real analysis - Evaluate $int_{-infty}^infty xexp(-x^2/2)sin(xi x) mathrm dx$

Evaluate $\int_{-\infty}^\infty x\exp(-x^2/2)\sin(\xi x)\ \mathrm dx$

The answer given by Wolfram Alpha is $\sqrt{2\pi}\xi\exp(-\xi^2/2)$.
Observe how this is related to the Fourier transform of $x\exp(-x^2/2)$:

the part $\int_{-\infty}^{\infty}x\exp(-x^2/2)\cos\xi x \ \mathrm dx=0$ since the integrand is odd.

In addition, what are the Fourier transforms of $x^k\exp(-x^2/2)$ for $k=2,3$?

Related:
How do I compute $\int_{-\infty}^\infty e^{-\frac{x^2}{2t}} e^{-ikx} \, \mathrm dx$ for $t \in \mathbb{R}_{>0}$ and $k \in \mathbb{R}$?

Answer

Using contour integration, then differentiating, we get
$$\begin{align} \int_{-\infty}^\infty e^{-x^2/2}e^{-i\xi x}\,\mathrm{d}x &=e^{-\xi^2/2}\int_{-\infty}^\infty e^{-(x+i\xi)^2/2}\,\mathrm{d}x\\ &=e^{-\xi^2/2}\int_{-\infty}^\infty e^{-x^2/2}\,\mathrm{d}x\\ &=\sqrt{2\pi}e^{-\xi^2/2}\tag{1}\\ \int_{-\infty}^\infty (-ix)^ke^{-x^2/2}e^{-i\xi x}\,\mathrm{d}x &=\left(\frac{\mathrm{d}}{\mathrm{d}\xi}\right)^k\sqrt{2\pi}e^{-\xi^2/2}\tag{2} \end{align}$$
By taking real and imaginary parts, for $k=1$, $(2)$ gives
$$\int_{-\infty}^\infty x\,e^{-x^2/2}\sin(\xi x)\,\mathrm{d}x =\sqrt{2\pi}\,\xi\,e^{-\xi^2/2}\tag{3}$$
and
$$\int_{-\infty}^\infty x\,e^{-x^2/2}\cos(\xi x)\,\mathrm{d}x\tag{4} =0$$
For larger $k$, $(2)$ says that the Fourier Transforms of $x^ke^{-x^2/2}$ are polynomials in $\xi$, with integer coefficients, times $\sqrt{2\pi}e^{-x^2/2}$.