# Probability integral

1. Mar 24, 2017

### Mr Davis 97

1. The problem statement, all variables and given/known data
$\displaystyle \int_{- \infty}^{\infty} \frac{1}{\sqrt{2 \pi}} x e^{- \frac{x^2}{2}} dx$

2. Relevant equations

3. The attempt at a solution
So first off, obviously the answer is 0, because the integrand is odd and we have symmetrical limits of integration. However, when I make the subsitution $u = - \frac{x^2}{2}$, I get $\displaystyle \frac{1}{\sqrt{2 \pi}} \int_{- \infty}^{\infty} e^u du$, which doesn't even converge. Why is there this difference? Am I doing something wrong?

2. Mar 24, 2017

I think the best way to do this integral is just to look at it being odd (editing... see comments in paragraph 2), but using your substitution, what do you get for lower and upper limits of u? I get $- \infty$ for both, so the resulting $e^u$ will vanish for both limits. $\\$ By your substitution, it would also make sense to do it in two parts=from $x=-\infty$ to $0$ and then a second integral from $x=0$ to $x=+\infty$. (Any time you get identical limits on an integral, it pays to take a second look at it.) In this case the integral from $-\infty$ to $0$ will get a negative sign and be opposite that of the second integral. This second method is perhaps mathematically better than just looking at even/oddness, because there are odd integrals where the two halves don't separately converge, so that assigning these odd integrals the value zero is not completely justified.