I want to find the normalization constant for the function $f(x)=e^{-\frac{x^{2}}{2}}$.
I start by calculating the norm of $f(x)$ over $\mathbb{R}$:$$ \newcommand{\dx}{\text{d}x}||f||^{2} = \int\limits_{-\infty}^{\infty}\left[e^{-\frac{x^2}{2}}\right]^{2} \dx = \int\limits_{-\infty}^{\infty}e^{\frac{-2x^{2}}{2}}\dx = \int\limits_{-\infty}^{\infty}e^{-x^{2}}\dx = \sqrt{\pi}.$$
However, I know that the answer should be $||f||=\sqrt{2\pi}$, and thus $||f||^{2}=2\pi$.
What am I missing here?..
$\endgroup$1 Answer
$\begingroup$Normalization ensures $\int f(x)\ dx = 1$, not that the integral of the square of the functions equals 1.
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