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Problem

I'm trying to derive the rule that $$\mathcal{F}\left[f'(x)\right] = iw \mathcal{F}(f),$$ where $\mathcal{F}(f)$ is the Fourier transform of $f$.

Partial Solution

By definition (taking the simplest convention for this proof), $$\mathcal{F}\left[f'(x)\right] = \int_{-\infty}^{\infty} f'(x) e^{-iwx} \ dx. $$ We note that $$ \int_{-\infty}^{\infty} \frac{d}{dx} \left( f(x) e^{-iwx} \right) \ dx = \int_{-\infty}^{\infty} f'(x) e^{iwx} \ dx - \int_{-\infty}^{\infty} iwf(x)e^{-iwx} \ dx. $$ It follows that $$ \int_{-\infty}^{\infty} f'(x) e^{-iwx} \ dx = \int_{-\infty}^{\infty} \frac{d}{dx} \left( f(x) e^{-iwx} \right) \ dx + \int_{-\infty}^{\infty} iwf(x)e^{-iwx} \ dx.$$ $$ =\int_{-\infty}^{\infty} \frac{d}{dx} \left( f(x) e^{-iwx} \right) \ dx+ iw\ \mathcal{F}(f). $$

So it seems that we need $$ \int_{-\infty}^{\infty} \frac{d}{dx} \left( f(x) e^{-iwx} \right) \ dx = 0, $$ for the rule to hold. This does not seem true in general. For example, $f(x) = 1$ gives $$ \int_{-\infty}^{\infty} \frac{d}{dx} \left( f(x) e^{-iwx} \right) \ dx = \int_{-\infty}^{\infty} -iw e^{-iwx} \ dx = -iw \int_{-\infty}^{\infty}e^{-iwx} \ dx. $$

I don't think this integral converges.

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For the formula to be valid you need indeed some assumptions on $f$. Being $C^1$ allows for differentiation to make sense. Then at infinity your function better go to zero.

But also $f'$ better go to zero at infinity in order for its Fourier transform to be well-defined. In any case if $|f(x)|\rightarrow 0$ as $x\rightarrow \pm \infty$ then your integral in the second last formula indeed vanishes.

Recall that an integral over ${\Bbb R}$ is interpreted as the limit of $\int_{R_1}^{R_2}$ as $R_1\rightarrow -\infty$ and $R_2\rightarrow +\infty$.

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Note that $$\mathcal{F}\left[f(x)\right] = \int_{-\infty}^{\infty} f(x) e^{-iwx} \ dx. $$ In order to make the improper integral converge, one has to add some condition on $f$, for example, $f\in L^1(\mathbb{R})$. Clearly $f(x)=1\not\in L^1(\mathbb{R})$ and hence $\mathcal{F}[1]$ is meaningless.

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