# Fourier transform of the Heaviside function

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.

• Maybe take a look at this: math.stackexchange.com/questions/73922/… There are some relevant links in the answers. – Bernhard Jul 8 '15 at 8:50
• 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. – 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 $f(x)=e^{-x^2}$.
• 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 ? – Nunzio Damino Jul 8 '15 at 10:21