# The inverse Fourier transform of $1$ is Dirac's Delta

From the definition of the Dirac delta $\delta_0$ one can infer that its Fourier transform is identically equal to $1$. But going in the other direction is not as straightforward. How can one show that $f(x) = \frac{1}{2\pi}\int_{-L}^{L}e^{ikx}dk$ behaves like a Dirac delta function, taking $L \to \infty$ after integration?

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Let me give a calculus-based "flavour" of the proof that is OK on the physics level. Denote your function by $f_L(x)$ (since it depends on $L$), so

$$f_L(x) = \frac{ \sin L x}{\pi x}.$$

This function is even, peaked around $x=0$ and decays as $O(1/x)$ to both sides. Around the origin,

$$f_L(x) = \frac{L}{\pi} \left[1 - \frac{L^2 x^2}{3!} + \frac{L^4 x^4}{5!} + \ldots\right].$$ This looks like a Gaussian (just plot it). In fact, if you write $$h_L(x) = \frac{L}{\pi} \exp\left[- \frac{x^2}{2 \sigma^2} \right]$$ then you can show that $$f_L(x) - h_L(x) = O(x^4).$$ if you pick $\sigma = \sqrt{3}/L$. So up to $O(x^4)$ terms, $f_L(x)$ is bell-shaped and has a width $\sim 1/L.$ But this is precisely what we want from a delta function: it sniffs out the value of the function you integrate it with at the origin.

We still need to check the prefactor. It suffices to do this for any function we like, say $$Z(x) = \exp(-x^2).$$ Now $$\int Z(x) f_L(x) dx = \text{Erf} \frac{L}{2}$$ where $\text{Erf}$ is the error function. But $$\lim_{L \rightarrow \infty} \int Z(x) f_L(x) dx = \lim_{L \rightarrow \infty} \text{Erf} \frac{L}{2} = 1 = Z(0)$$ so we're done.

[Edit: you should also check that

$$\int f_L(x) dx = 1$$ for all $L$.]

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Well, I think you would also have to discuss that the negative regions of $(\sin x)/x$ don't combine to other distributions like $\delta'$ etc. But otherwise OK, I upvote it. – Luboš Motl Sep 20 '13 at 11:12

One way to show this might be along the lines of the following. Recognize that

$f(x) = \frac{1}{2\pi}\int_{-L}^{L}e^{ikx}dk = \dfrac{L}{\pi}sinc(Lx)$

is the inverse Fourier transform of a rectangular function of amplitude 1, width L, and symmetric about k=0.

As L goes to infinity, the rectangular function goes to 1 for all k.

But this is just the Fourier transform of the Dirac delta function.

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This is meant as a constructive remark, but you might want to re-read your argument. We write $f$ as the inverse Fourier transform of the box function of height 1 (or $1/2\pi$) and width $L$, calculate what $f$ looks like in real space, transform back to $k$-space and conclude. If you are confident that the box function is the FT of the Dirac delta, you don't need all these intermediate steps. – Vibert Sep 20 '13 at 14:26
@Vibert, I've edited my answer to reflect your suggestion. – Alfred Centauri Sep 20 '13 at 14:48

There is a simple proof without having to take a limit (therefore I don't know how much this helps you)

Performing a Fourier transformation on a function and then an inverse Fourier transformation should yield the same function: $$f(x)=\iint\frac{dkdx'}{2\pi}e^{ik(x-x')}f(x')$$ now substitute $f(x)=\delta(x)$ and you obtain the desired result $$\delta(x)=\int\frac{dk}{2\pi}e^{ikx}$$

P.S.:
Of course it is assumed that it is already proven that the Fourier transformations works thit way and that a delta-function exists which behaves in exactly that manner.
Also the delta-function is not a function but a distribution, so maybe this proof is mathematically not 100% waterproof. Feel free to try to use the Dirac-measure instead ☺

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