Computation of $\int_0^\infty xe^{-\frac{x^2}{2}}\sin(x \xi) \, dx$ How to compute $$\int_0^\infty xe^{-\frac{x^2}{2}}\sin(x \xi) \, dx$$
I tried to integrate by part but it was bad idea.

Fourier transform, contour shifting, differential equation and other nice techniques work nicely here. Since they are already explained by other users, let me show a brutal-force computation.
Using the substitution $u = x^2/2$,
\begin{align*}
\int_{0}^{\infty} x \sin(\xi x) e^{-x^2/2} \, dx
&= \int_{0}^{\infty} \sin(\xi\sqrt{2u}) e^{-u} \, du \\
&= \int_{0}^{\infty} \left( \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \xi^{2n+1} (2u)^{n+\frac{1}{2}} \right) e^{-u} \, du \\
&= \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}\xi^{2n+1} 2^{n+\frac{1}{2}} \Gamma\left(n + \frac{3}{2}\right) \\
&= \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}\xi^{2n+1} 2^{n+\frac{1}{2}} \cdot \frac{\sqrt{\pi}}{2}\prod_{k=1}^{n} \left(k + \frac{1}{2}\right) \\
&= \sqrt{\frac{\pi}{2}} \sum_{n=0}^{\infty} \frac{(-1)^n}{n! 2^n}\xi^{2n+1} \\
&= \sqrt{\frac{\pi}{2}} \xi e^{-\xi^2/2}.
\end{align*}
Here, we utilized the following identities
$$ \Gamma(s) = \int_{0}^{\infty} u^{s-1}e^{-u} \, du, \qquad \Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}. $$
Also, interchanging the summation and integral is justified by Fubini's theorem together with the estimation
$$\int_{0}^{\infty} \left( \sum_{n=0}^{\infty} \frac{|\xi|^{2n+1}}{(2n+1)!} (2u)^{n+\frac{1}{2}} \right) e^{-u} \, du
= \int_{0}^{\infty} \sinh(|\xi|\sqrt{2u}) e^{-u} \, du < \infty. $$
