Can anyone provide or give an expression in the sense of distribution theory for the functions $x^{s} , \logx $? 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.

1Do you want the Fourier transform of $x^s$ or $x^s\log x$? (the first is in the body, the second in the title) – Davide Giraudo Apr 28 '12 at 10:17

1i am looking for the fourier transform of all $ x^{s} 4 and $logx $ although by differntiation with respect to 'x' i supsect they are all related. – Jose Garcia Apr 28 '12 at 10:50

In higher dimension, you might want to look here https://math.stackexchange.com/questions/3723136/thefouriertransformof1p3/3724502#3724502 – LL 3.14 Apr 27 '21 at 12:45

@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. – Mark Viola May 21 '21 at 17:49

Related to [Calculate the Fourier transform of $\log x $](https://math.stackexchange.com/questions/2340049/calculatethefouriertransformoflogx). – robjohn Aug 07 '21 at 18:22

@robjohn Hi Rob. While this is related to the posted question in the link you provided in your comment, this expands considerably by asking for the FT of $x^s$ also. I have posted two answers on this page. The first develops the FT for $\log(x)$ in a way that is distinct from the methods used (including my own) on the linked page you provided. The second develops the FT for $x^\alpha$ for all real values of $\alpha$. If you have time, I'd appreciate reading your review. – Mark Viola Aug 07 '21 at 19:37

