$$\int_{0}^{+\infty} \frac{e^{-r^2}}{r^2-i\gamma^2} dr = ?$$
I tried the normal semicircular contour integrals, but there is always a problem with the exponential when I close the contour.
This post is related, but I can't apply the method used to this case.
Did you try a Fourier Transform approach? The Fourier transform of $e^{-r^2}$ is still a gaussian function while the inverse Fourier Transform of $\frac{1}{r^2-i\gamma^2}$ is a Laplace distribution, hence the problem boils down to computing $$ I(z)=\int_{0}^{+\infty}\exp\left(zx-x^2\right)\,dx $$ for some fixed $z\in\mathbb{C}$. By completing the square we get: $$ I(z) = \frac{\sqrt{\pi}}{2}e^{-z^2/4}\left(1+\text{Erf}\left(\frac{z}{2}\right)\right)$$ that agrees with Guilherme Thompsons's computation in the comments.