Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
1  
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. –  Dilip Sarwate Apr 30 '12 at 1:55

2 Answers 2

You can write $2^{-x^2}$ as $$2^{-x^2}=e^{(\ln 2)({-x^2})}=e^{-x^2\ln 2}$$Using the error function you can calculate the integral $$\int_{-\infty}^{+\infty}e^{-x^2\ln 2}=\sqrt{\frac{\pi}{\ln 2}}$$ The rest is trivial.

share|improve this answer

Hint: Rewrite $$f(x) = c \,[e^{\ln(2)}]^{-x^2} = c\, e^{-x^2\ln(2)}$$ and try to exploit the following integral together with some change of variable: $$ \int^{\infty}_0 e^{-x^2} \,dx = \frac{\sqrt{\pi}}{2} $$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.