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 $k \in \mathbb{R}$?
 A: 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.
A: 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 $\sqrt{2\pi}e^{-x^2/2}$.
A: For the second integral, complete the square and use cauchy's theorem. For the first integral, notice that $\int _{-\infty} ^{\infty} xe^{-\frac{x^{2}}{2}}Sin(x \xi)=-\partial _{\xi}\int _{-\infty} ^{\infty} e^{-\frac{x^{2}}{2}}Cos(x \xi)=-\partial _{\xi}\int _{-\infty} ^{\infty} e^{-\frac{x^{2}}{2}-i\xi x}dx$
