Evaluation of integral $\int_{-\infty}^{\infty} e^{itx} \frac{1- e^{-\frac 1 2 t^2}}{\frac 1 2 t^2} \text d t$ / specific characteristic Function

Question

I want to calculate the value of $$I(x) :=\int_{-\infty}^{\infty} e^{itx} \frac{1- e^{-\frac 1 2 t^2}}{\frac 1 2 t^2} \text d t$$ where $x\in \Bbb R$. Of course we can write $$I(x) = \int_{-\infty}^{\infty} \cos (tx) \frac{1- e^{-\frac 1 2 t^2}}{\frac 1 2 t^2} \text d t + i \int_{-\infty}^{\infty} \sin (tx) \frac{1- e^{-\frac 1 2 t^2}}{\frac 1 2 t^2} \text d t$$ But I have no approach to this integrals.

OR

Does someone know a distribution with characteristic function $\frac{1- e^{-\frac 1 2 t^2}}{\frac 1 2 t^2} $ ?

Answer 1

For $x>0$, \begin{align} I(x)&=\int_0^1\int_{-\infty}^\infty e^{itx-yt^2/2}\,dt\,dy \\ &=\int_0^1\sqrt{\frac{2\pi}{y}}e^{-x^2/2y}\,dy &&\color{gray}{\text{[known inner integral]}} \\ &=2x\sqrt{2\pi}\int_x^\infty\frac{e^{-z^2/2}}{z^2}\,dz &&\color{gray}{\text{[substitution $y=x^2/z^2$]}} \\ &=2\sqrt{2\pi}\left(e^{-x^2/2}-x\int_x^\infty e^{-z^2/2}\,dz\right) &&\color{gray}{\text{[integration by parts]}} \\ &=2\sqrt{2\pi}e^{-x^2/2}-2\pi x\operatorname{erfc}(x/\sqrt{2}). \end{align}