Integrate $ x\cdot e^{i\omega x - x^2}$ from $0$ to $\infty$ Can anybody help me in solving the following integral:
$$\int_0^\infty x\cdot e^{i \omega x-x^2}\,dx\quad?$$
Any help/hints will be highly appreciated.
Thanks
 A: So let's call our integral $I$ and write it like this
$$
I=\int_0^\infty xe^{ax-x^2}
$$
where in our case $a=j\omega$. 
Now let's consider the following integral $K$
$$
K=\int_0^\infty (a-2x)e^{ax-x^2}
$$
Now is easy to see (if there are no mistakes in my calculation) that
$$
K=-2I+a\int_0^\infty e^{ax-x^2}   \, \, \, (1)
$$
now is easy to solve $K$ by doing the variable substitution $y=ax-x^2$. 
Then in (1) the last term is simply a generic gaussian integral (also should be easy to solve).
Then from (1) one can get $I$. Sorry again for the mistaken answer.
A: Assuming that $j$ is the immaginary unity and $\omega\in\mathbb{R}^+$ we have:
$$\int_{0}^{+\infty}x\cdot e^{i\omega x - x^2}\,dx = e^{-\omega^2/4}\cdot\int_{0}^{+\infty}x\cdot e^{-(x-i\frac{\omega}{2})^2}\,dx$$
and:
$$\int_{0}^{+\infty}x\cdot e^{-(x-i\frac{\omega}{2})^2}\,dx = \int_{0}^{+\infty}(x-i\omega/2)\cdot e^{-(x-i\frac{\omega}{2})^2}\,dx + \int_{0}^{+\infty}(i\omega/2)\cdot e^{-(x-i\frac{\omega}{2})^2}\,dx,$$
so:
 $$\int_{0}^{+\infty}x\cdot e^{-(x-i\frac{\omega}{2})^2}\,dx = \frac{1}{2}e^{\omega^2/4}+\frac{i\omega}{2}\int_{-i\omega/2}^{-i\omega/2+\infty}e^{-x^2}\,dx,$$
and by moving the integration path in the second integral we get:
$$\int_{0}^{+\infty}x\cdot e^{i\omega x - x^2}\,dx = \frac{1}{2}+\frac{i\omega}{2}e^{-\omega^2/4}\left(\frac{\sqrt{\pi}}{2}+i\int_{0}^{\omega/2}e^{x^2}\,dx\right),$$
or:

$$\int_{0}^{+\infty}x\cdot e^{i\omega x - x^2}\,dx = \left(\frac{1}{2}-\frac{\omega\sqrt{\pi}}{4}\int_{0}^{\omega/2}e^{x^2-\omega^2/4}\,dx\right)+i\frac{\omega\sqrt{\pi}}{4}e^{-\omega^2/4}.$$

Notice that the real part depends on the non-elementary imaginary error function, or Dawson's integral.
