On the Fourier transform of $f(x)=\ln(x^2+a^2)$ I would like to derive the Fourier transform of $f(x)=\ln(x^2+a^2)$, where $a\in \mathbb{R}^+$ by making use of the properties:
\begin{equation}
\mathcal{F}[f'(x)]=(ik)\hat{f}(k)\\
\mathcal{F}[-ixf(x)]=\hat{f}'(k)
\end{equation}
For the Fourier transform I use the definition given by:
\begin{equation}
\hat{f}(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x)e^{-ikx}dx, k \in \mathbb{R}
\end{equation}
Until now I found out that by taking the derivative of $f$ and finding the Fourier transform of $f'$ I can then use the relation $\mathcal{F}[f'(x)]=(ik)\hat{f}(k)$ and find $\hat{f}$. The derivative of $f$ would be:
\begin{equation}
f'(x)=\frac{2x}{x^2+a^2}
\end{equation}
and by considering $g(x)=1/(x^2+a^2)$, I then have:
\begin{equation}
f'(x)=2xg(x)
\end{equation}
Now I know that the Fourier transform of $g$ is given by:
\begin{equation}
\hat{g}(k)=\frac{1}{a}\sqrt{\frac{\pi}{2}}e^{-a|k|}, a \in \mathbb{R}, k\in \mathbb{R}
\end{equation}
Now I must find the Fourier transform of $xg(x)$ which would be given by the derivative of $\hat{g}$ right? But how can this possible since $\hat{g}$ has no derivative? 
I think I am really close now but I need that extra tip.
Thank you!
 A: If $\left| x \right|^{-1}$ is the distribution defined as
$$\left( \left| x \right|^{-1}, \phi \right) =
\int_{\left| x \right| < 1} \frac {\phi(x) - \phi(0)} {\left| x \right|} dx +
 \int_{\left| x \right| > 1} \frac {\phi(x)} {\left| x \right|} dx,$$
then
$${\mathcal F}\!\left[ \ln\left( x^2 +a^2 \right) \right] =
\frac 1 {\sqrt{2 \pi}}
 \int_{-\infty}^{\infty} \ln\left( x^2 +a^2 \right) e^{-i k x} dx =
-\sqrt{2 \pi} \left( \frac { e^{-a \left| k \right|}} {\left| k \right|} +
 2 \gamma \delta(k) \right),$$
where $\gamma$ is Euler's constant. Probably the easiest way to prove it is to compute the inverse transform directly from the definition of $\left| x \right|^{-1}$.
A: If $1/|x|$ means even and homog degree $-1$, the only such distn is $\delta$. Attempting to alter this fact by arbitrary truncations can indeed produce several different seeming answers. That’s not good.
I’d prefer to continue the approach in the question: up to the constant there, the desired Fourier transform is the derivative of $e^{-|ak|}$ in $k$. Without loss of generality, $a>0$. Then the deriv is $-sgn(x)ae^{-|ak|}-2a\delta$. Then mult by the ${1\over a}\sqrt{\pi/2}$.
A: 
I thought it might be instructive to present an approach that develops the Fourier Transform directly and is distinct and complements the approach I adopted in THIS ANSWER.  To that end, we now proceed.


Let $f(x)=\log(x^2+a^2)$ and let $F(k)$ denote its Fourier Transform.  Then, we write
$$F(k)=\mathscr{F}\{f\}(k)\tag 1$$
where $(1)$ is interpreted as a Tempered Distribution.  That is, for any $\phi \in \mathbb{S}$, we can write for any $\nu>0$
$$\begin{align}
\langle \mathscr{F}\{f\}, \phi\rangle &=\langle f, \mathscr{F}\{\phi\}\rangle\\\\
&=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty \log(x^2+a^2)\int_{-\infty}^\infty \phi(k)e^{-ikx}\,dk\,dx\\\\
&=\frac2{\sqrt{2\pi}}\int_0^\infty \log(x^2+a^2)\int_{-\infty}^\infty \phi(k)\cos(kx)\,dk\,dx\\\\
&=\frac4{\sqrt{2\pi}}\phi(0)\int_0^\infty \frac{\sin(\nu x)}{x}\,\log(x^2+a^2)\,dx\\\\
&+\frac2{\sqrt{2\pi}}\int_0^\infty \log(x^2+a^2) \left(\int_{-\infty}^\infty (\phi(k)-\phi(0)\xi_{[-\nu,\nu]}(k))\cos(kx)\,dk\right)\,dx\\\\
&=C_\nu(a) \phi(0)\tag2\\\\
&+\frac2{\sqrt{2\pi}}\lim_{L\to \infty } \int_{-\infty}^\infty \frac{\phi(k)-\phi(0)\xi_{[-\nu,\nu]}(k)}k\left(\log(L^2+a^2)\sin(kL)\\-\int_0^L \frac{2x\sin(kx)}{x^2+a^2}\,dx\right)\,dk\\\\
&=C_\nu(a) \phi(0)-\sqrt{2\pi}\int_{-\infty}^\infty \left(\phi(k)-\phi(0)\xi_{[-\nu,\nu]}(k)\right)\frac{e^{-|ka|}}{|k|}\,dk\tag3
\end{align}$$
where the constant $C_\nu(a)$ is given by
$$\begin{align}
C_\nu(a)&=-2\sqrt{2\pi} \log(\nu)+\frac4{\sqrt{2\pi}}\int_0^\infty \frac{\sin(x)}{x}\,\log(x^2+\nu^2a^2)\,dx\\\\
&=2\sqrt{2\pi} (\log(a)-\text{Ei}(-\nu a))
\end{align}$$


