Convergence of $\int_{-\infty}^{\infty}p(x)e^{-x^2}dx$ How to show the convergence of $\int_{-\infty}^{\infty}p(x)e^{-x^2}dx$?
Where $p(x)$ is a polynomial. 
This is a special case of the Gaußian Integral. Any help on this?
 A: Hint: $x^n(1+x^2) < M_n e^{x^2}$ for some $M_n > 0$. Does that help?
Edit: This implies that $x^n  e^{-x^2}< M_n/ (1+x^2)$. So for any polynomial $P$ there is a constant $M_P > 0$ such that $|P(x) e^{-x^2}| < M_p/ (1+x^2)$. So you can compare and use that $\int 1/1+x^2 < \infty$. 
A: Near $+\infty $,  for $\alpha>1$,We have always
$$\lim_{x\to+\infty}\color {red}{x^\alpha }p (x)e^{-x^2}=0$$
because exponential is faster.
thus for enough great $x $,
$$|\color {red}{x^\alpha}. p (x)e^{-x^2}|<1$$
or
$$|p (x)e^{-x^2}|<\frac {1}{\color {red}{x^\alpha}} $$
and by comparison criterion,
$$\int^{+\infty}p (x)e^{-x^2}dx $$ is convergent.
idem for $-\infty $.
A: $$\int_{-\infty}^{\infty}P_n(x)\exp(-x^2)dx = \int_{-\infty}^{\infty}\exp(-x^2)\left(\sum_{k=0}^n a_k x^k\right) dx$$
Because we have finite summation, we can interchange the integral
$$\sum_{k=0}^n \left(a_k \int_{-\infty}^{\infty}\exp(-x^2)x^k \,dx\right)$$
The integral inside can be shown to be $\frac{1}{2} \left((-1)^k+1\right) \Gamma \left(\frac{k+1}{2}\right)$, which is clearly convergent for any positive integers $k$. We conclude that the finite summation of convergent values (multiplied by some finite scalars) must be convergent.
