Fourier transform of unit step?

Question

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?

Answer 1

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).

Answer 2

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.

Answer 3

I thought that it might be instructive to present a way forward that complements the solutions that have already been posted. To that end, we proceed.

---

Let $u$ be the unit step function and $\phi\in \mathbb{S}$. Then, the Fourier transform of the unit step, $\mathscr{F}\{u\}$, is defined by the functional

$$\begin{align} \langle \mathscr{F}\{u\}, \phi\rangle &=\langle u,\mathscr{F}\{\phi\},\rangle\\\\ &=\int_0^\infty \int_{-\infty}^\infty \phi(t)e^{-i\omega t}\,dt\,d\omega\\\\ &=\lim_{L\to\infty}\int_{-\infty}^\infty \phi(t) \,\frac{1-e^{-itL}}{it}\,dt\tag1 \end{align}$$ where the Fubini-Tonelli theorem justifies the interchange of integrals that resulted in $(1)$.

---

For any $\delta>0$, we write the right-hand side of $(1)$ as $$\begin{align} \langle \mathscr{F}\{u\}, \phi\rangle&=\lim_{L\to\infty}\left(\int_{|t|\le \delta} \phi(t) \,\frac{1-e^{-itL}}{it}\,dt+\int_{|t|\ge \delta} \phi(t) \,\frac{1-e^{-itL}}{it}\,dt\right)\\\\ &=\lim_{L\to\infty}\left(\int_{|t|\le \delta} \phi(t) \,\frac{1-e^{-itL}}{it}\,dt\right)+\int_{|t|\ge \delta} \frac{\phi(t)}{it}\,dt\tag2\\\\ \end{align}$$

We applied the Riemann-Lebesgue Lemma to arrive at $(2)$.

---

Next, we write the first integral on the right-hand side of $(2)$ as $$\begin{align} \int_{|t|\le \delta} \phi(t) \,\frac{1-e^{-itL}}{it}\,dt&=\phi(0)\int_{|t|\le \delta} \frac{1-e^{-itL}}{it}\,dt+\int_{|t|\le \delta} (\phi(t)-\phi(0)) \,\frac{1-e^{-itL}}{it}\,dt\\\\ &=\phi(0)\int_{|t|\le \delta} \,\frac{\sin(Lt)}{t}\,dt+O(\delta)\tag3 \end{align}$$

where we exploited the odd symmetry of $\frac{1-\cos(Lt)}{it}$ and the smoothness of $\phi$ to arrive at $(3)$.

---

Letting $L\to \infty$ in $(3)$, and substituting into $(2)$, reveals $$\begin{align} \langle \mathscr{F}\{u\}, \phi\rangle&=\pi \phi(0)+O(\delta)+\int_{|t|\ge \delta} \frac{\phi(t)}{it}\,dt\tag 4 \end{align}$$

---

Finally, letting $\delta\to0^+$ yields the coveted result $$\langle \mathscr{F}\{u\}, \phi\rangle=\pi \phi(0)+\text{PV}\int_{-\infty}^\infty \frac{\phi(t)}{it}\,dt$$

from which we deduce that $$\bbox[5px,border:2px solid #C0A000]{\mathscr{F}\{u\}(\omega)=\pi \delta(\omega)+\text{PV}\left(\frac1{i\omega}\right)}$$

Answer 4

$ \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) $

Answer 5

TL;DR: Division by $i\omega$ commits a division by zero error when $\omega = 0$.
The result is only correct in the region where we didn't divide by zero (i.e. the region where $\omega \neq 0$), and eliminates our ability to deduce anything about the case where we divided by zero.

That's all!

---

Notice this is exactly the same error as taking $(x-1)(x-2) = 0$, dividing both sides by $(x-2)$, and then concluding that $x = 1$, when in fact we may have divided by zero, and $x$ might have been $2$ instead.

Basically, we've reversed the direction of our logic.
The correct direction of the implication is $$i\omega\ \mathcal{F}(u)(\omega) = 1 \ \Leftarrow \ \mathcal{F}(u)(\omega) = 1/(i\omega)$$ and NOT $$i\omega\ \mathcal{F}(u)(\omega) = 1 \ \Rightarrow \ \mathcal{F}(u)(\omega) = 1/(i\omega)$$.

The only reason this is more confusing than basic algebra is that we already have discontinuities at the origin that we "define" in some sense, but that generalization doesn't magically make division by zero well-defined! Division by zero is still just as undefined as it ever was.

