Fourier transform (logarithm) question 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 ??
 A: 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
$$
 \begin{eqnarray}
   \int \mathcal{J}_0(u) f(u) \, \mathrm{d} u &=& \int \left( s \mathcal{J}_1(u) - \mathcal{J}_1^{\prime\prime}(u) \right) f(u) \, \mathrm{d} u \\ &=&  \int \left( s f(u) - f^{\prime\prime}(u) \right) \mathcal{J}_1(u) \, \mathrm{d} u
\end{eqnarray}
$$
A: 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
$$\mathcal F\!\left[ \ln\left( x^2 + s \right) \right] =
-\sqrt{2 \pi} \left( \left| w \right|^{-1} \cos \left( \sqrt {-s} w \right) +
 2 \gamma \delta(w) \right).$$
