It is known that $$\int_{-1}^1 (1-x^2)^\lambda e^{ixs}dx = \sqrt{\pi}\Gamma(\lambda+1)\left(\frac{s}{2} \right)^{-\lambda-\frac{1}{2}}J_{\lambda+\frac{1}{2}}(s). \qquad (1)$$ One also has $$\int_{-\infty}^\infty (1+x^2)^\lambda e^{ixs}dx = \frac{2\sqrt{\pi}}{\Gamma(-\lambda)}\left|\frac{s}{2} \right|^{-\lambda-\frac{1}{2}}K_{-\lambda-\frac{1}{2}}(|s|). \qquad (2)$$ This can be found, for example, in the 1st volume of Gel'fand and Shilov on distributions.
I am wondering what is the inverse Fourier transform of $$\sqrt{\pi}\Gamma(\lambda+1)\left(\frac{s}{2} \right)^{-\lambda-\frac{1}{2}}I_{\lambda+\frac{1}{2}}(s)? \qquad (3)$$ Clearly, it is related to the right hand side of (1) by $s\to is$. So, one can just try to make a substitution $s\to is$, $x\to -ix$, but then one is lead to integrate over $x$ from $-i$ to $+i$. I do not see how this contour can be replaced with the real one, so I do not understand how to get the Fourier transform along these lines.
Of course, I understand that $I$ is exponentially growing, so this integral does not converge in the normal sense. Still, I guess, Fourier transforms can be defined for a large class of functions in the sense of distributions (e.g. in Gel'fand Shilov one evaluates Fourier transforms of exp[ax] with real a).
Any ideas?