Can we think, at least in the sense of distribution, about the Fourier transform of $\log(s+x^{2})$? Here '$s$' is a real and positive parameter
However $\int_{-\infty}^{\infty}dx\log(s+x^{2})\exp(iux)$ is not well defined.
Can the Fourier transform of logarithm be evaluated ??
Throughout, it is assumed that $s>0$ and $u \in \mathbb{R}$.
Define: $$ \mathcal{I}_\nu(u) = \int_{-\infty}^\infty \left(s+x^2\right)^{-\nu} \mathrm{e}^{i u x} \,\,\mathrm{d} x = \int_{-\infty}^\infty \left(s+x^2\right)^{-\nu} \cos\left(u x\right) \,\,\mathrm{d} x $$ The integral above converges for $\nu > 0$. We are interested in computing the (distributional) value of $\lim_{\nu \uparrow 0} \left( -\partial_\nu \mathcal{I}_\nu(u)\right)$. Let $\mathcal{J}_\nu(u) = -\partial_\nu \mathcal{I}_\nu(u)$.
Notice that $$ \begin{eqnarray} \mathcal{I}_{\nu-1}(u) &=& s \cdot \mathcal{I}_\nu(u) - \partial_u^2 \mathcal{I}_\nu(u) \\ \mathcal{J}_{\nu-1}(u) &=& s \cdot \mathcal{J}_\nu(u) - \partial_u^2 \mathcal{J}_\nu(u) \end{eqnarray} $$ whenever integrals are defined.
It's not hard to compute $\mathcal{I}_\nu(u)$ explicitly: $$ \mathcal{I}_\nu(u) = \sqrt{\pi} \cdot \frac{ 2^{\frac{3}{2}-\nu } s^{\frac{1}{4}-\frac{\nu }{2}} }{\Gamma (\nu )} \cdot |u|^{\nu -\frac{1}{2}} K_{\frac{1}{2}-\nu }\left(\sqrt{s} |u|\right) $$ One can also compute $J_1\left(u\right)$ by using known expressions for index derivatives of Bessel functions at half-integer order: $$ \mathcal{J}_1\left(u\right) = \pi \, \frac{\mathrm{e}^{\sqrt{s} |u|} }{\sqrt{s}} \cdot \operatorname{Ei}\left(-2 \sqrt{s} |u|\right) -\pi\, \frac{ \mathrm{e}^{-\sqrt{s} |u|} }{\sqrt{s}} \left(\frac{1}{2} \log \left(\frac{u^2}{4 s}\right)+\gamma \right) $$ Hence $J_1(u)$ is a continuous function of real argument $u$, and has the following series expansions: $$ \begin{eqnarray} \mathcal{J}_1(u) &=& \frac{\pi \log \left(16 s^2\right)}{2 \sqrt{s}}+\pi |u| (\log (u^2)+2 \gamma -2)+\mathcal{o}\left(u\right) \\ \mathcal{J}_1(u) &=& -\frac{\pi}{2} \mathrm{e}^{-\sqrt{s} |u| } \left(\frac{ \left( \log \left(\frac{u^2}{4s}\right) + 2 \gamma \right)}{\sqrt{s}} + \frac{1}{s |u|}+\mathcal{o}\left(|u|^{-1}\right)\right) \end{eqnarray} $$ They show that $\mathcal{J}_1^\prime(u)$ is discontinuous.
In order to express $\mathcal{J}_0(u)$ in terms of distributions we use $$ \begin{eqnarray} \int \mathcal{J}_0(u) f(u) \, \mathrm{d} u &=& \int \left( s \mathcal{J}_1(u) - \mathcal{J}_1^{\prime\prime}(u) \right) f(u) \, \mathrm{d} u \\ &=& \int \left( s f(u) - f^{\prime\prime}(u) \right) \mathcal{J}_1(u) \, \mathrm{d} u \end{eqnarray} $$
The answer is given here; $a$ is assumed to be positive, so $a=\sqrt s$: $$\mathcal F\!\left[ \ln\left( x^2 + s \right) \right] = -\sqrt{2 \pi} \left( \left| w \right|^{-1} e^{-\sqrt s \left| w \right|} + 2 \gamma \delta(w) \right).$$ Also, with $\left| x \right|^{-1}$ defined in the same way, for negative $s$ we have $$\mathcal F\!\left[ \ln\left( x^2 + s \right) \right] = -\sqrt{2 \pi} \left( \left| w \right|^{-1} \cos \left( \sqrt {-s} w \right) + 2 \gamma \delta(w) \right).$$
