20:38
  • $\begingroup$ 1. Since $g(x)=e^{-\pi x^2}$ we have $g(0) = 1$ so $f(0) - f(0) g(0) = f(0) - f(0) \cdot 1= 0$. 2. We have $f = f - f(0)g + f(0)g$ and with $u$ linear this implies $u(f) = u(f - f(0)g + f(0)g) = u(f - f(0)g) + u(f(0)g) = u(f - f(0)g) + f(0) u(g)$. $\endgroup$ – md2perpe Jul 3 '17 at 6:16
  • $\begingroup$ @paulgarrett Hi Paul, my friend. I hope that you are doing well and staying safe and healthy. I've posted a solution that permits test functions that don't vanish at the origin. Please let me know your thoughts. $\endgroup$ – Mark Viola Apr 19 at 20:02
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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 $\phi\in \mathbb{S}$,

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

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  • $\begingroup$ Our methods are similar in that we use the standard distributional definition of the Fourier Transform (modified Plancherel). You used $e^{-|x|}$, which works fine, but requires more ancillary functions than using $e^{-x^2/2}$. I'll have to look more closely at $(2)$ on. It looks as if we both got the same result, so that's good confirmation. $\endgroup$ – robjohn Apr 23 at 22:30
  • $\begingroup$ @robjohn I've posted another solution to develop the Fourier Transform of $\log(|x|)$ HERE. Instead of regularizing $\log(|x|)$ as we did herin, I directly evaluate the functional $$\langle \mathscr{F}\{\log(|x|)\}, \phi\rangle =\langle \log(|x|), \mathscr{F}\{\phi\} \rangle =\int_0^\infty \log(x) \int_{-\infty}^\infty \phi(k) e^{ikx}\,dk\,dx$$Soiler alert ... we get the same result! $\endgroup$ – Mark Viola Apr 23 at 22:37
  • $\begingroup$ I had just noticed that post and was about to mention that here, but you beat me to it. $\endgroup$ – robjohn Apr 23 at 22:43
  • $\begingroup$ @robjohn It's unfortunate that MSE cannot consolidate distinct solutions posted as responses to similar questions. I could post both of my solutions to both questions, but unsure about the protocols on MSE. $\endgroup$ – Mark Viola Apr 23 at 22:47
  • $\begingroup$ Rather than copying different answers, it might be better to provide links to other answers (as long as they’re on StackExchange). $\endgroup$ – robjohn Apr 23 at 23:45
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\ln\pars{\verts{x}} = \int_{-\infty}^{\infty}\hat{\mrm{f}}\pars{k} \expo{\ic kx}\,{\dd k \over 2\pi}}$.


\begin{align} \hat{\mrm{f}}\pars{k} & = \int_{-\infty}^{\infty} \ln\pars{\verts{x}}\expo{-\ic kx}\,\dd x \\[5mm] & = 2\int_{0}^{\infty}\ln\pars{x} \cos\pars{\verts{k}x}\,\dd x \\[5mm] & = 2\,\Re\int_{0}^{\infty}\ln\pars{x} \expo{\ic\verts{k}x}\,\dd x \\[5mm] & = \left.2\,\Re\,\partiald{}{\nu}\int_{0}^{\infty} x^{\nu - 1}\expo{\ic\verts{k}x}\,\dd x \,\right\vert_{\ \nu\ =\ 1} \end{align} Note that $$ \expo{\ic\verts{k}x} = \sum_{n = 0}^{\infty} {\pars{\ic\verts{k}x}^{n} \over n!} = \sum_{n = 0}^{\infty} \color{red}{\expo{-\ic \pi n/2}\,\,\verts{k}^{n}}\, {\pars{-x}^{n}\over n!} $$

With Ramanujan's Master Theorem: \begin{align} \hat{\mrm{f}}\pars{k} & = \left.2\,\Re\,\partiald{}{\nu}\int_{0}^{\infty} x^{\nu - 1}\,\expo{\ic\verts{k}x}\,\dd x \,\right\vert_{\ \nu\ =\ 1} \\[5mm] & = \left.2\,\Re\,\partiald{}{\nu}\bracks{% \Gamma\pars{\nu}\expo{\ic\pi\nu/2} \,\verts{k}^{-\nu}} \,\right\vert_{\ \nu\ =\ 1} \\[5mm] & = \bbx{-\,{\pi \over \verts{k}}} \\ & \end{align}

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  • $\begingroup$ Hi Felix my friend. I hope that you are doing well and staying safe and healthy. I've posted a solution that permits test functions that don't vanish at the origin. Please let me know your thoughts. $\endgroup$ – Mark Viola Apr 19 at 20:01

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