Clear explanation of heaviside function fourier transform I know that fourier transform of Heaviside function is : $\hat{H}(x) = \pi \delta(\omega) + i (v.p. \frac{1}{\omega})$
How can i proof this result?
 A: The Heaviside function $H$ (seen as a tempered Distribution) is the weak limit
of the function
$$
H(x) = \mathcal{S}'-\lim_{\epsilon \rightarrow 0} H(x) e^{-\epsilon x}.
$$
To see this let $\varphi$ be an arbitrary function of rapid decay.  Split
$$
\lim_{\epsilon \rightarrow 0} \int_0^{\infty} dx e^{-\epsilon x} \varphi(x) \,
$$
the integral into positive and negative part and use monotone convergence to
take the limit under the integral.
The Fourier transform of a tempered distribution is continuous and coincides
with the usual Fourier transform on, well anything, but lets say $L^{2}$.  Thus
$$
\hat{H} = \mathcal{S}'-\lim_{\epsilon \rightarrow 0}
  \int_{0}^{\infty} dx e^{i\,k\,x -\epsilon x}
= \mathcal{S}'-\lim_{\epsilon \rightarrow 0} \frac{i}{k + i \epsilon}
$$
Now for all for all test functions $\varphi$ we have
$$
\hat{H}[\varphi] = \lim_{\epsilon \rightarrow 0}
\int_{-\infty}^{+\infty} dk \frac{i}{k + i \epsilon} \varphi(k)
$$
Let $\eta > 0$ be some small fixed constant and split the Integral
$$
\hat{H}[\varphi] = \left\{
  \int_{|k| > \eta} dk + \int_{|k| < \eta} dk
\right\} \frac{i}{k + i \epsilon} \varphi(k)
$$
In the first Integral you can take the limit under the integral, when letting
$\eta \rightarrow 0$ this becomes the principal value term.  In the second
Integral write $\varphi(k) = \varphi(0) + O(k)$.  The first Term gives the
$\delta$ function:
$$
\int_{|k| < \eta} dk \frac{i}{k + i \epsilon} \varphi(0)
= \varphi(0) (\log (i \epsilon + \eta)  - \log(i \epsilon - \eta))
\rightarrow \varphi(0) (\log(\eta) - \log(-\eta)) = -i \pi.
$$
The second term vanishes (left as an exercise :).
