gaussian integral from zero to infinty I want to calculate the following integral:
\begin{align} I_1'&amp;= \int^\infty_0 e^{-\alpha^2r^2/2 + iwr} \ dr \\
    &amp; = e^{-w^2/2\alpha^2} \ \int^\infty_0 e^{-\alpha^2\big{(} \frac{r}{\sqrt{2}} - \frac{iw}{\sqrt{2} \alpha^2} \big{)}^2} \ dr.\end{align}
Here I used completion of the square technique to get a Gaussian integral. However, if I substitute $u = \frac{r}{\sqrt{2}} - \frac{iw}{\sqrt{2} \alpha^2}$ the bounds go from $\frac{-iw}{\sqrt{2} \alpha^2}$ to $+\infty$ which I cannot calculate further using (half of) the formula
\begin{align} \int^{+ \infty}_{- \infty} e^{a(x+b)^2} dx = \sqrt{\frac{\pi}{a}}. \end{align}
Am I doing something wrong or is there a way ?
Matching your integral to your formula, you want $u = r / \sqrt{2}$.  Then your bounds are $0$ and $\infty$, $a = -\alpha^2$ (This minus sign makes me think something is missing from your formuala.), and $b = \mathrm{i}w/(\sqrt{2} \alpha^2)$.
Matching your integral to your formula, you want $u = r / \sqrt{2}$.  Then your bounds are $0$ and $\infty$, $a = -\alpha^2$ (This minus sign makes me think something is missing from your formuala.), and $b = \mathrm{i}w/(\sqrt{2} \alpha^2)$.