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 $G = 1$ ?