2
$\begingroup$

so the Fourier transform of the Kronecker Delta function is (up to sign conventions / normalisation)

$$\int_{-\infty}^\infty dt\; e^{i t \omega} = \delta(\omega).$$

Can one say anything about the half-Fourier transform

$$\int_0^\infty dt\; e^{i t \omega}$$

and its relation to the Kronecker Delta function?

Specifically, I have come across the relation

$$\int_0^\infty dt\; \textrm{Re}[e^{i t \omega}] \;\;\Big(=\int_0^\infty dt \cos( t \omega)\Big) \;\;= \delta (\omega),$$

but cannot seem to prove this. Any ideas?

$\endgroup$
  • $\begingroup$ It is strange to see \textrm{cos} with manually added space before it. If you just write \cos then it appears without italics and with the conventional spacing before and after it in expressions like $a\cos b$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Oct 16 '14 at 16:50
  • 1
    $\begingroup$ I believe that this question and its answer completely answer your question. $\endgroup$ – Matt L. Oct 17 '14 at 7:14
  • $\begingroup$ @MattL,Thanks for the link. Much appreciated. $\endgroup$ – Can't integrate Oct 20 '14 at 7:42
2
$\begingroup$

I think a factor $\frac 12$ is missing in the last equation.

$$\int_{\mathbb R}e^{itw}dt=\delta(w),$$hence $$\Re \int_{\mathbb R}e^{itw}dt= \int_{\mathbb R}\Re e^{itw}dt=\Re \delta(w) = \delta(w).$$

Then again, $\Re e^{itw} = \cos (tw) = \cos (-tw) =\Re e^{-itw} $, hence $$ \int_{\mathbb R}\Re e^{itw}dt=2\int_{0}^\infty\Re e^{itw}dt = \delta(w),$$therefore $$ \int_{0}^\infty\Re e^{itw}dt = \frac 12 \delta(w).$$

$\endgroup$
  • $\begingroup$ Thank you... That clears it up. $\endgroup$ – Can't integrate Oct 16 '14 at 16:14

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.