$$\int_{0}^{+\infty} \frac{e^{-r^2}}{r^2-i\gamma^2} dr = ?$$

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.

  • 1
    $\begingroup$ I can't really help you directly. All I did was to use Mathematica and I got: $$\frac{1}{2} \pi e^{-i \gamma ^2} \left(\frac{\sqrt[4]{-1} \, \text{Erf}\left((-1)^{3/4} \gamma \right)}{\gamma }+\sqrt{\frac{i}{\gamma ^2}}\right)$$ for $\Im(\gamma) + \Re(\gamma) \neq 0$ $\endgroup$ – Guilherme Thompson Apr 23 '16 at 12:53

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.


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