How to integrate $\int_0^\infty e^{-\frac{1}{2\sigma^2}x^2}\sin \omega x~dx$

Question

How to deal with this integral without Taylor series expansion: $$\int_0^\infty e^{-\frac{1}{2\sigma^2}x^2}\sin \omega x~dx$$

Answer 1

Let $a=-\frac{1}{2 \sigma ^2}$

\begin{align} I &= \int\limits_{0}^{\infty} \mathrm{e}^{-ax^2 + iwx} dx \\ \tag{a} &= \mathrm{e}^{-\frac{w^2}{4a}} \int\limits_{0}^{\infty} \mathrm{e}^{-a(x-\frac{iw}{2a})^2} dx \\ &= \mathrm{e}^{-\frac{w^2}{4a}} \int \mathrm{e}^{-az^2} dz \\ &= \mathrm{e}^{-\frac{w^2}{4a}} \frac{1}{2}\sqrt{\frac{\pi}{a}} \mathrm{erf}(z\sqrt{a}) \\ &= \mathrm{e}^{-\frac{w^2}{4a}} \frac{1}{2}\sqrt{\frac{\pi}{a}} \mathrm{erf}\left(x\sqrt{a} - \frac{iw}{2\sqrt{a}}\right) \Big|_{0}^{\infty} \\ \tag{b} &= \mathrm{e}^{-\frac{w^2}{4a}} \frac{1}{2}\sqrt{\frac{\pi}{a}} \left[ 1 + i\mathrm{erfi}\left(\frac{w}{2\sqrt{a}} \right) \right] \end{align}

a. Complete the square.

b. $\lim_{ x \to \infty} \mathrm{erf}(x + ic) = 1$ for finite $c$, $c \in \mathbb{C}$ and $\mathrm{erf}(-iz) = -i\mathrm{erfi}(z)$

Now we have \begin{equation} \int\limits_{0}^{\infty} \mathrm{e}^{-ax^2}\sin(wx) dx = \mathfrak{I}(I) = \mathrm{e}^{-\frac{w^2}{4a}} \frac{1}{2}\sqrt{\frac{\pi}{a}} \mathrm{erfi}\left(\frac{w}{2\sqrt{a}} \right) = \frac{1}{\sqrt{a}} \mathrm{F}\left(\frac{w}{2\sqrt{a}} \right) \end{equation}

Where \begin{equation} \mathrm{F}(x) = \frac{\sqrt{\pi}}{2} \mathrm{e}^{-x} \mathrm{erfi}(x) = \mathrm{e}^{-x^2} \int\limits_{0}^{x} \mathrm{e}^{z^2} dz \end{equation} is Dawson's integral.