Expected value of normal distribution given that distribution is positive
Let $f(x)$ be the density of $X$; let $F(x)$ be its CDF.
Then the density of $X$, conditional on it being positive, is $f(x)/P(X \ge 0)$ if $x \ge 0$, and $0$ otherwise.
Of course $P(X \ge 0) = 1/2$ by symmetry, so the density of $X$ conditional on $X \ge 0$ is $2f(x)$ (on $x \ge 0$).
So you need to do the integral $$ \int_0^\infty 2xf(x) \: dx $$ which is twice the integral you've written.