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Dirac's delta function represents a wave whose amplitude goes to infinity as its duration in time goes to zero. It is a pulse of infinite intensity but infinitesmal duration.

Please provide an exact description of what the Fourier transform of the delta function looks like.

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special-functions? – mrf Nov 22 '12 at 21:29

The Fourier transform of the delta distribution is the (distribution corresponding to) the constant function $1$ (or possibly some other constant depending on normalization factor - but usually one wants $\mathcal F\delta = 1$ such that $\delta$ is the identity for convolution).

The physical intuition is that the "delta function" is "infinitely concentrated" in the time domain, so its Fourier transform should be "completely spread out" in the frequency domain.

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The Fourier transform of any distribution is defined to satisfy the self-adjoint property with any function from the Schwartz's class, $\mathscr{S}$ i.e. if $\delta$ is the Dirac Delta distribution and $f \in \mathscr{S}$, we have $$\langle\delta, \tilde{f} \rangle = \langle\tilde{\delta}, f \rangle $$ where $\tilde{g}$ denotes the Fourier transform of $g$ and $$\langle h,k \rangle = \int_{-\infty}^{\infty} h(y) k(x-y) dy$$ for $h,k \in \mathscr{S}$ and $\langle\delta, \tilde{f} \rangle = \tilde{f}(0)$. We have that $$\int_{-\infty}^{\infty} f(x) \exp(2 \pi i \xi x) dx = \tilde{f}(\xi)$$ Hence, $$\tilde{f}(0) = \int_{-\infty}^{\infty} f(x) dx$$ Since $$\langle\delta, \tilde{f} \rangle = \langle\tilde{\delta}, f \rangle $$ we get that $$\langle 1,f \rangle = \int_{-\infty}^{\infty} f(x) dx = \langle \tilde{\delta},f \rangle$$ for all $f \in \mathscr{S}$. Hence, $\tilde{\delta} = 1$.

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To complete Marvis' answer, the Dirac comb $\delta_0$ is a measure, and there is a well developed theory of Fourier Transform for positive definite measures and also for finite measures ($\delta_0$ is both). In this setting the Fourier Transform of $\delta_0$ is actually the Lesbegue measure on $\hat{\mathbb R}$.

Also, in this context, the meaning of the relation

$$\widehat{\delta_0} = 1_{\mathbb R}$$

is just the Bochner's Theorem for positive definite functions: The Fourier Transform of a finite positive measure is just the Positive definite function given by Bochner's Theorem.

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You wrote "the Dirac comb $\delta_0$", isn't that Ш, an infinite sum of translated delta distributions? – kahen Nov 22 '12 at 21:20
@kahen In my research area we always use $\delta$ for a Dirac measure/comb. If $x$ is a point, the measure $\delta_x$ is just $\delta_x(A)= 0$ or $1$, depending if $x \in A$ or not. Also, if $F$ is a (locally countable) countable set, by $\delta_F$ we understand $\delta_F= \sum_{x \in F} \delta_x$. I never seen that notation before, but it is possible that it is used in other areas.... But also, I am only interested in $\delta_x$ and $\delta_F$ as measures, not distributions... – N. S. Nov 22 '12 at 21:31
It's just that calling $\delta_0$ a comb seems a bit odd. I'd expect something called a comb to have more than one tooth. – kahen Nov 22 '12 at 21:34
@kahen yea, technically $\delta_0$ is a Dirac measure not a comb, but is a trivial case of a Dirac comb $\delta_F$... maybe a broken comb :)... I got used to also call it a comb, because otherwise whenever I use $\delta_F$ I would need to say: "Let $\delta_F$ be a comb or a Dirac measure....." – N. S. Nov 22 '12 at 21:36

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