Thread: Series expansion View Single Post
 There is no standart way for those transforms. You have to proceed from the integral form: $$\frac{1}{2\pi}PV \int_{-\infty}^{\infty}{exp(st)s^{-3/2}ds}=\frac{1}{2\pi}2\sqrt{t}\int_{0}^{\infty}{exp(z)z^{-1/2-1}dz}=2\sqrt{t}\frac{1}{2\pi}\Gamma[-1/2]=-2\sqrt{\frac{t}{\pi}}$$ (for t>0)