How to integrate using known distributions I'm having trouble figuring this integration out using known distributions. I don't know which distribution to use to solve this problem. It looks like a gamma to me.
$$\int_{0}^{\infty} x^{3}e^{-x^2}dx$$
I tried substituting $u=x^{2}$ to simplify the problem but it doesn't seem to help.
 A: Substitution $u=x^2$ leads to:$$\int_{0}^{\infty}x^{3}e^{-x^{2}}dx=\frac{1}{2}\int_{0}^{\infty}ue^{-u}du=\frac{1}{2}\mathbb{E}U$$
where $U$ has exponential distribution with parameter $\lambda=1$ (so $\mathbb E(U)=\frac{1}{\lambda}=1$)
A: Alternatively, you can integrate by parts twice.  First use $u = x^2$, $v = e^{-x^2}$.
A: I would say you can integrate by parts:
$\displaystyle I=\int_{0}^{\infty} x^{3}e^{-x^2}dx,\quad u = x^2, v' = xe^{-x^2}, u'=2x, v = -\frac{1}{2}e^{-x^2}$
$\displaystyle I=\left[-x^2 \cdot \frac{1}{2}e^{-x^2}\right]_{0}^{\infty}+\int_{0}^{\infty}xe^{-x^2}dx=-\frac{1}{2}\left[e^{-x^2}\right]\int_{0}^{\infty}=\frac{1}{2}$
A: If you substitute $u = x^2$, you get $\displaystyle \frac{1}{2}\int_0^\infty ue^{-u} du$.
Here, $\displaystyle \int_0^\infty u e^{-u} du = 1$ because $ue^{-u}$ is the pdf of $\Gamma(2,1)$.
A: The term $e^{-x^2}$ does not come up in a Gamma distribution but in a Normal distribution with mean $μ=0$ and variance $σ^2$ such that $$2σ^2=1 \iff σ^2=\frac12$$ The term $x^3$ links to the $E[X^3]$, but the integration limits are not over the entire $\mathbb R$ but only from $0$ to $\infty$. However, knowing that $$E[X^k]=0$$ for every odd $k$, where $X \sim N(μ=0, σ^2=\frac12)$, and due to the symmetry of the normal distribution of $X$ over $μ=0$ you can conclude that $$\int_{0}^{\infty}x^3e^{-x^2}dx=\frac12$$ 
