Evaluating the definite integral $\int_{-\infty}^{+\infty} \mathrm{e}^{-x^2}x^n\,\mathrm{d}x$ I recognize that the $\int_0^\infty \mathrm{e}^{-x}x^n\,\mathrm{d}x = \Gamma(n+1)$ and $\int_{-\infty}^{+\infty} \mathrm{e}^{-x^2}\,\mathrm{d}x = \sqrt{\pi}$. I am having difficulty, however with $\int_{-\infty}^{+\infty} \mathrm{e}^{-x^2}x^n\,\mathrm{d}x$. By the substitution $u=x^2$, this can be equivalently expressed as $\frac{1}{2} \int_{-\infty}^{+\infty} \mathrm{e}^{-u}u^{\frac{n-1}{2}}\,\mathrm{d}u$. This integral is similar to the first one listed (which equates to the $\Gamma$ function), except that its domain spans $\mathbb{R}$ like the second integral (which equates to $\sqrt{\pi}$). Any pointers on how to evaluate this integral would be helpful.
 A: The function $x\mapsto x^n e^{-x^2}$  is absolutely integrable on the real line.  If $n$ is odd, the integrand is odd, and we have
$$\int_0^\infty x^n e^{-x^2}\, dx = 0$$
Now consider the even case.  We first use symmetry to get the integral onto $[0,\infty)$ and then use the subsitution $x \rightarrow \sqrt{x}$ as follows
$$\int_{-\infty}^\infty  x^n e^{-x^2}\, dx= 2\int_0^\infty x^n e^{-x^2}\, dx
= 2\int_0^\infty x^{n/2} e^{-x}{dx\over2\sqrt{x}} = \Gamma\left({n + 1\over 2}\right).$$
Invoking the factorial property of the $\Gamma$ function relates this solution to the other posted solution.
A: Let $I_n:=\int_{-\infty}^{+\infty}e^{-x^2}x^ndx$. If $n$ is odd then $I_n=0$ and for $p\geq 1$:
\begin{align}
I_{2p}&=\int_0^{+\infty}e^{-x^2}x^{2p}dx+\int_{-\infty}^0e^{-x^2}x^{2p}dx\\
&=\int_0^{+\infty}e^{-t^2}t^{2p}dt+\int_0^{+\infty}e^{-t^2}(-t)^{2p}dt\quad (\mbox{left: } t=x,\mbox{right: } t=-x)\\
&=2\int_0^{+\infty}e^{-t^2}t^{2p}dt\\
&=2\int_0^{+\infty}e^{-s}s^p\frac 1{2\sqrt s}ds \quad (s=t^2)\\
&=\int_0^{+\infty}e^{-s}s^{p-1/2}ds\\
&=\left[-e^{-s}s^{p-1/2}\right]_0^{+\infty}+\int_0^{+\infty}e^{—s}\left(p-\frac 12\right)s^{p-1-1/2ds}\\
&=\left(p-\frac 12\right)I_{2(p-1)}.
\end{align}
Finally we get  $I_{2p+1}=0$ and $I_{2p}=\sqrt \pi\prod_{j=1}^p\left(j-\frac 12\right)$ for all $p\geq 0$.