In general dimension, you have the Fourier transform of $\ln x$ here https://math.stackexchange.com/questions/3723136/thefouriertransformof1p3 – LL 3.14 Apr 21 '22 at 05:06
4 Answers
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, $\S2.5$) $$ F\left[{\cal P}\frac1{x}\right]=2\gamma2\log\xi, $$ 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 $\logx$: $$ F[\logx](\xi)=2\pi\gamma\delta(\xi)\frac\pi{\xi}, $$ taking into account that FT is defined in this book as $$ F[f](\xi)=\int_{\mathbb R}f(x)e^{ix\xi}\,dx. $$
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PRELIMINARIES FOR EVALUATING THE FOURIER TRANSFORM OF $x^\alpha$:
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 $\phi\in \mathbb{S}$
$$\langle \left(\psi_\alpha\right)_{D_2}, \phi\rangle = \int_{\infty}^\infty\frac{\phi(x)\sum_{m=0}^{n1} \frac{\phi^{m}(0)}{m!}x^m}{x^{\alpha}}\,dx\tag3$$
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
$$\begin{align} \mathscr{F}\{\psi_\alpha\}(k)&=\int_{\infty}^\infty x^\alpha e^{ikx}\,dx\\\\ &=2\text{Re}\left(\int_0^\infty x^\alpha e^{ikx}\,dx\right)\\\\ &=\frac{2}{k^{1\alpha}}\text{Re}\left(\int_0^\infty x^\alpha e^{ix}\,dx\right)\\\\ &=\frac{2\sin(\pi \alpha/2)\Gamma(1\alpha)}{k^{1\alpha}} \end{align}$$
NOTE:
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
$$\begin{align} \langle \mathscr{F}\{\left(\psi\right)_{D_1}\},\phi\rangle &=2\text{Re}\left(e^{i(n+1)\pi/2}\int_0^\infty x^{\alphan1} \int_{\infty}^\infty \phi^{(n+1)}(k)e^{ikx}\,dk\,dx\right)\\\\ &=2\text{Im}\left(e^{in\pi/2} \int_{\infty}^\infty \phi^{(n+1)}(k)\int_0^\infty x^{\alphan1}e^{ikx}\,dx\,dk\right)\tag6 \end{align}$$
NOTE:
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
$$\begin{align} \int_0^\infty x^{\alphan1}e^{ikx}\,dx=e^{i\text{sgn}(k)\pi(\alphan)/2}\frac{\Gamma(\alpha  n)}{k^{\alpha n}}\tag7 \end{align}$$
Using $(7)$ in $(6)$, we obtain
$$\begin{align} \langle \mathscr{F}\{\left(\psi\right)_{D_1}\},\phi\rangle &=2\Gamma(\alphan)\text{Im}\left(e^{in\pi/2} \int_{\infty}^\infty \frac{\phi^{(n+1)}(k)}{k^{\alphan}}e^{i\text{sgn}(k)\pi(\alphan)/2}\,dk\right)\\\\ &=2\sin(\pi \alpha/2)\Gamma(\alphan)\int_{\infty}^\infty \frac{\phi^{(n+1)}(k)}{k^{\alphan}}\left(\text{sgn}(k)\right)^{n+1}\,dk\tag8 \end{align}$$
Integrating by parts $n+1$ times the integral on the righthand 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
$$\begin{align} \text{PV}\left(\frac1{k^{n+1}}\right)_{D_2}&=\lim_{\delta\to 0^+}\left(\int_{x\ge \delta} \frac{\phi(k)\sum_{m=0}^{n}\frac{\phi^{(m)}(0)}{m!}k^m }{k^{n+1}}\,dk\right)\\\\ &=\lim_{\delta\to 0^+}\left(\int_{x\ge \delta} \frac{\phi(k)\sum_{m=0}^{n1}\frac{\phi^{(m)}(0)}{m!}k^m }{k^{n+1}}\,dk\right)\tag{12} \end{align}$$
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 $\alpha\in ((n+1),n)$, $n>1$
$$\begin{align} \bbox[5px,border:2px solid #C0A000] {\mathscr{F}\{\left(\psi\right)_{D_2}\}(k)=2\sin(\pi \alpha/2)\Gamma(1\alpha)\left(k^{\alpha1}\right)_{D_1}}\tag{13} \end{align}$$
We can extend the result in $(13)$ to include $\alpha=n$ for even values of $n$. Thus, for $\alpha=n$, $n$ even we have
$$\begin{align} \bbox[5px,border:2px solid #C0A000] {\mathscr{F}\{\left(\psi_n\right)_{D_2}\}(k)=\frac{\pi \cos(n\pi/2)}{\Gamma(n)}\left(k^{n1}\right)_{D_1}}\tag{14} \end{align}$$
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
$$\begin{align} \langle \mathscr{F}\{\left(\psi_{n}\right)_{D_2}\},\phi\rangle &=\langle \left(\psi_{n}\right)_{D_2},\mathscr{F}\{\phi\}\rangle \\\\ &=2\text{Re}\int_0^\infty \int_{\infty}^\infty \phi(k) \frac{ e^{ikx}\sum_{m=0}^{n2}\frac{(ikx)^m}{m!}\frac{(ikx)^{(n1} \xi_{[0,1]}(x)}{(n1)!}}{x^{n}}\,dk\,dx\\\\ &=\frac2{(n1)!}\text{Re} \int_0^\infty \frac1x \int_{\infty}^\infty \phi(k)(ik)^{n1}(e^{ikx}\xi_{[0,1]}(x))\,dk\,dx\\\\ &+\frac{2\sin(n\pi/2)H_{n1}}{(n1)!}k^{n1}\\\\ &=\frac{2\sin(n\pi/2)}{(n1)!}\int_{\infty}^\infty k^{n1}\phi(k)\int_0^\infty \frac{\cos(kx)1\xi_{[0,1]}(x)}{x}\,dx\,dk\\\\ &+\frac{2\sin(n\pi/2)H_{n1}}{(n1)!}k^{n1}\tag{15} \end{align}$$
In the question posted HERE, I showed using both real analysis and complex analysis that the inner integral on the righthand side of $(15)$ is given by
$$ \int_0^\infty \frac{\cos(kx)1\xi_{[0,1]}(x)}{x}\,dx=\gamma\log(k)\tag{16}$$
Using $(16)$ in $(15)$ reveals
$$\begin{align} \langle \mathscr{F}\{\left(\psi_n\right)_{D_2}\},\phi\rangle &=\frac{2\sin(n\pi/2)}{(n1)!}\int_{\infty}^\infty k^{n1}(\gamma\log(k)+H_{n1})\phi(k)\,dk\tag{17} \end{align}$$
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
$$\begin{align} \bbox[5px,border:2px solid #C0A000] {\mathscr{F}\{\left(\psi_{n}\right)_{D_2}\}(k)=2\frac{\sin(n\pi/2)}{\Gamma(n)}\,k^{n1}(\gamma+\log(k)H_{n1})}\tag{18} \end{align}$$
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 righthand side of $(A2)$ is $2\pi \phi^{(n)}(0)$. For the second integral on the righthand 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 righthand side of $(A3)$, we see that
$$\int_0^\infty \int_{\infty}^\infty \phi^{(n)}(k)\sin(kx)\,dk\,dx=n!\lim_{\delta \to 0^+}\int_{k\ge \delta}\frac{\phi(k)\sum_{m=0}^{n1}\frac{\phi^{(m)}(0)k^{m}}{m!}}{k^{n+1}}\,dk \tag{A4}$$
Using $(A4)$ in $(A2)$ we obtain
$$\begin{align} \langle \mathscr{F}\{\psi\},\phi\rangle &=\cos(n\pi/2)2\pi \phi^{(n)}(0)\\\\&2\sin(n\pi/2)n!\lim_{\delta \to 0^+}\int_{k\ge \delta}\frac{\phi(k)\sum_{m=0}^{n1}\frac{\phi^{(m)}(0)k^{m}}{m!}}{k^{n+1}}\,dk\tag{A5} \end{align}$$
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 $(12)$.
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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](https://math.meta.stackexchange.com/questions/4666/sandboxfordraftsoflongcomplexposts). – Asaf Karagila May 07 '21 at 21:25

