Calculate the Fourier transform of $\log |x| $ How can one prove that the Fourier transform of $\log |x|$ is 
$$-\pi \mathrm{pf} \frac{1}{|\xi|} +C \delta,$$
where $\mathrm{pf}\frac{1}{|x|} = D(\mathrm{sign}(x)\log|x|)$ (in the sense of distributions) and how can I compute the constant $C$? 
 A: 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.
A: 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.
A: 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 $\phi\in \mathbb{S}$,
$$\int_{-\infty}^\infty  \left(\frac1{|x|}\right)_\nu \phi(x)\,dx=\int_{|x|\le \nu}\frac{\phi(x)-\phi(0)}{|x|}\,dx+ \int_{|x|\ge \nu}\frac{\phi(x)}{|x|}\,dx$$
A: 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.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\ln\pars{\verts{x}} =
\int_{-\infty}^{\infty}\hat{\mrm{f}}\pars{k}
\expo{\ic kx}\,{\dd k \over 2\pi}}$.

\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}
