# what is the neccessity of the 2pi coefficients in the fourier transform?

I notice that in whatever kind of expression of fourier transformation, we always have some coefficients like 2pi in the equation. And I can't see any reason there. Is there anyone who may give me some enlightment about that? And fourier tranform can be used to represent delta function. If I give up the coefficients, will that affect the deduced delta function?

• This would be more appropriate for math.se – ClassicStyle Sep 27 '16 at 4:11
• Because $\exp(i x)$ has a period of $2\pi$. – Count Iblis Sep 27 '16 at 4:13
• I believe it's an arbitrary convention, which is convenient in many contexts but fundamentally not derivable. Of course, if you omit it, certain other properties of the Fourier transform will have ugly $2 \pi$'s in them. I'm not even sure it's a universal convention. It fundamentally depends on how you want the Fourier transform to behave. – QuantumFool Sep 27 '16 at 4:31

## 1 Answer

The reason usually given is that we should want the following equation to hold,

$$\int_{\mathbb{R}}\mathcal{F}\mathcal{F}^*=\int_{\mathbb{R}}ff^*$$

Where $\mathcal{F}$ is the Fourier transform of $f$, and ${}^*$ is complex conjugation. If you work out the integrals you will see the need for a normalization factor.