I am currently working with Fourier integrals and transformations. Now, I have a task that I have done, but my answer is different by one single detail, which is a minus sign. I will post what I have done and the problem itself. There are parts (a) and (b), and my problem is (b). For (b) I will post all of my calculations,
(a) Find the function $f(x)$ that satisfies $$\int_{-\infty}^{\infty}f(x-t)\,\mathrm{e}^{-2t^2}\,\mathrm{d}t=\mathrm{e}^{-x^2}$$ for $-\infty <x<\infty$ by using the Fourier transform.
I did the calculations and ended up with $f(x)=\dfrac2{\sqrt\pi}\mathrm{e}^{-2x^2}$. According to my book this is the correct answer.
(b) Let the functions $f(\cdot)$ and $g(\cdot)$ be defined as $f(x)=\mathrm{e}^{-x^2}$ and $g(x)=x\mathrm{e}^{-x^2}$ for $-\infty <x<\infty$. Use the Fourier tranform to show that $$(f\star g)(x)=-\frac{\mathrm{i}}4\int_{-\infty}^\infty\omega\mathrm{e}^{\frac{-\omega^2}2}\,\mathrm{e}^{\mathrm{i}\omega x}\,\mathrm{d}x.$$
My calculations:
I use the following formula that I have in my book. $$(f\star g)(x)=-\frac{\mathrm{i}}4\int_{-\infty }^\infty f(\omega)g(\omega)\,\mathrm{e}^{iwx}\,\mathrm{d}x.$$
$f(\omega)$ and $g(\omega)$ are the Fourier transforms to $f$ and $g$. Now, I use Fourier tables in my book and Wolfram Alpha and get: $$(f\star g)(x)=\frac{\mathrm{i}}4\int_{-\infty}^\infty\frac1{\sqrt2}\,\mathrm{e}^{\frac{-\omega^2}4}\frac{\mathrm{i}\omega}{2\sqrt2}\,e^{\frac{-\omega^2}4}\,\mathrm{e}^{\mathrm{i}\omega x}\,\mathrm{d}\omega=\frac{\mathrm{i}}4\int_{-\infty}^\infty\omega\,\mathrm{e}^{\frac{-\omega^2}2}\,\mathrm{e}^{\mathrm{i}\omega x}\,\mathrm{d}\omega.$$
As you can see the only difference (that I can see) is the minus sign in front of the integral. I really need help to see what I am doing wrong here. Thank you so much to all of you!
Let me see the Fourier transform operation (which I indicate with $\mathcal{F}$). You have (just to recall):
$$f(x) = e^{-x^2}$$
$$\mathcal{F}(e^{-x^2}) = \frac{e^{-\frac{\omega^2}{4}}}{\sqrt{2}}$$
Then your function $g$ is
$$g(x) = xe^{-x^2}$$
To calculate its Fourier transform, you can observe that
$$g(x) = -\frac{1}{2}\frac{\text{d}}{\text{d}x}e^{-x^2}$$
Namely it's related to $f$, hence
$$\mathcal{F}(g(x)) = \mathcal{F}\left(-\frac{1}{2}\frac{\text{d}}{\text{d}x}e^{-x^2}\right) = -\frac{1}{2}\mathcal{F}\left(\frac{\text{d}}{\text{d}x}e^{-x^2}\right) = -\frac{-i\omega e^{-\frac{\omega^2}{4}}}{2\sqrt{2}}$$
Indeed by definition:
$$\mathcal{F}\left(\frac{\text{d}}{\text{d}x}h(x)\right) = -i\omega\mathcal{F}(h(x))$$
BUT according to that, the minus sign should vanish. Probably it's a book error, because... maths works! :D