Evaluation of a formula about error function

Question

I am reading a physics paper and encountered the following formula which I can't derive. \begin{equation} \int_{-\infty}^{\infty}\frac{\mathrm{d}z}{z-i\alpha}e^{-\beta^2z^2}=i\pi \,sgn(\alpha)\,e^{\alpha^2\beta^2}\, erfc(|\alpha\beta|), \end{equation} where $erfc(x)$ is complementary error function and is defined as \begin{equation*} erfc(x)=\frac{2}{\sqrt{\pi}}\int_x^\infty e^{-t^2}\mathrm{d}t. \end{equation*} The formula is valid for $\alpha,\beta\in\mathbb{R}$ and nonzero. Any comment is appreciated.

Answer 1

Write $\gamma = \alpha\beta$ and notice that

$$ \int_{-\infty}^{\infty} \frac{\mathrm{d}z}{z-i\alpha}e^{-\beta^2z^2} \stackrel{(w=\beta z)}= \int_{-\infty}^{\infty} \frac{\mathrm{d}w}{w-i\gamma}e^{-w^2} = \int_{-\infty}^{\infty} \frac{i\gamma}{w^2+\gamma^2}e^{-w^2} \, \mathrm{d}w. \tag{*} $$

Now we want to apply Plancherel's theoreom. To this end, recall that the unitary Fourier transformation $\mathcal{F}f(\xi) = \mathcal{F}[f(\bullet)](\xi) := \int_{\mathbb{R}} f(x)e^{-2\pi i \xi x} \, \mathrm{d}x$ satisfies the following identity

$$ \int_{\mathbb{R}} f(x)\overline{g(x)} \, \mathrm{d}x = \int_{\mathbb{R}} \mathcal{F}f(\xi) \overline{\mathcal{F}g(\xi)} \, \mathrm{d}\xi $$

Together with the following well-known computations

$$ \mathcal{F}[e^{-\bullet^2}](\xi) = \sqrt{\pi}e^{-\pi^2\xi^2}, \qquad \mathcal{F}\left[\frac{\gamma}{\bullet^2+\gamma^2}\right](\xi) = \pi \operatorname{sign}(\gamma) e^{-2\pi \lvert \gamma \xi \rvert},$$

it follows that

$$ \text{(*)} = i \pi \operatorname{sign}(\gamma) \int_{\mathbb{R}} \sqrt{\pi}e^{-\pi^2\xi^2}e^{-2\pi \lvert \gamma \xi \rvert} \, \mathrm{d}\xi = i \pi \operatorname{sign}(\gamma) e^{\gamma^2}\operatorname{erfc}(\gamma). $$