# Half-Fourier transform, relation to Delta function

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 7:14
• @MattL,Thanks for the link. Much appreciated. – Can't integrate Oct 20 '14 at 7:42

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