@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. – Mark Viola May 08 '21 at 02:25

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. – Asaf Karagila May 08 '21 at 06:20

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 – Asaf Karagila May 08 '21 at 06:27

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 $\mathcal F[x^{3}](k) = k^2 (\gamma + \ln k)$. – Maxim Jul 13 '21 at 17:22

Do we agree that $x^{n}$ is defined as in my comment? Then take a simple test function and verify that $(\mathcal F[x^{n}], \phi) = (x^{n}, \mathcal F[\phi])$ doesn't hold. Your derivation implicitly uses different regularizations of $x^{n}$, losing $\mathcal F[\delta^{(m)}]$ terms. It's not obvious whether or not $PV(k^{n})$ in (13) and $\psi$ in (17) agree with $\lambda_1$ in (1). Also, $\mathcal F[x^{3}](k)$ should have a $+k^2 \ln k$ term. – Maxim Jul 13 '21 at 18:45

But the first condition for (20) is $n < 0$, so we should take $n = 3$, which gives $k^2 \ln k$. I'm saying you're using different regularizations of the ordinary function $x^{n}$ for the same $n$. The question is not whether or not $x^{n}$ in my first comment is a valid regularization (it is) but rather whether or not it's the one that you're using. If it is, take $\phi(x) = e^{x^2}$, then $(\mathcal F[x^{n}], \phi)$ and $(x^{n}, \mathcal F[\phi])$ can be found in closed form. – Maxim Jul 13 '21 at 19:50

Since you do not clearly state whether or not the regularization you're using is the same as the one in my first comment (for $n \in \mathbb N$), all I can say is that if it is, then the derivation cannot be correct because with $x^{n}$ as in my first comment and $\mathcal F[x^{n}]$ as in your answer, we get $(\mathcal F[x^{n}], \phi) \neq (x^{n}, \mathcal F[\phi])$, which is impossible. – Maxim Jul 13 '21 at 20:07

If the discussion in the comments in the post about $\arctan(1/x)/x$ clarified things, then, for the distribution that you call $\psi(x) = x^\alpha$, $\alpha \leq 1$, can you write out $(\psi, \phi)$ explicitly without employing multiple integrals or $\mathcal F[\phi]$? The difference is that here, unless you define $x^{3}$ in a very contrived way, $k^2 (\gamma + \ln k)$ will be unambiguously wrong. – Maxim Jul 18 '21 at 01:24

