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.
Best Answer
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 $f(x)=e^{-x^2}$.