TITLE: Proving convergence of $ \int \limits_0^{\infty} \cos\left(x^2\right) dx $
QUESTION [7 upvotes]: How would one prove the convergence of $$ \int_0^{\infty} \cos\left(x^2\right) \,\mathrm dx $$
I tried using the integral test for convergence by noting that making the substitution $u = x^2$ means you can write the integral in the form:
$$ \int_0^{\infty} \frac{1}{2\sqrt{u}}\cos(u) \space \mathrm{du}$$
I'm not sure what to do here - I though of rewriting the cosine as a Taylor series, and then applying the integral test, but I don't think that would work.

REPLY [2 votes]: I can give you a fast way of treating this question:
First you notice that your function is continuous on $[0,+\inf[  $
So your integral is defined on [0,1] for instance:
I(X) = $\int_{[0,X]} cos(t^2)dt = \int_{[0,1]} cos(t^2)dt + \int_{[1,X]} cos(t^2)dt = A + I_1(X)$
$ I_1(X) = \int_{[1,X]} 2t*\frac{cos(t^2)}{2t}dt $ = $[\frac{sin(t^2)}{2t}]_1^X + \int_{[1,X]} \frac{sin(t^2)}{2t^2}dt $
The first term is a constant and a term that ->0 when x->inf.
The second verifies: 
$|g(t)| = |\frac{sin(t^2)}{2t^2}| \leq \frac{1}{2t^2} $ => g integrable on [1,+inf[ , because it is dominated by an integrable function on [1, +inf[
Hence $I_1(X)$ has a limit when X->inf, and so does your integral