Finding a constant to make a valid pdf

Question

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

Answer 1

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.

Answer 2

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: $$ \int^{\infty}_0 e^{-x^2} \,dx = \frac{\sqrt{\pi}}{2} $$