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

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 $k \in \mathbb{R}$?

Solutions Collecting From Web of "Evaluate $\int_{-\infty}^\infty x\exp(-x^2/2)\sin(\xi x)\ \mathrm dx$"

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 $\sqrt{2\pi}e^{-x^2/2}$.

First, let us notice that, due of the the oddity of the sine function, we have $(-x)\sin(-x)=(-x)\cdot$ $(-\sin x)=x\sin x\iff I=2\int_0^\infty xe^{-\frac{x^2}2}\sin kx\,dx$. Secondly, since $\sin kx=\Im(e^{ikx})$, we have $I=2\cdot\Im\left[\int_0^\infty xe^{-\left(\frac{x^2}2-ikx\right)}dx\right]$, where $\Im(z)=\Im(a+bi)=b$. Now, let us pay a closer look at the exponent : $\frac{x^2}2-ikx=\frac12(x^2-2ikx)=\frac12\Big[(x-ik)^2+k^2\Big]=\frac{t^2+k^2}2$, where $t=x-ik$, and $dt$ $=dx$. Then:

$$I=2\,\Im\left[\int_{0-ik}^{\infty-ik}(t+ik)e^{-\frac{t^2+k^2}2}dt\right]=2\,\Im\left[\int_{0-ik}^{\infty-ik}te^{-\frac{t^2+k^2}2}dt+ik\int_{0-ik}^{\infty-ik}e^{-\frac{t^2+k^2}2}dt\right]=$$

$$=2\,\Im\left[\int_0^\infty e^{-u}du+ike^{-\frac{k^2}2}\int_{0-ik}^{\infty-ik}e^{-\frac{t^2}2}dt\right]=2\,\Im\left[1+ike^{-\frac{k^2}2}\sqrt\frac\pi2\left(1+\text{Erf}\left(\tfrac{ki}{\sqrt2}\right)\right)\right]=$$

$=ke^{-\frac{k^2}2}\sqrt{2\pi}$ , since the error function of purely imaginary argument is purely imaginary as well, meaning that i times itself possesses no imaginary part, and hence does not ultimately influence the final result.

For the second integral, complete the square and use cauchy’s theorem. For the first integral, notice that $\int _{-\infty} ^{\infty} xe^{-\frac{x^{2}}{2}}Sin(x \xi)=-\partial _{\xi}\int _{-\infty} ^{\infty} e^{-\frac{x^{2}}{2}}Cos(x \xi)=-\partial _{\xi}\int _{-\infty} ^{\infty} e^{-\frac{x^{2}}{2}-i\xi x}dx$