Integral of Polynomial Series

Relevant wiki: Gamma Function

The integral can be written into:

$\begin{aligned} I & = \int_0^\infty x^3 \sum_{n=0}^\infty \frac {(-2x^2)^n}{n!} dx \\ & = \int_0^\infty x^3 e^{-2x^2} dx & \small \color{#3D99F6} \text{Let }u = x^2 \implies du = 2x \ dx \\ & = \frac 12 \int_0^\infty u e^{-2u} du & \small \color{#3D99F6} \text{Let } t = 2u \implies dt = 2 \ du \\ & = \frac 18 \int_0^\infty t^{2-1} e^{-t} dt & \small \color{#3D99F6} \text{Gamma function }\Gamma (z) = \int_0^\infty t^{z-1}e^{-t} dt \\ & = \frac {\Gamma (2)}8 = \frac {1!}8 & \small \color{#3D99F6} \text{Since }\Gamma (n) = (n-1)! \\ & = \frac 18 = \boxed{0.125} \end{aligned}$

At some level, we have to simply guess what the given power series is, so the first step in the solution is rather unsatisfying: We just have to note that $x^3e^{-2x^2} = \sum_{n=0}^\infty \frac{(-1)^n \cdot 2^n}{n!} x^{3+2n} = x^3 - 2x^5 + 2x^7 - \frac{4}{3}x^9 + \frac{2}{3}x^{11} - \frac{4}{15}x^{13} + \cdots$

That is, we need to evaluate $\int_0^\infty x^3e^{-2x^2}\,dx$ To do this, use the substitution $u=2x^2$ to rewrite it as $\frac{1}{8} \int_0^\infty u e^{-u}\,du = \frac{1}{8} \Big[-(u+1)e^{-u}\Big]_0^\infty = \frac{1}{8} = \boxed{0.125}$