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Where does the $\frac{1}{2 \pi}$ come from in this pair?

Please try to explain the Plancherel's theorem and the Parseval's theorem!

$ X(j \omega)=\int_{-\infty}^\infty x(t) e^{-j \omega t}d t$

$ x(t)=\frac{1}{2 \pi} \int_{-\infty}^{\infty} X(j \omega)e^{j \omega t}d \omega $

user88004
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    It comes from the fact that $\int_{-\infty}^\infty e^{-x^2/2},dx = \sqrt{2\pi}$ (and thus from the fact that $2\pi$ is the circumference of a unit circle). So somewhere you must put the $2\pi$. You can put it on the inverse, as physicists do, or split it between the Fourier transform and the inverse, as is done in part of mathematics, or you can put it in the exponent of the Fourier-kernel, as is done in other parts of mathematics. Each has advantages and disadvantages. – Daniel Fischer Jul 28 '13 at 09:23
  • Why is '${-j \omega t}$' replace by ${-x^2/2}$? And if the result is $sqrt(2pi)$ why isn't it $1/sqrt(2pi)$ instead of $\frac{1}{2 \pi}$? – user88004 Jul 28 '13 at 11:27
  • It's not replacing $-j\omega t$ with $-x^2/2$. $\varphi(x) = e^{-x^2/2}$ is an eigenfunction of the Fourier transform (except for the variant with the $2\pi$ in the exponent, there it's $e^{-\pi x^2}$). $\varphi(0) = 1$, and $\mathcal{F_{phys}}\varphi = \int_\mathbb{R} \varphi(x), dx$, hence for the physicist's transform, the eigenvalue is $\sqrt{2\pi}$. The other integral gives a further factor of $\sqrt{2\pi}$, so altogether you need to eliminate a factor of $2\pi$. – Daniel Fischer Jul 28 '13 at 11:42
  • Physicists (often) do that by putting the entire factor on the inverse, mathematicians prefer putting $\frac{1}{\sqrt{2\pi}}$ on each integral so that the Fourier transform is an isometry on $L^2$. – Daniel Fischer Jul 28 '13 at 11:43
  • @DanielFischer Could you try to explain the Plancherel's theorem and the Parseval's theorem first? – user88004 Jul 31 '13 at 16:52
  • @DanielFischer: isn't $|x|^{-1/2}$ also an eigenfunction of the FT? – Ron Gordon Jul 31 '13 at 16:56
  • @RonGordon As a tempered distribution? Could be, I don't know. It's not in any $L^p(\mathbb{R}^n)$, so you'd need some extended theory for that. – Daniel Fischer Jul 31 '13 at 17:54

1 Answers1

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I am going to "derive" this heuristically, because the question is concerned with the origin of the $1/(2 \pi)$ factor. Questions about integrability, order of integration, limts, etc., are to be smoothed over here (unless, of course, I have erred somewhere in the derivation - then all bets are of course off).

Consider the FT

$$\hat{f}(k) = \int_{-\infty}^{\infty} dx \, f(x) \, e^{i k x}$$

Now let's assume that the inverse FT may be written as

$$f(x) = A \int_{-\infty}^{\infty} dk \, \hat{f}(k) \, e^{-i k x}$$

where we are to show, again heuristically, that $A = 1/(2 \pi)$. To do this, I am going to rewrite the IFT as follows:

$$f(x) = A \lim_{N \to \infty} \int_{-N}^{N} dk \, \hat{f}(k) \, e^{-i k x}$$

Now, substitute the above definition of the FT into the integral, and reverse the order of integration (to repeat, I am assuming that $f$ is such this step and all the others are OK). In doing this, we get

$$f(x) = A \lim_{N \to \infty} \int_{-\infty}^{\infty} dx' \, f(x') \, \int_{-N}^{N} dk \, e^{i k (x'-x)}$$

Evaluating the inner integral, we get a single integral back:

$$f(x) = A \lim_{N \to \infty} \int_{-\infty}^{\infty} dx' \, f(x') \frac{e^{i N (x'-x)}-e^{-i N (x'-x)}}{i (x'-x)} = 2 A \lim_{N \to \infty} \int_{-\infty}^{\infty} dx' \, f(x') \frac{\sin{N (x-x')}}{x-x'}$$

Now, as $N \to \infty$, the sinc kernel behaves as a distribution which is characterized by a sifting property. Thus, we may, in this limit, take $f$ out of the integral and replace it with the value at $x'=x$. Thus we have

$$f(x) = 2 A f(x) \lim_{N \to \infty} \int_{-\infty}^{\infty} dx' \frac{\sin{N (x-x')}}{x-x'}$$

or, simplifying things a bit and incorporating $N$ into the integral, we get that

$$1 = 2 A \int_{-\infty}^{\infty} dy \frac{\sin{y}}{y} = 2 A \pi$$

Thus, $A = 1/(2 \pi)$ if the above steps are valid.

Ron Gordon
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  • But why does the Inverse Discrete Fourier Transform have a '1/N' constant instead of 1/(2π)? – user88004 Aug 03 '13 at 00:40
  • If you make analogous manipulations, you get an inner sum $$\sum_{k=0}^{N-1} e^{i 2 \pi k (n-n')/N} = \frac{e^{i 2 \pi (n-n')}-1}{e^{i 2 \pi (n-n')/N}-1}$$ for integers $n$ and $n'$. This quantity on the RHS is zero when $n \ne n'$ and is equal to $N$ when $n=n'$. This is where the factor of $N$ comes from. – Ron Gordon Aug 03 '13 at 13:27
  • Could you go through step-by-step please? What does 'RHS' stand for? What do "n′" and "x′" stand for here? – user88004 Aug 10 '13 at 21:52
  • Please pose this in a separate question. – Ron Gordon Aug 21 '13 at 01:03