I'm a B.C. student. I took a test and a question was:
does $\int_0^\infty \cos x^2 dx$ converge?
- Yes
- No
I undoubtedly answered No because this is going to be somehow periodically, not exactly periodically but after any given number, there is a greater number $s$ that $\int_0^s \cos x^2 dx = 0$ and a little bit after $s$ the value is not zero. because it is $\cos$ you know!
But after the test, I found out that the right answer is Yes and I was wrong.
Now I'm thinking for hours and I really really can't understand it. please help.
To use some elementary methods to prove this integral converges, consider that $$\int_0^\infty \cos(x^2)\,dx=\int_0^1\cos(x^2)\,dx+\int_1^\infty\cos(x^2)\,dx.$$ The first integral obviously converges since $|\cos(x^2)|\leq1$. For the second integral, we have $$\int_1^\infty\frac{1}{x}\,x\cos(x^2)\,dx=\frac{\sin(x^2)}{2x}\bigg|_{x=1}^{x=\infty}+\int_1^\infty\frac{\sin(x^2)}{2x^2}\,dx.$$ This equality follows from integration by parts; the first term above is obviously convergent, while the integral can be bounded above, since $$\left|\frac{\sin(x^2)}{2x^2}\right|\leq\frac{1}{2x^2},$$and $\int x^{-2}\,dx$ converges as can be shown directly using the power rule for integration.