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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, 2017 at 19:26
  • $\begingroup$ Okay then what is the right answer? $\endgroup$
    – Chirag Arora
    May 6, 2017 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, 2017 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, 2017 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, 2017 at 21:41

4 Answers 4

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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, by linearity, we have $$\mathcal{F}(c) = c \mathcal{F}(1) = 2 \pi c \delta(x)$$

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You can derive the answer very easily with the general formula for the fourier series of a complex exponential:

$\mathcal F(e^{jw_0 t}) = 2\pi \delta(w-w_0)$

This identity is very intuitive: Since a complex exponential only has one frequency ($w_0$), its fourier transform only has one pulse at that frequency[1].

Set $w_0 = 0$ and you get:

$\mathcal F(e^{0}) = \mathcal F(1) = 2\pi \delta(w-0) = 2\pi \delta(w)$

[1] Here is a proof: https://staff.fnwi.uva.nl/r.vandenboomgaard/SignalProcessing/FrequencyDomain/CTNP.html#complex-exponential

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I know that this has been answered, but it's worth noting that the confusion between factors of $2\pi$ and $\sqrt{2\pi}$ is likely to do with how you define the Fourier transform in the first place.

Wolfram Alpha seems to be taking the definition that involves placing a factor of $1/\sqrt{2\pi}$ on both the transform and the inverse transform (as is sometimes done in physics), rather than placing the factor $1/2\pi$ entirely on the inverse transform.

If you take the former approach, I believe you end up with the desired result, $\mathcal{F}(1)=\sqrt{2\pi}\delta(k)$, since (compare LtSten's answer) then $\hat{f}(k)$ will originally contain a factor of $1/\sqrt{2\pi}$ (i.e. $\mathcal{F}(\delta(x))=1/\sqrt{2\pi}$), meaning that $\mathcal{F}(1)=\sqrt{2\pi}\delta(x)$ is consistent with $\mathcal{F}(\mathcal{F}(f(-x)))=2\pi f(x)$.

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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, 2019 at 14:06
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    $\begingroup$ See math.stackexchange.com/questions/1343859/… $\endgroup$
    – alvitawa
    Nov 15, 2020 at 17:43
  • $\begingroup$ Yes, all of this can be made very rigorous using distribution theory. But most people who use the FT in the "real world" (whatever that actually is) don't bother with that much rigor. Statics has a good answer on the linked question which explains the reasoning behind this definition in a simple way. $\endgroup$
    – JMJ
    Nov 16, 2020 at 18:33
  • $\begingroup$ This is a math forum ... people come here to understand what is the good rigorous explanation $\endgroup$
    – LL 3.14
    Nov 3 at 23:58
  • $\begingroup$ Not necessarily. Many people come just wanting the answer to or a way to think about a math problem they don't understand, and a rigorous answer by current standards might just make things more obscure. Many of the most useful results in mathematics were not discovered or first presented in a "rigorous" way. In the current case, an actually rigorous answer would require the first few chapters of a distribution theory book, which is probably not what the asker was looking for. $\endgroup$
    – JMJ
    Nov 4 at 0:44

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