I'm asking about the definition though, not elucidations and reinforcements. If you say that $\psi = \lambda_1$ for all real $\alpha \leq 1$, that will be good enough. But you haven't said that. – Maxim Jul 18 '21 at 04:00

I disagree. Where does it say that $\lambda_1$ and $\psi$ are the same distribution? If this is indeed what you mean, please clearly state that $\psi = \lambda_1$ (which, of course, can and will be used against you). – Maxim Jul 18 '21 at 04:28

That's literally what my very first comment said. We have $$(x^{3/2}, e^{x^2/4} \sqrt \pi) = 2 \sqrt {2 \pi} \, \Gamma(3/4) + 4 \sqrt \pi, \\ (2 \sqrt {2 \pi} \, k^{1/2}, e^{k^2}) = 2 \sqrt {2 \pi} \, \Gamma(3/4),$$ a contradiction (because the subtracted terms in (1) and in (6) do not match). – Maxim Jul 18 '21 at 05:04

Your derivation gives the Fourier transform of the distribution defined by the rhs of (6). $\psi$ is a different distribution. – Maxim Jul 19 '21 at 19:33


@Maxim Hi Max. I hope that you are doing well. I've made major changes to this post in order to address the inconsistencies you reported. Whenever you have some time, and if you are still interested, I'd appreciate your review. Hope to hear from you at your earliest. – Mark Viola Aug 06 '21 at 17:04

$k^2 (\gamma + \ln k)$ is still not the correct FT for the new $x^{3}$: $$(x^{3}, e^{x^2/4} \sqrt \pi) = \sqrt \pi (\gamma  \ln 4  1)/4, \\ (k^2 (\gamma + \ln k), e^{k^2}) = \sqrt \pi (\gamma  \ln 4 + 2)/4.$$ – Maxim Aug 07 '21 at 10:49

@Maxim You're right. I've corrected the integration by parts procedure I used to arrive at $(15)$ to include the missing term. – Mark Viola Aug 07 '21 at 17:24

@Maxim I used the distribution $D_1$ throughout to maintain consistency. And the $D_1$ distribution seems to be used more commonly. Yet the term with $\delta^{(n)}$ that appears in $(10)$ and the corresponding term that appears in $(13)$ seem awkward (and unnecessary if we modified the distribution to extend $\xi_{[1.1]}(x)$ to $1$). Would you have used a distribution different from $D_1$ for noninteger values of $\alpha<1$? – Mark Viola Aug 07 '21 at 19:11

Yes, the standard choice is $$(\operatorname {FP}(x^{\alpha}), \phi) = \int_{\mathbb R} x^{\alpha} \left( \phi(x)  \sum_{m = 0}^{\lfloor \alpha \rfloor  1} \frac {\phi^{(m)}(0)} {m!} x^m \right) dx$$ for nonintegral $\alpha > 1$. – Maxim Aug 07 '21 at 19:47


Yes, I think it would simplify things (besides, there is a typo somewhere in (13) which gives the wrong sign of the second term for $\alpha = 3/2$). And I wouldn't introduce (12) without clarifying whether or not it's the same distribution as in (3) for $\alpha = n$. Also, note that $x^\alpha$ is not in $L^1(\mathbb R)$ for $1 < \alpha < 0$. – Maxim Aug 07 '21 at 20:16

@Maxim Well, I've taken your advice and edited again. Thank you for all of your input. Hopefully, all is good now. – Mark Viola Aug 07 '21 at 21:39

$H_{n  1}$ should be $H_{n  1}$. For $1 < \alpha < 0$, it's true that the FT is equal to the distributional limit of $\int_{x < A} x^\alpha e^{i k x} dx$, what needs to be justified is that the improper integral $\int_{\infty}^\infty$, which is the pointwise limit, gives the same distribution. Similarly, we don't have absolute convergence in (6), so Fubini's theorem is not applicable. – Maxim Aug 08 '21 at 11:11

