Show that Improper integral converges

Question

Show that the following integral is convergent.

$$ \int_{1}^{\infty}\sin\left(1 \over x^{2}\right)\cos\left(x^{2}\right)\,{\rm d}x $$

Not sure how I can solve this using absolute convergence

$$ \int_{1}^{\infty}\left\vert\, \sin\left(1 \over x^{2}\right)\cos\left(x^{2}\right)\,\right\vert\,{\rm d}x $$

any ideas?

Answer 1

We have

$$\left|\sin\left(\frac1{x^2}\right)\cos(x^2)\right|\le \left|\sin\left(\frac1{x^2}\right)\right|\sim_\infty \frac1{x^2}$$ and the integral

$$\int_1^\infty\frac{dx}{x^2}$$ is convergent. Conclude.

Answer 2

got it. sin(1) is a constant so you can place it as a multiple on the outside of the integral. cosx^2=1/secx^2 since cos is in the denominator when you do the trig identity 1/secx^2 you flip it to the numerator. This keeps you from having the discontinuities that cos sign has ie 1/cos(pi/2)=1/0. sec is never zero and the new integral is sin(1)*integral of secx^2/x^2 from 1 to infinity which is continuous over its range and therefore converges.