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 ??

-
It might help to differentiate wrt $s$, and use $\mathcal{F}_x\left(\frac{1}{s+x^2}\right)(u) = \frac{\pi}{\sqrt{s}} \exp\left(- \sqrt{s} |u| \right)$ – Sasha Dec 22 '11 at 20:50
Umm of course Sasha.. then i could integrate over 's' to get the final result . HOwever in order to get rid of the divergent constant of integration i should try the Fourier transform of $log(s+x^{2})-log(e+x^{2}$ with $e \to 0$ – Jose Garcia Dec 23 '11 at 9:23

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 $$\begin{eqnarray} \int \mathcal{J}_0(u) f(u) \, \mathrm{d} u &=& \int \left( s \mathcal{J}_1(u) - \mathcal{J}_1^{\prime\prime}(u) \right) f(u) \, \mathrm{d} u \\ &=& \int \left( s f(u) - f^{\prime\prime}(u) \right) \mathcal{J}_1(u) \, \mathrm{d} u \end{eqnarray}$$