Prove that integral of $ 1-\cos\left(\frac{1}{x^2}\right) $ is finite I need to prove that 
$$ \int_0^{\infty} \left(1 - \cos\left(\frac{1}{x^2}\right)\right) dx < \infty $$
My attempt: 
$$ \forall x\in[0,\infty] \hspace{1cm} 0 < 1 - \cos\left(\frac{1}{x^2}\right) < 2 \tag{1}. $$ 
Using L'Hôpital's rule I can show that: 
$$ 1 - \cos\left(\frac{1}{x^2}\right) \underset{x \to \infty}{\sim} \frac{1}{2x^4} \tag{2} $$
Which means that $ 1 - \cos\left(\frac{1}{x^2}\right)$ behaves like $ \frac{1}{2x^4} $ when $ x \to \infty $.
So I think that, there exist $N \in \mathbb{N} $ such that: 
$$ \int_0^{\infty} \left(1 - \cos\left(\frac{1}{x^2}\right)\right)dx < \int_0^{N} 2dx + \int_N^{\infty} \frac{1}{x^4}dx < \infty $$
but I'm not sure. I would appreciate any tips or hints.
 A: $$I=\int_{0}^{+\infty}\left[1-\cos\left(\tfrac{1}{x^2}\right)\right]\,dx \stackrel{x\mapsto 1/x}{=} \int_{0}^{+\infty}\frac{1-\cos(x^2)}{x^2}\,dx $$
where the function $\frac{1-\cos(x^2)}{x^2}$ is continuous and bounded over $(0,1]$, non-negative and bounded by $\frac{2}{x^2}$ over $[1,+\infty)$. It follows that the above integral is finite. Its value can be found through the Laplace transform:
$$\int_{0}^{+\infty}\frac{1-\cos x}{2x\sqrt{x}}\,dx\!\stackrel{\text{IBP}}{=}\!\int_{0}^{+\infty}\frac{\sin x}{\sqrt{x}}\,dx\!\stackrel{\mathcal{L}}{=}\!\frac{1}{\sqrt{\pi}}\!\int_{0}^{+\infty}\frac{ds}{(s^2+1)\sqrt{s}}\!\stackrel{s\mapsto t^2}{=}\!\frac{1}{\sqrt{\pi}}\!\int_{0}^{+\infty}\frac{2\,dt}{t^4+1} $$
leads to $I=\color{red}{\sqrt{\frac{\pi}{2}}}$.
A: Use the substitution $1/x=t$, so the integral becomes
$$
\int_0^{\infty} \frac{1-\cos(t^2)}{t^2}\,dt
$$
Since
$$
\lim_{t\to0}\frac{1-\cos(t^2)}{t^2}=0
$$
convergence is not an issue at $0$. Since
$$
0\le 1-\cos(t^2)\le 2
$$
we have that, for $t\ge1$,
$$
0\le\frac{1-\cos(t^2)}{t^2}\le\frac{2}{t^2}
$$
Now show convergence of
$$
\int_1^\infty \frac{2}{t^2}\,dt
$$
A: $\int_0^\infty(1-\cos (1/x^2))dx= \int_0^1(1-\cos (1/x^2))dx+\int_1^\infty(1-\cos (1/x^2))dx\leq 2+\int_1^\infty2\sin^2 \left(\frac{1}{2x^2}\right)dx\leq 2+\int_1^\infty2\sin \left(\frac{1}{2x^2}\right)dx\leq2+\int_1^\infty2 \frac{1}{2x^2}dx \text{ (as } \sin x\leq x \text{ forall } x>0) =2+\int_1^\infty \frac{1}{x^2}dx .$
A: One way is to use $ 1 - \cos\theta = 2\sin^2\theta$ and $u = x^{-2}$ to get
\begin{multline}
\int_0^\infty\left[1-\cos(x^{-2})\right]dx = \int_0^1 2\sin^2(x^{-2})dx + \int_1^\infty 2\sin^2(u)\frac{du}{2u^{3/2}}
\\ = 2\int_0^1\sin^2(x^{-2})dx +\int_0^1 u^{1/2}\left[\frac{\sin(u)}{u}\right]^2 du
\end{multline}
Since both integrals are of bounded functions over a finite range, they must be finite.