@MaximThank you for catching the missing absolute value sign. For $(6)$ Fubini's was not applied directly. Rather, as the note following $(6)$ asserts, for $\beta\in (0,1)$, there exists a number $C_\beta$ such that for any $L>0$, $\left \int_0^L \frac{e^{it}}{t^\beta}\,dt\right
– Mark Viola Aug 08 '21 at 13:49 
I see, so I assume the meaning of $e^{i t}/t^\beta$ is that you're taking $t = k x$ and uniformly bounding $\int_{t < A k} t^{\beta} e^{i t} dt$ by a constant to show that $\phi(k) \int_{x < A} x^{\beta} e^{i k x} dx$ is uniformly bounded by an integrable function, which allows us to interchange the integral and the limit. – Maxim Aug 08 '21 at 17:13

@maxim Yes. The DCT guarantees that we can bring the limit (as $L\to \infty$) inside the integral – Mark Viola Aug 08 '21 at 18:15
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 nonunique 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
$$\begin{align} \langle \mathscr{F}\{\psi\}, \phi\rangle &=\langle \psi, \mathscr{F}\{\phi\}\rangle\\\\ &=\int_{\infty}^\infty \log(x)\int_{\infty}^\infty \phi(k)e^{ikx}\,dk\,dx\\\\ &=2\int_0^\infty \log(x)\int_{\infty}^\infty \phi(k)\cos(kx)\,dk\,dx\\\\ &=4\phi(0)\int_0^\infty \frac{\sin(x)}{x}\,\log(x)\,dx\\\\ &+2\int_0^\infty \log(x) \left(\int_{\infty}^\infty (\phi(k)\phi(0)\xi_{[1,1]}(k))\cos(kx)\,dk\right)\,dx\\\\ &=2\pi \gamma \phi(0)\tag2\\\\ &+2\lim_{L\to \infty } \int_{\infty}^\infty \frac{\phi(k)\phi(0)\xi_{[1,1]}(k)}k\left(\log(L)\sin(kL)\int_0^L \frac{\sin(kx)}{x}\,dx\right)\,dk\\\\ &=2\pi \gamma \phi(0)\pi\int_{\infty}^\infty \frac{\phi(k)\phi(0)\xi_{[1,1]}(k)}{k}\,dk\tag3 \end{align}$$
NOTES:
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 righthand side of $(3)$.
From $(3)$, we deduce the Fourier Transform of $\psi$ in distribution
$$\bbox[5px,border:2px solid #C0A000] {\mathscr{F}\{\psi\}(x)=2\pi \gamma \delta(x)+\left(\frac1{x}\right)_1}$$
where the distribution $\left(\frac1{x}\right)_1$ is defined by its action on and $\phi\in \mathbb{S}$ as
$$\int_{\infty}^\infty \left(\frac1{x}\right)_1\phi(x)\,dx=\int_{\infty}^\infty \frac{\phi(x)\phi(0)\xi_{[1,1]}(x)}{x}\,dx$$
NOTE:
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 $\phi\in \mathbb{S}$,
$$\int_{\infty}^\infty \left(\frac1{x}\right)_\nu \phi(x) \,dx=\int_{x\le \nu}\frac{\phi(x)\phi(0)}{x}\,dx+ \int_{x\ge \nu}\frac{\phi(x)}{x}\,dx$$
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I don't disagree with the content of your answer, but there is significant overlap with [this answer on MO](https://mathoverflow.net/questions/300024/aboutthefouriertransformofthelogarithmfunction/390647#390647), [this answer on MSE](https://math.stackexchange.com/questions/2340049/calculatethefouriertransformoflogx/4108819#4108819) and [this answer on MSE](https://math.stackexchange.com/questions/73922/fouriertransformofunitstep/4111722#4111722), maybe adding a link to the best one would suffice – Calvin Khor Apr 23 '21 at 05:14

@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. – Mark Viola Apr 23 '21 at 13:49

This is a considerably nonstandard notion of "principal value integral"... ?!? – paul garrett May 05 '21 at 22:24

@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](https://terrytao.wordpress.com/2009/04/19/245cnotes3distributions/#1xr) 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. – Mark Viola May 06 '21 at 00:46

