Why does $\int_0^\infty \cos x^2 dx$ converge?

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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.

3

There are 3 best solutions below

3
On BEST ANSWER

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.

2
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The usual idea of a function decaying sufficiently fast for convergence of the improper integral does not apply to sign-changing functions. The reason for the convergence in this case is the highly oscillatory character of the function $\cos x^2$ at infinity. The positive and negative "areas" cancel each other and the integral converges.

3
On

Don't be deceived; although $\cos x^2$ oscillates, it does at a frequency which increases as $x,$ which makes the argument go to $+\infty$ as $x$ becomes arbitrarily large. Therefore, although it oscillates, it does so in such a way that it adds and subtracts sufficiently smaller and smaller areas, and does so faster than $x$ is running away, in order to converge.


I have given a heuristic explanation in the spirit of OP