Show that $\sin(x^2)$ is integrable around $\infty$. I have to show that $f(x)=\sin(x^2)$ is integrable on $[1, \infty[$. This is French terminology, so "intégrable" specifically means that the integral of $|f|$ exists.
The only method I know is to compare it to functions of the form $\frac{1}{x^\alpha}$, but it's not eventually smaller or larger than any of these. I can't imagine how it could be asymptotically equivalent to anything useful either, seeing as it oscillates like crazy.

If you mean that
$$
\lim_{N\to\infty}\int_0^N\sin(x^2)\,\mathrm{d}x
$$
exists, then change variables $x\mapsto\sqrt{x}$ and integrate by parts:
$$
\begin{align}
&\lim_{N\to\infty}\int_0^N\sin(x^2)\,\mathrm{d}x\\
&=\int_0^1\sin(x^2)\,\mathrm{d}x
+\lim_{N\to\infty}\frac12\int_1^{N^2}\frac{\sin(x)}{x^{1/2}}\,\mathrm{d}x\\
&=\int_0^1\sin(x^2)\,\mathrm{d}x
+\lim_{N\to\infty}\frac12\left[\frac{1-\cos(x)}{x^{1/2}}\right]_1^{N^2}
+\lim_{N\to\infty}\frac14\int_1^{N^2}\frac{1-\cos(x)}{x^{3/2}}\,\mathrm{d}x\\
\end{align}
$$
Now each piece has a limit as $N\to\infty$ since
$$
\int_0^1\sin(x^2)\,\mathrm{d}x
$$
is constant
$$
\lim_{N\to\infty}\frac12\left[\frac{1-\cos(x)}{x^{1/2}}\right]_1^{N^2}=\frac{\cos(1)-1}2
$$
and
$$
\left|\frac{1-\cos(x)}{x^{3/2}}\right|\le\frac2{x^{3/2}}
$$
which is integrable over $[1,\infty]$ since $\frac32\gt1$.
