I'm supposed to find if the the following integral converges or diverges by comparison:
$$\int_0^\infty\sin(x^2)\cdot e^{-x} dx$$
It would be easy to take away $e^{-x}$ and solve using Fresnel but I'm not supposed to use that. The integral converges. Any suggestions?
Since $|\sin(x^2)|\leq 1$, you have $$|\sin(x^2)\cdot e^{-x}|\leq e^{-x} $$ and $$\int_0^{\infty}e^{-x}\,\mathrm{d}x=1 $$ By the comparison test, your integral is convergent.