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Wolfram alpha gives the answer to be

$$F(\omega)=\sqrt{2\pi}\delta(\omega)$$

Does that mean that the function is valued $\sqrt{2\pi}$ at all points in the frequency domain? I think this is reasonable because such function i.e. $f(t)=1$ in the time domain would be sum of all the harmonics of a sinusoid and hence would contain all the frequencies. Maybe no, the function isn't varying at all and hence the frequency is $0$. But then the Fourier transform should have been $\delta(0)$ instead of $\delta(\omega)$.

Someone please shed some light on this!

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  • $\begingroup$ "Does that mean that the function is valued 2π−−√2π at all points in the frequency domain?" No, completely different. $\endgroup$ – Cardinal May 6 '17 at 19:26
  • $\begingroup$ Okay then what is the right answer? $\endgroup$ – Chirag Arora May 6 '17 at 19:27
  • $\begingroup$ If you think that dirac or delta function is discrete, an imaginary situation, then it will have a value on 0 and it will be zero for the other points. In fact, the continuous form of the delta function has no definite value at zero. I think you may want to have a look on delta function definition. $\endgroup$ – Cardinal May 6 '17 at 19:29
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    $\begingroup$ There is something mentioned as "Dirac Comb" which represents the way I think DiracDelta(w) should be represented. The only difference is that it is defined at discrete points. Now this has made me even more confused ;___; $\endgroup$ – Chirag Arora May 6 '17 at 19:50
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    $\begingroup$ Here's another way of putting it. The Dirac delta distribution $\delta(x)$ has a spike whenever $x = 0$. That means that $\delta(\omega)$ has a spike only at the single point $\omega = 0$. On the other hand, $\delta(0)$ (if there were such a thing) would have a spike whenever $0 = 0$; it would have a spike everywhere! We want it to spike only at $\omega = 0$, so we use $\delta(\omega)$ instead of $\delta(0)$. $\endgroup$ – Tanner Swett May 6 '17 at 21:41
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Just use the definition. $F(w) = \int_{-\infty}^\infty e^{-iwt}dt =\delta(w)$ by how the dirac delta is defined. This is normalized to 1at zero and 0 otherwise.

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  • $\begingroup$ I don't think anyone defines $\delta(\omega) $ as the limit in the sense of (tempered) distributions $\lim_{T \to \infty}\frac1{2\pi} \int_{-T}^T e^{-i\omega t}dt=\lim_{T \to \infty} \frac{\sin(\omega T)}{\pi \omega }$, which is a theorem equivalent to the Fourier inversion theorem for tempered distributions. $\endgroup$ – reuns Aug 11 at 14:06
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I think the clearest way to see this is by noting that we have (depending on your convention for the placement of $2 \pi$ in Fourier transforms) that $$\mathcal{F}(\mathcal{F}(f(x))) = 2 \pi f(-x)$$ Taking the convention that $$\tilde{f}(k) = \int_{-\infty}^\infty e^{ikx} f(x) \; dx$$ so $$f(x) = \frac{1}{2\pi} \int_{-\infty}^\infty e^{ikx} \tilde{f}(k) \; dk$$ we get $$f(-x) = \frac{1}{2\pi} \int_{-\infty}^\infty e^{-ikx} \tilde{f}(k) \; dk = \frac{1}{2\pi} \mathcal{F(\tilde{f}(k))} = \frac{1}{2\pi}\mathcal{F}(\mathcal{F}(f(x)))$$ Note we have $$\mathcal{F}(\delta(x)) = \int_{-\infty}^\infty e^{ikx} \delta(x) \; dx = 1$$ So then $$\mathcal{F}(1) = \mathcal{F}(\mathcal{F}(\delta(x))) = 2 \pi \delta(-x) = 2 \pi \delta(x)$$ For other constants, note by linearity we have $$\mathcal{F}(c) = c \mathcal{F}(1) = 2 \pi c \delta(x)$$

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