Derivation of a fourier transfrom of a derivative multiplied by a linear function

Question

How can the following identity be derived $$F\left\{\frac{d(xf(x))}{dx}\right\}=-k\frac{d\hat{f}(k)}{dk}$$

where $$\hat{f}(k)=F\left\{f(x)\right\}=\frac{1}{2\pi}\int_{-\infty}^{\infty}f(x)e^{-ikx}dx$$

Answer 1

I assume here that $f$ is a continuous function such that $x\,f(x)\to0$ sufficiently quickly as $x\to\pm\infty$. We have $$\begin{align} F\left\{\frac{\text{d}}{\text{d}x}\,\big(x\,f(x)\big)\right\}(k)&=\frac{1}{2\pi}\,\int_{x=-\infty}^{x=+\infty}\,\exp(-\text{i}kx)\,\text{d}\big(x\,f(x)\big) \\ &=-\frac{k}{2\pi}\,\int_{-\infty}^{+\infty}\,(-\text{i}x)\,f(x)\,\exp(-\text{i}kx)\,\text{d}x\,, \end{align}$$ where we have applied the technique of integration by parts. In this part, continuity of $f$ ensures that the integral $\int_{-R}^{+R}\,(-\text{i}x)\,f(x)\,\exp(-\text{i}kx)\,\text{d}x$ can be evaluated at each $R>0$, and the condition that $x\,f(x)\to 0$ quickly as $|x|\to\infty$ guarantees that the limit of this integral as $R\to\infty$ exists and is finite for each $k$.

Next, observe that $$\begin{align} F\left\{\frac{\text{d}}{\text{d}x}\,\big(x\,f(x)\big)\right\}(k)&=-\frac{k}{2\pi}\,\int_{-\infty}^{+\infty}\,(-\text{i}x)\,f(x)\,\exp(-\text{i}kx)\,\text{d}x\\ &=-\frac{k}{2\pi}\,\int_{-\infty}^{+\infty}\,\frac{\text{d}}{\text{d}k}\big(f(x)\,\exp(-\text{i}kx)\big)\,\text{d}x \\ &=-k\,\frac{\text{d}}{\text{d}k}\,\left(\frac{1}{2\pi}\,\int_{-\infty}^{+\infty}\,f(x)\,\exp(-\text{i}kx)\,\text{d}x\right) \\&=-k\,\frac{\text{d}}{\text{d}k}\,\hat{f}(k)\,, \end{align}$$ where we have implemented Leibniz Integral Rule. Continuity of $f$ means that the derivative commutes with the integral, and it also means that $\hat{f}$ is differentiable.

However, you may still have a meaningful question even if $f$ is not continuous (using weak derivatives instead) or does not vanish very quickly, but the Fourier transform can end up being a distribution (i.e., not a function). Nonetheless, I am not certain which class of $f$ will be a good choice.