How to evaluate this Fourier Transform $A\int_{-\infty}^{\infty} \frac{e^{ikx}}{(1+x^{2})^{\frac{\nu+1}{2}}}dx$ This is basically the Fourier transform of a Student´s T pdf. How do we compute it?
$$A\int_{-\infty}^{\infty} \frac{e^{ikx}}{(1+x^{2})^{\frac{\nu+1}{2}}}dx$$
for $\nu$ any number greater than zero
Thank you in advance.
 A: Consider
$$\oint_C dz \frac{e^{i k z}}{(1+z^2)^{(1+\nu)/2}} $$
where $k \gt 0$ and $C$ is a semicircular contour in the upper half plane of radius $R$ with a detour up and down the imaginary axis from the circular arc with a small circular path around the branch point at $z=i$ of radius $\epsilon$.
The contour integral is then equal to
$$\int_{-R}^R dx \frac{e^{i k x}}{(1+x^2)^{(1+\nu)/2}} + i R \int_0^{\pi/2} d\theta \, e^{i \theta} \frac{e^{i k R e^{i \theta}}}{\left (1+R^2 e^{i 2 \theta} \right )^{(1+\nu)/2}} \\ + i \int_R^{1+\epsilon} dy \, \frac{e^{-k y}}{e^{i \pi (1+\nu)/2} (y^2-1)^{(1+\nu)/2}} + i \epsilon \int_{\pi/2}^{-3 \pi/2} d\phi \, e^{i \phi} \frac{e^{i k \epsilon e^{i \phi}}}{\left (1+\epsilon^2 e^{i 2 \phi} \right )^{(1+\nu)/2}} \\ + i \int_{1+\epsilon}^R dy \, \frac{e^{-k y}}{e^{-i \pi (1+\nu)/2} (y^2-1)^{(1+\nu)/2}} + i R \int_{\pi/2}^{\pi} d\theta \, e^{i \theta} \frac{e^{i k R e^{i \theta}}}{\left (1+R^2 e^{i 2 \theta} \right )^{(1+\nu)/2}}$$
We take the limits as $R \to \infty$ and $\epsilon \to 0$; in these limits, the second, fourth, and sixth integrals vanish.  Because the contour integral is zero by Cauchy's theorem, we have, in this limit,
$$\int_{-\infty}^{\infty} dx \frac{e^{i k x}}{(1+x^2)^{(1+\nu)/2}} + i \left (i 2 \sin{\left [\frac{\pi}{2} (1+\nu)\right ]} \right ) \int_1^{\infty} dy \, \left (y^2-1 \right )^{-(1+\nu)/2} e^{-k y} = 0$$
In the second integral, sub $y=\cosh{t}$ and the FT is, for $k \gt 0$,
$$\int_{-\infty}^{\infty} dx \frac{e^{i k x}}{(1+x^2)^{(1+\nu)/2}} = 2 \sin{\left [\frac{\pi}{2} (1+\nu)\right ]} \int_0^{\infty} dt \sinh^{-\nu}(t) \, e^{-k \cosh{t}} $$
This can be expressed in terms of a Modified Bessel function:
$$\int_{-\infty}^{\infty} dx \frac{e^{i k x}}{(1+x^2)^{(1+\nu)/2}} =   \frac{ \sqrt{\pi} 2^{1+\nu/2}}{ \Gamma \left (\frac{1+\nu}{2} \right )} k^{-\nu/2} K_{\nu/2} (k)$$
where $k \gt 0$.  You can do a similar analysis for $k \lt 0$ for a contour below the real axis.
