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?


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 :).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.