Prove that integral of $ 1-\cos\left(\frac{1}{x^2}\right) $ is finite

Question

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.

Answer 1

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 $$ \int_1^\infty \frac{2}{t^2}\,dt $$

Answer 2

$\int_0^\infty(1-\cos (1/x^2))dx= \int_0^1(1-\cos (1/x^2))dx+\int_1^\infty(1-\cos (1/x^2))dx\leq 2+\int_1^\infty2\sin^2 \left(\frac{1}{2x^2}\right)dx\leq 2+\int_1^\infty2\sin \left(\frac{1}{2x^2}\right)dx\leq2+\int_1^\infty2 \frac{1}{2x^2}dx \text{ (as } \sin x\leq x \text{ forall } x>0) =2+\int_1^\infty \frac{1}{x^2}dx .$

Answer 3

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.

Answer 4

$$I=\int_{0}^{+\infty}\left[1-\cos\left(\tfrac{1}{x^2}\right)\right]\,dx \stackrel{x\mapsto 1/x}{=} \int_{0}^{+\infty}\frac{1-\cos(x^2)}{x^2}\,dx $$ 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 $I=\color{red}{\sqrt{\frac{\pi}{2}}}$.