What's wrong with my calculation? $\mathrm E[X^3]$ where $X\sim N(0,1)$ Let $X\sim N(0,1)$, compute $\mathrm E[X^3]$:
My attempt:
$$
\mathrm E[X^3]=\int_{-\infty}^{\infty}x^3\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\mathrm d x \\
x^2=t \Rightarrow \mathrm dx=\frac{\mathrm dt}{2x}\\
\mathrm E[X^3]=\frac{1}{\sqrt{2\pi}}\int_{0}^{\infty} t\frac{1}{2}e^{-\frac{t}{2}}\mathrm dt=\frac{1}{\sqrt{2\pi}}E[Y] \\ Y\sim \exp(\frac{1}{2})
\\ \Rightarrow \mathrm E[X^3]=\frac{1}{\sqrt{2\pi}}\frac{1!}{0.5^1}=\frac{2}{\sqrt{2\pi}}
$$
I used the fact that if $X\sim \exp(\lambda)$ then $\mathrm E[X^n]=\dfrac{n!}{\lambda^n}$
But the actual is result is $0$! Why is that? what am I missing here??
Thanks.
 A: Your change of variables is off. If you want to reason on integral (though drhab reasonning is much simpler) : 
$$ E[X^3]=\int_{-\infty}^{+\infty} x^3\frac{1}{\sqrt{2\pi}}e^{-x^2/2}dx$$
$$ E[X^3]=\int_{-\infty}^{0} x^3\frac{1}{\sqrt{2\pi}}e^{-x^2/2}dx + \int_{0}^{+\infty} x^3\frac{1}{\sqrt{2\pi}}e^{-x^2/2}dx$$
Applying the change of variable $x\mapsto -x$ in the first integral, step by step
$$ E[X^3]=\int_{+\infty}^{0} (-x)^3\frac{1}{\sqrt{2\pi}}e^{-x^2/2}(-dx) + \int_{0}^{+\infty} x^3\frac{1}{\sqrt{2\pi}}e^{-x^2/2}dx$$
$$ E[X^3]=\int_{+\infty}^{0} x^3\frac{1}{\sqrt{2\pi}}e^{-x^2/2}dx + \int_{0}^{+\infty} x^3\frac{1}{\sqrt{2\pi}}e^{-x^2/2}dx$$
$$ E[X^3]=-\int_{0}^{+\infty} x^3\frac{1}{\sqrt{2\pi}}e^{-x^2/2}dx + \int_{0}^{+\infty} x^3\frac{1}{\sqrt{2\pi}}e^{-x^2/2}dx$$
$$E[X^3]=0$$
A: Your change of variable is valid, but not as regards the integration range.
If $x$ runs from $-\infty$ to $\infty$, $t:=x^2$ runs from $\infty$ to $0$ then back to $\infty$, and the two contributions cancel each other.
The good news is that your one-sided integral, from $0$ to $\infty$ is correct.
