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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?

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possible duplicate of Fourier transform of unit step? – Fabian Mar 19 '12 at 21:03

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 :).

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