Show that $\mathcal{F}(f'(x)) = i\alpha\mathcal{F}(f(x))$

Question

Using that the Fourier transform of a function $f$ is given by

$$\mathcal{F}(f(x)) = \int_{-\infty}^{\infty}f(x)e^{-i\alpha x}dx$$

and that the inverse transform is given by

$$f(x) = \frac{1}{2\pi}\int_{-\infty}^{\infty}\mathcal{F}(f(x))e^{i\alpha x}d\alpha,$$

I want to show that $\mathcal{F}(f'(x)) = i\alpha\mathcal{F}(f(x))$.

I started with $f'(x) = \frac{1}{2\pi}\mathcal{F}(f(x))e^{i\alpha x}$ since differentiation undoes the results of integration. I'm not sure where to go from here, however.

Answer 1

$$\mathcal{F}(f')(\alpha) = \int_{-\infty}^{\infty} \frac{df}{dx}(x) e^{-i\alpha x}dx=f(x)e^{-i\alpha x}\big|_{x=-\infty}^{\infty}-\int_{-\infty}^{\infty}f(x)\frac{d}{dx}e^{-i\alpha x}=i\alpha \int_{-\infty}^{\infty}f(x)e^{-i\alpha x}dx=i\alpha \mathcal{F}(f)(\alpha)$$

Note the third equality holds when $f$ is a Schwartz function.

You should have considered $\mathcal{F}(f(x))$ in your notation as a function in $\alpha.$