# Applying Plancherel's theorem to a simple function

Let $f(t) = \frac{1}{25}e^{-(t-11)^2}-\frac{1}{36}e^{-(t-13)^2}$.

Using the Wiki definition of the Fourier transform pair, I calculated $\hat{f}$ in Mathematica as $$\hat{f}(\omega) = \frac{\sqrt{\pi}}{900}e^{-\frac{1}{4}\omega(52i+\omega)}(36e^{2i\omega}-25).$$

The point was just to see an example of Plancherel's theorem, and so I calculated,

$\int_{-\infty}^{\infty} |f|^2 = \frac{1}{\alpha}$ and reasoned that if $g = \sqrt{\alpha} f$ then both $g$ and $\hat{g}$ are normalized, i.e., that $G = \int_{-\infty}^{\infty} |\hat{g}|^2 = 1$.

Now in Mathematica I set the FourierParameters to $\{{1,-1}\}$. But when I calculate the integral $G$ (using $\hat{g}$), I get something that looks suspiciously like $2\pi$.

So two questions. First, is there anything about the function that prevents us from applying Plancherel? And (if not) second, is there anything in principle about the calculation that might prevent us from getting $G = 1$ ?

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I also calculated the transform of f using result 206 on the same Wiki page, which gives me the same result as Mathematica for the transform. –  daniel Dec 5 '11 at 18:46
I verified it by Wolfram Alpha. f is both integrable and square integrable, so Plancherel applies. –  niyazi Dec 5 '11 at 19:10
The setting FP->{0,-2Pi} seems to work. This seems to correspond to the Wiki page definition of the transform so it may be the answer to my question. –  daniel Dec 5 '11 at 20:22

Under certain conventions, the Fourier transform is not unitary; that is, it doesn't satisfy Plancherel's theorem, but a version with a $2\pi$ stuck in there. The convention $(1,-1)$ is one in which $\mathcal{F}$ is not unitary, so you will get an undesirable $2\pi$ floating around. If you want a unitary Fourier transform, you found $(0,\pm 2\pi)$ on Wikipedia, and $(0,\pm 1)$ will also work.