Physics help?

A free particle has the initial wave function:

phi(x,0) = Ae^(-a|x|)

A and a are constants,

Normalise phi(x,0)

I got A = root a, but I doubt this is right somehow...

>>3
\int_{0}^{\infty} A^2 e^{-2ax}\,dx = \frac{A^2e^{-2ax}}{-2a}|_{0}^{\infty} = (\frac{A^2}{-2a})(e^{-\infty} - e^0) = (\frac{A^2}{-2a})(0 - 1) = \frac{A^2}{2a}

You probably flipped your limits; e^x is positive definite.  Of course, the integral from -infty to 0 is the same thing by symmetry.