Evaluate the Integral \(\int_{0}^{\infty} \mathrm{e}^{-ax^2}\sin(wx) dx\)

This problem was posed at Mathematics Stack Exchange, here is my solution.

\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 \(\mathrm{erf}(-iz) = -i\mathrm{erfi}(z)\)

\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.

4 thoughts on “Evaluate the Integral \(\int_{0}^{\infty} \mathrm{e}^{-ax^2}\sin(wx) dx\)”

    1. my approach was like this i took $$ f(t)=e^{-a.t^{2}}.\sin(\omega.t) $$ then took it’s fourier transform and put $$ \omega=0 $$

      1. yeah i think it also works $$ \mathcal{F}[f(t)]=\dfrac{1}{2 \sqrt{a}}.j.\pi \left[ e^{-\dfrac{(\omega+\omega_{0})^{2}}{4a}}-e^{-\dfrac{(\omega-\omega_{0})^{2}}{4a}} \right] $$ , then if i put $$\omega_{0} =0 $$, it gives the desired result

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