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I want to calculate the Fourier transform of the unit step function (given by $\phi(x) = 1$ for $x \geq 0$ and $\phi(x) = 0$ for $x < 0$) regarded as a tempered distribution. Note that I don't care about the answer (one can basically look it up on Wikipedia) but rather the method. I believe I am making a mistake in some more involved calculations, and I think I can trace the basic error back to this calculation.

Anyway, here is my attempt so far. By the definition of the Fourier transform of a tempered distribution, $\hat{\phi}$ is the tempered distribution whose pairing with a Schwartz class function is given by:

$\langle \hat{\phi}, f \rangle = \langle \phi, \hat{f} \rangle$

The latter pairing is given by

$\int_{-\infty}^{\infty} \phi(x) \hat{f}(x) dx = \int_0^\infty \hat{f}(x)dx$

Define $F(x) = \frac{1}{2\pi} \int_{-\infty}^\infty \frac{-1}{i\omega} f(\omega) e^{-i \omega x} \, \mathrm{d}\omega$ (in the Principal Value sense), so that $F'(x) = \hat{f}(x)$ and $F(x) \to 0$ as $x \to \infty$. By the fundamental theorem of calculus:

$\int_0^\infty \hat{f}(x)dx = -F(0) = \int_{-\infty}^\infty \frac{1}{2\pi i \omega} f(\omega) d\omega$

This shows that the pairing of $\hat{\phi}$ with any test function $f$ agrees with the pairing of the PV distribution $g(\omega) = \frac{1}{2\pi i \omega}$ and $f$, so it seems to me that $\hat{\phi} = g$.

Unfortunately, this is the wrong answer. I know this because $\phi$ differs from the sign function by an additive constant, and the distributional Fourier transform of the sign function is $2g$. So I'm missing a delta distribution somewhere, and I don't see where it's supposed to come from. Can anyone help?

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1 Answer 1

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As you have defined it, it is not true that $F(x) \to 0$ as $x\to \infty$. For example, if $f(x) = \text{exp}\bigl(-x^2\bigr)$, then $$ F(x) \;=\; \frac{1}{2} \text{erf}\left(\frac{x}{2}\right) $$ so $F(x) \to 1/2$ as $x\to\infty$. As a general rule, $F(x) \to f(0)/2$ as $x \to \infty$, so $$ \widehat{\phi}(\omega) \;=\; \frac{1}{2}\delta(\omega) \,+\, \frac{1}{2\pi i\omega}. $$

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Good point - that's exactly what I'm missing! Thanks! –  Paul Siegel Jun 10 '11 at 15:31

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