Fourier transform of $g(x)=x\frac{\partial f}{\partial x}$?

Question

I have a problem with the Fourier transform of the function $g(x)=x\frac{\partial f}{\partial x}$. I need the transform to be itself a function of the Fourier transform of $f(x)$ and I don't know how to go about this problem. I know that if $g(x)=\frac{\partial f}{\partial x}$, the FT would be $\hat{g}=is\hat{f}$. I also know that if $g(x)=xf(x)$, the FT would be $\hat{g}=i\frac{\partial {\hat{f}}}{\partial x}$. But what is the FT of $g(x)=x\frac{\partial f}{\partial x}$ as a function of $\hat{f}$? I would really appreciate if anyone could give any hints. Thank you.

Answer 1

Let

$$\hat{g}(k) = \int_{-\infty}^{\infty} dx \: f'(x) e^{i k x} = -i k \hat{f}(k)$$

Then

$$\begin{align}\frac{d}{dk} \hat{g}(k) &= i \int_{-\infty}^{\infty} dx \: x f'(x) e^{i k x}\\ &= -i \left [ k \frac{d}{dk} \hat{f}(k) + \hat{f}(k) \right ]\end{align}$$

Therefore

$$ \int_{-\infty}^{\infty} dx \: x f'(x) e^{i k x} = -\left [ k \frac{d}{dk} \hat{f}(k) + \hat{f}(k) \right ]$$

This of course assumes we can differentiate under the integral sign, which places restrictions on $f$ (i.e. that it vanish at least as $1/|x|$ as $x \rightarrow \pm \infty$).

Answer 2

Let $f_1(x)= f'(x)$, $f_2(x) = x f_1(x)$. Denote the Fourier transform by $\hat{f} = {\cal F} (f)$.

Then $\hat{f_1}(\omega) = {\cal F} (f_1)(\omega) = {\cal F} (f')(\omega) = i \omega \hat{f}(\omega)$, and $\hat{f_2}(\omega) = {\cal F} (f_2)(\omega) = i \hat{f}_1'(\omega)$.

Since $\hat{f_1}(\omega) = i \omega \hat{f}(\omega)$, we have $\hat{f}_1'(\omega) = i \hat{f}(\omega) + i \omega \hat{f}'(\omega)$.

So, we have $\hat{f_2}(\omega) = i \hat{f}_1'(\omega) = i(i \hat{f}(\omega) + i \omega \hat{f}'(\omega)) = -(\hat{f}(\omega) + \omega \hat{f}'(\omega))$.

This gives $\hat{g} (\omega) = -(\hat{f}(\omega) + \omega \hat{f}'(\omega))$. (Remember that $\hat{f}'$ is the derivative of $\hat{f}$ with respect to $\omega$.)

Answer 3

Perhaps it is easier if we let $h(x)=\frac{d f}{dx}$. Then, as you stated, $\hat{g}(s)=i\frac{d \hat{h}}{ds}(s)$. However since, $\hat{h}(s)=\widehat{\frac{d f}{dx}}(s)=is\hat{f}(s)$ we immediately obtain the answer $ \hat{g}(s)=-\frac{d (s\hat{f})}{ds}(s)=-\hat{f}(s)-s\frac{d \hat{f}}{ds}(s)$