Fourier Series Transformation

Question

I have a question regarding the following:

Compute the Fourier transform of $f(x)=xe^{-2x^2}$, $x\in\mathbb{R}$.

The Fourier Transform of $f(x)$ is given by $\mathcal{F}\{f(x)\}=\mathbb{F}(\omega) = \int_{-\infty}^{\infty}f(x)e^{-i\omega x}dx$. As a result, this is what I have:

\begin{eqnarray} \mathcal{F}\{f(x)\}=\mathbb{F}(\omega) & = &\int_{-\infty}^{\infty}f(x)e^{-i\omega x}dx\\ & =& \int_{-\infty}^{\infty}xe^{-2x^2}e^{-i\omega x}dx\\ & =& \int_{-\infty}^{\infty}xe^{-2x^2-i\omega x}dx\\ \end{eqnarray}

This is where I am stuck. Going about solving for the frequency equation, $\mathbb{F}(\omega)$.

Thank you for your time and thanks in advance for any feedback.

Answer 1

\begin{eqnarray} \mathbb{F}(\omega) & =& \int_{-\infty}^{\infty}xe^{-2x^2}e^{-i\omega x}dx\\ & =&-2i \int_{0}^{\infty}xe^{-2x^2}\sin(\omega x)dx\\ & =&(-i/4) (\pi/2)^{1/2}\omega \exp(-\omega^2/8) \end{eqnarray}

Answer 2

\begin{eqnarray} \mathcal{F}\{e^{-ax^2}\}=\mathbb{F}(\omega) & = &\int_{-\infty}^{\infty}e^{-ax^2}e^{-i \omega x}dx\\ & =& \frac{1}{\sqrt{2a}}e^{\frac{-\omega^2}{4a}}\\ \end{eqnarray}

$$\frac{d}{d\omega}(\mathcal{F}\{e^{-ax^2}\})=\frac{d}{d\omega}(\int_{-\infty}^{\infty}e^{-ax^2}e^{-i \omega x} dx)=-i\int_{-\infty}^{\infty}xe^{-ax^2}e^{-i \omega x} dx$$

Thus $$-i\int_{-\infty}^{\infty}xe^{-ax^2}e^{-i \omega x} dx=\frac{d}{d\omega}(\frac{1}{\sqrt{2a}}e^{\frac{-\omega^2}{4a}})$$