Half-Fourier transform, relation to Delta function

Question

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?

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 $a\cos b$. ${}\qquad{}$ — Michael Hardy, Oct 16 '14 at 16:50
I believe that this question and its answer completely answer your question. — Matt L., Oct 17 '14 at 07:14

score 4 · Accepted Answer · answered Oct 16 '14 at 16:03

4

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 $$ \int_{0}^\infty\Re e^{itw}dt = \frac 12 \delta(w).$$

answered Oct 16 '14 at 16:03

TZakrevskiy

22,980

Thank you... That clears it up. – Can't integrate Oct 16 '14 at 16:14

Half-Fourier transform, relation to Delta function

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