# Fourier transform normalization

I can't understand the following integral, someone can help? $$\int dk e^{ikx} = \delta(x)2\pi$$

-
 To be more precise: I can't understand where the $2\pi$ comes from and where the delta function is located (where does the $2\pi$ "concentrate" along the x axis?) – Lorenzo Feb 28 at 11:42 It is concentrated near $x=0$. – Vladimir Kalitvianski Feb 28 at 12:50

## migrated from physics.stackexchange.comFeb 28 at 15:28

It means this integral behaves as a delta-function when integrated over $x$ with another regular function.

EDIT: Factor $2\pi$ does not depend on "normalization". It is a strict value. Integrate this function over $x$ within $\pm\varepsilon$ and you will obtain: $$2\int_{-\infty}^{\infty} \frac{\sin(z)}{z}dz=2\pi$$

-
 does this means that the integral has a result only if x=k? If so, why this become $$2\pi$$? – Lorenzo Feb 28 at 11:57 @Lorenzo: No, $x=0$, not $k$. If you integrate, for simplicity this integral over $x$ within $\pm\epsilon$, you will obtain this $2\pi$. – Vladimir Kalitvianski Feb 28 at 12:25 I knew this 2pi wasn't due to normalization, but the normalization is due to this integral! Thank you for clearing my mind! – Lorenzo Feb 28 at 13:14

Do the integral putting in some limits $\pm a$ that we'll later take to infinity. Then we get:

$$\begin{split} \int_{-a}^{a} e^{ikx}dx &= \frac{1}{ik}[e^{ikx}]_{-a}^{a} \\ &= \frac{1}{ik}\left(e^{ika}-e^{-ika}\right) \\ &= \frac{1}{ik}2i\sin{ka} \\ &= 2a\frac{\sin{ka}}{ka} \\ &= 2a \space \text{sinc}(ka) \end{split}$$

where sinc(x) is $sin(x)/x$. As x goes to infinity $sinc(x)$ goes to a delta function, so when we take our integration limits to $\pm\infty$ we'll end up with a delta function.

As for the $2\pi$, there are various conventions for writing Fourier transforms and they tend to scatter factors of $\pi$ around. For example, Wikipedia gives the Fourier transform as:

$$\hat{f(k)} = \int f(x) \space e^{i \space 2\pi kx} dx$$

For example this makes intuitive sense if $k$ is a frequency, and of course Fourier transforms frequently involve time/frequency analyses. Anyhow the $2\pi$ in your expression can be justified by making the substitution $k = 2\pi l$ to put the integral into the standard form, in which case we get:

$$\int dk \space e^{ikx} = 2\pi \int dl \space e^{i2\pi l x}$$

and then taking the standard integral to be $\delta(x)$.

-
Ah, I see I and Kitchi have just both been downvoted. Are you going to say why you downvoted, or is this just another drive by downvote? – John Rennie Feb 28 at 12:41
It is I who downvoted because $2\pi$ is a strict result rather than a matter of convention. – Vladimir Kalitvianski Feb 28 at 12:48

Actually, as you can see the delta function is defined to be the following integral -

$$\delta(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty}e^{ikx}dx$$

Why the integral of $e^{ikx}$ goes to the delta is expressed in other answers to this question. The $2\pi$ is just a normalization constant, presumably one that is put there for conventional reasons.

The normalization itself isn't that important, as long as the same convention is stuck to throughout whatever calculation is being done. For example, the fourier transform of the delta function is $1$. But if your normalization is different, it may well turn out to be $2\pi$ or something else. The physics of the situation will still remain the same regardless of your convention.

-
 The delta-function cannot be "defined" this way if the integral over $x$ including the point $x=0$ is not equal to unity. Normalization of the Fourier image $g(k)$ (convention, definition) has nothing to do with calculations of integrals. – Vladimir Kalitvianski Feb 28 at 13:14