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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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3 Answers

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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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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:

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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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