# Fourier Transform of Absolute Value

If you ask Mathematica to provide the Fourier Transform of a singular functions it is likely to provide an answer that while nearly correct, is technically incorrect and it will do so without a word of warning. Are there any ways to guard for this other than telling the user to beware?

Below I show what happens when you ask for the Fourier Transform of the absolute value function.

Mathematica defines the Fourier transform, $F(\omega)$, of the function $f(t)$, to be: $$F(w) = \frac{1}{\sqrt{2 \pi}}\int_{-\infty}^\infty Exp({i w t})\ f(t)\ dt$$ which is a standard normalization. If you ask for the Fourier Transform of $|t|$,

FourierTransform[Abs[t], t, w]


Mathematica returns: $$F(w)=-\frac{\sqrt{2/\pi}}{w^2}$$ which is fine except at $w=0$ where it gives $F(0)=-\infty$. For $w=0$ we have, by definition, $$F(0) = \frac{1}{\sqrt{2 \pi}}\int_{-\infty}^\infty |t| \ dt = +\infty \ .$$

Thus it is WAY OFF at $w=0$. The correct transform is of the form:

$$A \delta(w) - \frac{\sqrt{2/\pi}}{w^2}$$ where $A$ is a divergent integral. I expect that $|t|$ is not the only function for which Mathematica returns an incorrect Fourier Transform with no warning.

• Usually GenerateConditions should work for this kind of thing. However, at least on V9 FourierTransform[Abs[t], t, w, GenerateConditions -> True] returns unevaluated Jan 8 '15 at 4:37
• With v10: SameQ @@ FourierTransform[{Abs[t], t Sign[t], Sqrt[t^2], Piecewise[{{-t, t < 0}, {t, t >= 0}}]}, t, w] returns True. Although, using GenerateConditions -> True evaluates with t Sign[t]; however, returns the same result, i.e., no conditions. Jan 8 '15 at 5:56
• Using the identity $|t|=\int_0^t Sign[s] ds$ with GenerateConditions also returns unevaluated. I am using Version 10.0.1.0.
– JEP
Jan 8 '15 at 16:44
• The identity $$|t| = \int_{-\infty}^\infty \frac{1-\cos{\alpha t}}{\alpha^2}\ d\alpha$$ with GenerateConditions also returns unevaluated. It gives the same result as $|t|$ for the transform.
– JEP
Jan 8 '15 at 17:24
• The last identity can be used to obtain the correct transform. Insert it into the definition of the transform and switch the order of integration will yield an additional term $\propto \delta(\omega) \int_{-\infty}^\infty \frac{d\omega'}{\omega'^2}$
– JEP
Jan 8 '15 at 19:19