Fourier transform of $ |x|^{s} $ and $\log|x| $

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 $$ F[f](\xi)=\int_{\mathbb R}f(x)e^{ix\xi}\,dx. $$

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 $\phi\in \mathbb{S}$

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

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

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

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

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

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

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

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 $(12)$.

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.

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 right-hand side of $(3)$.

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

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

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 $\phi\in \mathbb{S}$,

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