NOTES:
In arriving at $(2)$, 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 $\log(L^2+a^2)$ 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 $(3)$.
For $\nu=1$ and $a\to 0$,  $C_1(a\to 0)=-2\sqrt{2\pi}\gamma$, which agrees with the results in the previously referenced answer (modulo the scale factor $1/\sqrt{2\pi}$ here from the definition of the Fourier Transform).


From $(3)$, we deduce the Fourier Transform of $\psi$ in distribution
$$\bbox[5px,border:2px solid #C0A000] {\mathscr{F}\{f\}(k)=C_\nu(a) \delta(k)-\sqrt{2\pi}\left(\frac{e^{-|ka|}}{|k|}\right)_\nu}\tag4$$
where the distribution $\left(\frac{e^{-|ka|}}{|k|}\right)_\nu$ is defined by its action on any $\phi\in \mathbb{S}$
$$\int_{-\infty}^\infty \left(\frac{e^{-|ka|}}{|k|}\right)_\nu\phi(k)\,dk=\int_{-\infty}^\infty \left(\phi(k)-\phi(0)\xi_{[-\nu,\nu]}(k)\right)\frac{e^{-|ka|}}{|k|}\,dk$$


SOLUTION VERIFICATION:
In this section, we take the inverse Fourier Transform of the distribution in $(6)$ and show that it is indeed equal to $\log(x^2+a^2)$.  To begin, we write for any $\phi\in \mathbb{S}$
$$\begin{align}
\langle \mathscr{F}^{-1}\{F\},\phi\rangle&=\langle F,\mathscr{F}^{-1}\{\phi\}\rangle \\\\
&=C_\nu(a)\mathscr{F}^{-1}\{\phi\}(0)-\int_{-\infty}^\infty \frac{e^{-|ka|}}{|k|}\int_{-\infty}^\infty \phi(x)\left(e^{ikx}-\xi_{[-\nu,\nu]}(k)\right)\,dx\,dk\\\\
&=C_\nu(a)\mathscr{F}^{-1}\{\phi\}(0)\\\\
&-2\int_{-\infty}^\infty \phi(x)\int_{0}^\infty \frac{e^{-ka}}{k}\left(\cos(kx)-\xi_{[-\nu,\nu]}(k)\right)\,dk\,dx\tag5
\end{align}$$
Using "Feynman's Trick," we find that the inner integral on the right-hand side of $(5)$ is given by
$$\begin{align}
\int_{0}^\infty \frac{e^{-ka}}{k}\left(\cos(kx)-\xi_{[-\nu,\nu]}(k)\right)\,dk&=\int_\infty^a \left(\frac{1-e^{-\nu a'}}{a'}-\frac{a'}{x^2+a'^2}\right)\,da'\\\\
&=\log(a)+\int_{\nu a}^\infty \frac{e^{-a' }}{a'}\,da'-\frac12\log(x^2+a^2))\\\\
&=-\frac12\log(x^2+a^2)+\log(a)+\text{Ei}(-\nu a)\tag6
\end{align}$$
Finally, substituting $(6)$ in $(5)$ yields the anticipated result that
$$\langle \mathscr{F}^{-1}\{F\},\phi\rangle=\int_{-\infty}^\infty \phi(x)\log(x^2+a^2)\,dx$$
from which we conclude that
$$f(x)=\mathscr{F}^{-1}\{F\}(x)=\log(x^2+a^2)$$
