Convergence of improper integral: $x^2\cos(e^x)$

Question

I am pretty sure the following converges, but I'm having trouble proving it. $$\int_0^\infty x^2\cos(e^x)dx$$ I tried breaking up the integral into pieces, where each piece corresponds to a period of $\cos(e^x)$, but I'm stuck.

Any hints?

Edit: Later I'd like to extend the argument to other exponents, e.g. $\int_0^\infty x^{1000} \cos(e^x)$. I know that not every exponent works, but I think that exponents > -1 work.

Answer 1

A start: Rewrite the integrand as $x^2e^{-x}e^x\cos(e^x)$, and integrate by parts, with $u=x^2e^{-x}$ and $dv=e^x\cos(e^x)\,dx$. Then $du=(-x^2e^{-x}+2xe^{-x})\,dx$ and we can take $v=\sin(e^x)$.

Answer 2

Let $x= \ln y.$ Your question is then the same as asking if

$$ \int_1^\infty (\ln y)^2 \cos y \frac{dy}{y}$$

converges. Because $(\ln y)^2/y$ is eventually decreasing, Dirichlet's test shows the answer is yes.