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Let $f(x) = c\cdot 2^{-x^2}$. How do I find a constant $c$ such that the integral evaluates to $1$?

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    Since you already have two answers showing that $f(x) = c\, e^{-x^2\ln(2)}$, I will suggest that rather than the error function, you simply use what I hope you already know: $$\frac{1}{\sigma \sqrt{2\pi}}e^{-x^2/(2\sigma^2)}~~\text{is the density function of a}~N(0,\sigma^2)~ \text{random variable}.$$ Now compare constants and deduce the value of $c$. As a side benefit, you also get the mean and variance of the random variable for free.2012-04-30

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