If the Fourier Transform of $f(x)$ is known, can one deduce the Fourier Transform of $|x|f(x)$ ?
I've been trying to find the Fourier Transform of $|x|^{7/6}K_{-1/6}(x)$. I know the transform of $|x|^{a}K_{-a}(x)$ and thought I could make use of this fact.
Note that $|x|f(x)=\text{sgn}(x)xf(x)$. Therefore, the convolution theorem shows that
$$ \mathcal{F}(|x|f(x))(\omega)=\mathcal{F}(\text{sgn}(x))(\omega)\ast\mathcal{F}(xf(x))(\omega)=\mathcal{F}(\text{sgn}(x))(\omega)\ast \frac{d}{d\omega} \mathcal{F}(f)(\omega)$$
I don't know how much you know about Fourier transforms, but the Fourier transform of $sgn(x)$ is a constant times $1/\omega$. This function is not locally integrable, but the convolution with it can still be defined, and is called Hilbert-transform in case you want to look it up.