Existence of kth moment Degroot makes a statement in his textbook:
It can be shown that for every positive integer $k$,
$$\int_{-\infty}^{\infty}|x|^ke^{-(x-3)^2}\,dx<\infty$$
Can someone show me how I might prove this statement?
 A: The short answer is "because $|x|^k$ is way smaller than $e^{x^2}$, if $|x|$ is large".  If you need more details. . . 
For any $k$, the power $x^k$ grows much more slowly than the exponential.  More precisely, there exists a positive real number $a$ such that

if $x\ge a$ then $x^k<e^x$.

So if $x\ge a$ we have
$$|x|^ke^{-(x-3)^2}\le e^{-x^2+7x-9}\ .$$
If also $x\ge8$ then $-x^2+7x-9<x(7-x)\le-x$ and so
$$|x|^ke^{-(x-3)^2}\le e^{-x}\ .$$
Hence
$$\int_{\max(a,8)}^\infty |x|^ke^{-(x-3)^2}\,dx\le\int_{\max(a,8)}^\infty e^{-x}\,dx<\infty\ .$$
This takes care of the positive "tail" of the integral, the negative tail is similar, and the bit in between is finite because it is the integral over a closed interval of a continuous function.
A: For all $x$, $x\lt1+x\le e^x$. Therefore, substituting $x\mapsto x^2/n$ and raising to the power $n$ yields
$$
\frac{x^{2n}}{n^n}\le e^{x^2}\implies e^{-x^2}\le\frac{n^n}{x^{2n}}\tag{1}
$$
Note that although not proven above, the inequality is true for $n=0$.
Therefore, applying $(1)$ with $n=0$ and $n=k+1$, yields
$$
\begin{align}
\int_{-\infty}^\infty|x|^ke^{-(x-3)^2}\,\mathrm{d}x
&=\int_{-\infty}^\infty|x+3|^ke^{-x^2}\,\mathrm{d}x\\
&\le\int_{-1}^1|x+3|^k\,\mathrm{d}x+\int_{|x|\gt1}|x+3|^k\frac{(k+1)^{k+1}}{x^{2k+2}}\,\mathrm{d}x\tag{2}
\end{align}
$$
both integrals in $(2)$ are obviously finite.
