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!

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 $\left| x \right|^{-1}$.

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 ${1\over a}\sqrt{\pi/2}$.