---

Note: The previously accepted answer was great, and I had accepted it for a decade. However, it had initially left me with the impression that an understanding of this issue requires careful calculus, which is not the case. Given that this question has proven to remain quite popular, I thought it would be worth showing that the fundamental problem with the logic is in fact rooted (no pun intended) in basic algebra.

Answer 6

Supplementing other good answers, there is a bit of structure that (I think) helps explain some details. Specifically, parity and (positive-) homogeneity.

Heaviside's step function $H$ is ${1\over 2}(1+\mathrm{sgn})$, with sign function. The identically-one function is even, and homogeneous of degree $0$ (in one normalization, anyway). The sign function is odd, homogeneous of degree $0$ in that same convention.

Fourier transform preserves parity, and on $\mathbb R^n$ converts positive-homogeneity of degree $s$ to pos-homog of degree $-(s+n)$. (This can be seen by looking at the defn of Fourier transform, and also by examining the commutation of FT and the Euler operator $\sum_i x_i{\partial\over \partial x_i}$.

Thus, the FT of the sign fcn is a constant multiple of an odd, degree $-1$ distribution. This is not quite integration-against-$1/x$, because of the not-local-$L^1$-ness, but must be the principal value (PV), because of uniqueness results. The constant can be determined by application to $xe^{-\pi x^2}$, whose FT is $-i$ times itself.

We certainly know that the FT of $1$ is $\delta$...

Answer 7

In addition to the other good answers, here is another way to get/check the Fourier transform of the step function. \begin{align*} \tilde{u}(\omega) = \int_{\infty}^\infty u(t) e^{-i \omega t} dt \end{align*}

For $\epsilon > 0$, we can use contour integration to evaluate the following. \begin{align*} \int_{-\infty}^\infty \frac{e^{i\omega t}}{\epsilon + i \omega} d \omega = 2\pi e^{-\epsilon t} u(t) \end{align*}

Taking the limit of $\epsilon \to 0^+$, \begin{align*} \frac{1}{2\pi} \lim_{\epsilon \to 0^+} \int_{-\infty}^\infty \frac{e^{i\omega t}}{\epsilon + i \omega} d \omega = u(t) \end{align*}

Interpreting the above as an inverse Fourier transform, \begin{align*} \tilde{u}(\omega) = \lim_{\epsilon \to 0^+} \frac{1}{\epsilon + i \omega} \end{align*}

You can use the Sokhotski-Plemelj formula to simplify further \begin{align*} \tilde{u}(\omega) = \pi \delta(\omega) - i \mathcal{P}(1/\omega) \end{align*}

which is what you have in the other answer.

Answer 8

Also, note that if $\mathcal{F}(\mathcal{f}) = F(\omega)$, then it is not necessary that $\mathcal{F}(\mathcal{f}^{'}) = j\omega F(\omega)$.

Proof: If $$\mathcal{F}(\mathcal{f}) = F(\omega) = \int_{-\infty}^{+\infty}f(t) e^{-j \omega t}\,\mathrm{d}t$$

Then, by simple integration by parts... $$\mathcal{F}(\mathcal{f^{'}}) = \int_{-\infty}^{+\infty}e^{-j \omega t}f'(t)\,\mathrm{d}t = |e^{-j \omega t}f(t)|_{-\infty}^{+\infty} + j\omega\int_{-\infty}^{+\infty}e^{-j \omega t}f(t)\,\mathrm{d}t = |e^{-j \omega t}f(t)|_{-\infty}^{+\infty} + j\omega F(\omega)$$ $$\implies \mathcal{F}(\mathcal{f^{'}}) = |e^{-j \omega t}f(t)|_{-\infty}^{+\infty} + j\omega F(\omega)$$

The convergence of $\mathcal{F}(\mathcal{f^{'}})$ to $j \omega F(\omega)$ only happens in the case when $|e^{-j \omega t}f(t)|_{-\infty}^{+\infty} = 0$. A weaker modification of this expectation is expressed as $f(+\infty) = f(-\infty) = 0$. These criterion are not followed in your case, if you consider $f(t) = u(t)$, since $u(+\infty) = 1$.

Hence $$\mathcal{F}(\delta) = \mathcal{F}(\mathcal{u}^{'}) \neq j\omega\mathcal{F}(\mathcal{u})$$