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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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    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.2011-12-05

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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.

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    I can't read your answer at the moment because of a computer issue, but my last comment above was in fact the answer. I just had the wrong settings in Mathematica. I think this is what you are conveying, so I will accept the answer.2012-04-25