Question

What is `int_(0)^(oo) 1/x^2sin(1/x) dx`?

Answer 1

integral does not converge.

Explanation:

there's an obvious sub here in `q = 1/x` so `dq = - 1/ x^2 dx \implies dx = - x^2 dq`

mixing and matching, the integral becomes

`int_(q=oo)^(0) 1/x^2 sin(q)( - x^2) \ dq = int_(q=0)^(oo) sin(q) \ dq`

the sine function is periodic ie it doesn't converge to anything as `q \to \infty`. so the integral does not converge. there is no answer.