I don't understand what's wrong with my derivation below...

$\delta(t) = u'(t)$

$\mathcal{F}(\delta)(\omega) = 1 = \mathcal{F}(u')(\omega) = i\omega \times \mathcal{F}(u)(\omega)$ (since the Fourier transform of a delta is 1)

$\Rightarrow \mathcal{F}(u)(\omega) = 1/(i\omega)$

That's clearly wrong because the answer is $1/(i\omega) + \pi \delta(\omega)$, but where do I go wrong?

Isn't the Fourier transform of a derivative of a function just $i \omega$ times the Fourier transform of the function itself?

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    $\begingroup$ The problem is always summarized as "The dirac delta and alike does not mean anything outside an integral". See the lecture 12 from here Take all you can take from this. It has taught me a lot. $\endgroup$ – user13838 Oct 19 '11 at 7:24

To complement Dirk's answer:

Your inversion of differentiation can't work like that. There's a family of functions that differ by additive constants and all have the same derivative. Their Fourier transforms differ by deltas at the origin (proportional to the additive constants), so it can't be the case that you get the Fourier transform of all of them by dividing the transform of the derivative by $\mathrm i\omega$.

The catch is that multiplying by $\mathrm i\omega$ multiplies by $0$ at the origin and thus annihilates any delta that might be sitting there, and you can't reconstruct it by dividing by $\mathrm i\omega$. What you get by dividing by $\mathrm i\omega$ (if you interpret the pole appropriately) is the unique function with that derivative with zero average. In your case, that would be a step function that jumps from $-1/2$ to $1/2$.

Here's another way of looking at it: The function

$$H_\alpha(t)=\begin{cases}\mathrm e^{-\alpha t}&t\ge0\\0&t\lt0\end{cases}$$

defined in the note Dirk linked to decays at infinity and converges to the Heaviside function pointwise as $\alpha\to0$. Its Fourier transform is $\hat H(\omega)=1/(\alpha+\mathrm i\omega)$, which converges to $1/(\mathrm i\omega)$ pointwise as $\alpha\to0$, except at $\omega=0$. What you're doing corresponds to directly setting $\alpha$ to zero and using the pointwise limit as the Fourier transform. However, that ignores the fact that while the imaginary part goes to $1/(\mathrm i\omega)$, the real part has a spike at the origin, which gets narrower as $\alpha\to0$ but whose integral is independent of $\alpha$:

$$ \begin{align} \int_{-\infty}^{\infty}\frac1{\alpha+\mathrm i\omega}\mathrm d\omega &=\int_0^{\infty}\left(\frac1{\alpha+\mathrm i\omega}+\frac1{\alpha-\mathrm i\omega}\right)\mathrm d\omega \\ &=\int_0^{\infty}\frac{2\alpha}{\alpha^2+\omega^2}\mathrm d\omega \\ &=2\left[\arctan\frac \omega\alpha\right]_0^{\infty} \\ &=\pi\;. \end{align} $$

It's this spike of size $\pi$ that you're dropping when you use the pointwise limit, which has average $0$ (if you interpret the pole appropriately).

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    $\begingroup$ Huh, so the derivative rule isn't as simple as I thought. :) Thanks! $\endgroup$ – user541686 Oct 19 '11 at 17:31

Well, the Fourier-transform of the heaviside function almost always leads to confusion. Instead of an answer I would like to point you to the nice note The Fourier transform of the Heaviside function: a tragedy`by Ed Buehler which hopefully will answer your question.

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$ \displaystyle \frac{d\operatorname u(t)}{dt}=\delta(t)\\ \rightarrow\mathcal F \frac{d\operatorname u(t)}{dt}=\mathcal F \delta(t)\\ \rightarrow j\omega\ \hat {\operatorname u} (\omega) = 1\\ \rightarrow \hat {\operatorname u} (\omega) = \left.\frac{1}{j\omega\ }\right|_{\omega\ne 0} + \left.\hat {\operatorname u}(\omega)\right|_{\omega=0}\\ \left.\hat {\operatorname u}(\omega)\right|_{\omega=0}=\mathcal{F}\operatorname{DC}(u(t))=\mathcal{F}\frac 12=\frac 12 \mathcal{F}1=\frac 12 2\piδ(ω)=\piδ(ω)\\ \rightarrow \hat {\operatorname u} (\omega) = \frac 1 {j\omega} + \piδ(ω) $

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