Here is an exercise about the properties of the Fourier transform.
Given $f(x)=\frac{e^{ -x^2}}{\sqrt{ \pi}} $ and $g(x)=xe^{ -2x^2}$, and knowing that the Fourier transform of $f(x)$ is $\hat{f}(k)=e^{ -k^2/4}$, how can I evaluate the Fourier transform $\hat{g}(k)$ of $g(x)$ exploiting the properties of the Fourier transform?
I came to see that $g(x)=x\sqrt{\pi}f(x\sqrt{2})$ but I cannot see how to go on from that as I would have to combine more than just one property of the FT.
The expected result is $\hat{f}(k)=i \sqrt{\pi} \left(\frac{1}{\sqrt{2}} e^{-k^2/8}\right)$.
Thank you in advance for the help and merry Christmas anybody!
I am suspicious about the "expected result".
We proceed by forst of all observing the function can be Fourier-Transformed for it is absolutely integrable, Lebesgue-measurable on the real line, indeed: $$\int_{\mathbb{R}} \vert x e^{-2x^2}\vert \text{d}x = \dfrac{1}{2}$$
Then proceed by calling the function $h_k(x) \equiv e^{-2kx^2}$, such that
$$g(x) \equiv -\dfrac{1}{4} \dfrac{\text{d}}{\text{d}x} h_k(x)\bigg|_{k = 1}$$
Now, let's recall the property of the Fourier Transform of a derivative:
$$\mathcal{F}\left(\dfrac{\text{d}}{\text{d}x} f(x), s\right) = - i s \mathcal{F}(f(x), s)$$
which in this case reads
$$\mathcal{F}\left(-\dfrac{1}{4} \dfrac{\text{d}}{\text{d}x} h_k(x)\bigg|_{k = 1}\right) = -\dfrac{1}{4} (- i s) \mathcal{F}\left(e^{-2kx^2}\right)\bigg|_{k = 1} = \dfrac{1}{8} i s \frac{e^{-\frac{s^2}{8 k}}}{2 \sqrt{k}}\bigg|_{k = 1} = \dfrac{1}{8} i s\ e^{-s^2/8}$$
There are different conventions which may or may not introduce additional $2\pi$ factors, though. Sometimes it's a $\sqrt{2\pi}$ factor, this depends on how you defined F.T.
Merry Christmas Ilaria!