@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. – Mark Viola May 08 '21 at 02:28

@MarkViola, I don't really have a suggestion for an optimal notation, just that saying "principal value" might confuse people... – paul garrett May 08 '21 at 17:15

@paulgarrett I agree. I've used a different notation herein. Hopefully it is clearer now. And thank you for your comment. – Mark Viola May 08 '21 at 17:18

@JoseGarcia Hi Jose. Please let me know how I can improve my answer on determining the Fourier Transform of $\log(x)$. I really want to give you the best answer I can. – Mark Viola May 21 '21 at 17:52

@MarkViola Great solutions, really appreciate the rigour of your answers. +1 – spaceman Jul 12 '21 at 10:29



@MarkViola That's strange. I have added a +1 to the solution. For some reason I am not even allowed to change the vote. I have given a +1 to another solution of yours elsewhere as an appreciation of your post. – spaceman Jul 12 '21 at 16:05

@Spaceman I do see a recorded up vote from 2 days ago. Perhaps that was yours. And your kind and flattering words are encouraging. Thank you again. :) – Mark Viola Jul 12 '21 at 16:13

@MarkViola Ah yes, I believe that was me. Just came today to show appreciation. I hope you have a nice morning/day/evening wherever you may be. – spaceman Jul 12 '21 at 17:09

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 $1/x^{ns}$. 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 bumpedup 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 $\logx$ on $\mathbb R^1$, while dodging some of the complications of the more general $x^s\cdot \logx$, we can proceed as follows. First, the derivative of $\logx$ 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 $x\cdot \widehat{\logx}={1\over 2}\,\mathrm{sgn}(x)$.
If we imagine that we can divide by $x$, then $\widehat{\logx}={1\over 2}\,{1\over x}$. However, integrationagainst$1/x$ does not extend to a homogeneous, even distribution on test functions. (Without the requirement of homogeneity, sure, HahnBanach gives lots of extensions...)
But it is true that $x\cdot \widehat{\logx}={1\over 2}\mathrm{sgn}(x)$. This implies that we can evaluate that Fourier transform by integrationagainst $f$ for Schwartz functions $f$ vanishing at $0$. In particular, and since we do already know that this Fourier transform exists, for general Schwartz functions $f$,
$$ \widehat{\logx}(f) \;=\; \widehat{\logx}(ff(0)\cdot e^{\pi x^2}) + \widehat{\logx} (e^{\pi x^2}) $$ $$ \;=\; x\cdot \widehat{\logx}\Big({ff(0)\cdot e^{\pi x^2}\over x}\Big) + \widehat{\logx}(e^{\pi x^2}) $$ $$ \;=\; {1\over 2}\int_{\mathbb R} \mathrm{sgn}(x)\Big({ff(0)\cdot e^{\pi x^2}\over x}\Big)\;dx + \delta(f)\cdot\widehat{\logx}(e^{\pi x^2}) $$ $$ \;=\; {1\over 2}\int_{\mathbb R} {ff(0)\cdot e^{\pi x^2}\over x}\;dx + \delta(f)\cdot \widehat{\logx}(e^{\pi x^2}) $$ The value of $\widehat{\logx}$ 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 $f$, $$ \widehat{\logx}(f) \;=\; {1\over 2}\int_{\mathbb R} {f(x)f(0)\cdot e^{\pi x^2}\over x}\;dx \;+\; \delta(f) \cdot \Big({\log\pi\over 2} + {\Gamma'({1\over 2})\over 2\sqrt{\pi}}\Big) $$
(Edit: corrected some dropped constants, etc.)
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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 $\widehat{\logx} (e^{\pi x^2})$. 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. ;). – Mark Viola Aug 06 '21 at 17:20


You're welcome Paul. Are you using the FT with a $\frac1{2\pi}$ scaling factor? – Mark Viola Aug 06 '21 at 20:38

@MarkViola, my intention, though I may of course fail, is to put the $2\pi$ in the exponent, and have no adjustment of measure... – paul garrett Aug 06 '21 at 20:39