Inverse Fourier transform of a specific function

Question

Is there an explicit formula for the inverse Fourier transfrom of the following function $$f(\xi)=\frac{1}{(1+\xi^2)^\alpha}\,,$$ where $\xi\in\mathbb{R}$ and $\alpha>1$?

Answer 1

There is an integral representation that shows the inverse transform to be a positive function. A derivation starts with the gamma function: \begin{align} \Gamma(\alpha)&=\int_{0}^{\infty}e^{-t}t^{-1+\alpha}dt,\;\; \Re\alpha > 0. \\ \int_{0}^{\infty}e^{-st}t^{-1+\alpha}dt&=s^{-\alpha}\int_{0}^{\infty}e^{-st}(st)^{-1+\alpha}d(st)=\frac{\Gamma(\alpha)}{s^{\alpha}}. \\ \frac{1}{(1+\xi^2)^{\alpha}} &= \frac{1}{\Gamma(\alpha)}\int_{0}^{\infty}e^{-(1+\xi^2)t}t^{-1+\alpha}dt \end{align} Therefore, \begin{align} f^{\vee}(x) & = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\frac{e^{i\xi x}}{(1+\xi^2)^{\alpha}}d\xi \\ &=\frac{1}{\Gamma(\alpha)}\int_{0}^{\infty}t^{-1+\alpha}e^{-t} \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-t\xi^2}e^{i\xi x}ds dt \\ &= \frac{1}{\Gamma(\alpha)}\int_{0}^{\infty}t^{-1+\alpha}e^{-t}\frac{1}{\sqrt{2t}}e^{-x^2/4t}dt \\ &=\frac{1}{\sqrt{2}\Gamma(\alpha)}\int_{0}^{\infty}t^{-3/2+\alpha}e^{-t}e^{-x^2/4t}dt. \end{align}