As you can see from the title I want to calculate the Fourier transform of the Heaviside function $u(t)$. Proven the the Heaviside function is a tempered distribution I must evaluate:

$$ \langle F(u(t)), \varphi \rangle \qquad \varphi \in S_{\xi} $$

Then I use the following property of the Fourier transform:

$$ F(T^{(n)}) = (2 \pi i)^n \xi^n F(T) $$

In my case, as we have that $u' = \delta$:

$$ F(\delta) = 2 \pi i \xi F(u) $$

In this way I proved that $F(u)$ it's a solution of the following division problem for tempered distribution:

$$ \begin{cases} \xi T = \frac{1}{2 \pi i} \\ T \in S' \end{cases} $$

If I find another solution of the problem, then the two solution will differ of $c \delta \ , c \in \mathbb{C}$. Let's prove that $p.v. \frac{1}{2 \pi i \xi}$ it's a solution for the problem.

$$ \langle p.v. \frac{1}{2\pi i \xi}, \varphi\rangle = \frac{1}{2\pi i}\ p.v. \int_{\mathbb{R}} \frac{\xi \varphi(\xi)}{\xi} d\xi = \frac{1}{2 \pi i} \int_{\mathbb{R}} \varphi(\xi) d\xi = \langle \frac{1}{2 \pi i} , \varphi \rangle $$

Then we conclude that:

$$ F(u) = p.v.\ \frac{1}{2\pi i \xi} + c \delta \qquad c \in \mathbb{C} $$

Now, there is the problem. How can I set the value of c ? Thanks in advance.

  • $\begingroup$ Maybe take a look at this: math.stackexchange.com/questions/73922/… There are some relevant links in the answers. $\endgroup$ – Bernhard Jul 8 '15 at 8:50
  • $\begingroup$ I've already seen the links, but I've only understood intuitively that $c = \frac{1}{2}$, and I don't know how to prove it through my reasoning. $\endgroup$ – Nunzio Damino Jul 8 '15 at 9:38

If you know your distribution up to a constant, a good way to fix the constant is to pair the distribution against a test function $f$. For simplicity, we can pick such an $f$ that both $f$ and $F(f)$ are real and symmetric (a Gaussian, for example). Now calculate $\langle F(u),{F(f)}\rangle$ in two ways: $$ \langle F(u),F(f)\rangle = p.v.\int\frac1{2\pi i\xi}F(f)(\xi)d\xi+c\langle\delta,F(f)\rangle = cF(f)(0) = c\int_{-\infty}^\infty f(x)dx. $$ The principal value integral vanishes because $F(f)(\xi)$ is symmetric and $1/\xi$ is antisymmetric. On the other hand, $$ \langle F(u),F(f)\rangle = \langle u,f\rangle = \int_0^\infty f(x)dx = \frac12\int_{-\infty}^\infty f(x)dx. $$ These two have to be equal, so $c=1/2$.

Note that we did not even need an explicit function $f$, just the knowledge that there is a function with suitable symmetries. If you prefer something more explicit, you can choose $f(x)=e^{-x^2}$.

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  • $\begingroup$ Only one question: the integral: $$ p.v \int_{\mathbb{R}} \frac{F(f)(\xi)}{2 \pi i \xi}\ d\xi $$ It's 0 for the Riemann-Lebesgue lemma ? $\endgroup$ – Nunzio Damino Jul 8 '15 at 10:21
  • $\begingroup$ @NunzioDamino, you can't have line breaks in a comment (or at least you can't produce them with enter). I'll add a clarification to my post that was meant to be there in the first place. $\endgroup$ – Joonas Ilmavirta Jul 8 '15 at 10:23
  • $\begingroup$ Ok, seems fine to me. Thanks ! $\endgroup$ – Nunzio Damino Jul 8 '15 at 10:27
  • $\begingroup$ @NunzioDamino, you are welcome! In problems like this, you only need one test function, and with a good choice the calculations are almost trivial. I didn't want to do an integral by hand... $\endgroup$ – Joonas Ilmavirta Jul 8 '15 at 10:29

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