calculus - Evaluating $int_1^infty sin frac{1}{x^2} dx$

Evaluating $$\int_1^\infty \sin \frac{1}{x^2} dx$$

Actually I'm quite curious whether we can have a good evaluation of this integral, because I found this equivalent to evaluate $$\int_0^1\cos(x^2) dx
$$

My Attempt

\begin{align}
\int_1^\infty \sin \frac{1}{x^2} dx
&= \frac{1}{2} \int_0^1 \frac{\sin u}{ u^{ \frac{3}{2} } }du \\
&= -\int_0^1 \sin u\, d\frac{1}{\sqrt u} \\
&= -\sin 1 + \int_0^1 \frac{\cos u}{\sqrt u} du \\
&= -\sin 1 + 2\int_0^1\cos(x^2) dx
\end{align}

But I got stuck on how to evaluate the integral mentioned above.

After a brief search I found that
$$
\int \cos(x^2) dx
$$
doesn't have a closed form.

So can we evaluate this well? I would highly appreciate it if you could share any thoughts. Thanks in advance!

Answer

For an integral such as this, without looking up the relevant special functions, we can always just use the series:

$$\int_0^1\cos(x^2) dx=\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} \int_0^1 x^{4n} dx=\sum_{n=0}^\infty \frac{(-1)^n}{(2n)! (4n+1)}$$

This series converges fast and it's quite easy to get the numerical value to any precision.

But, as per Michael's comment, Fresnel integrals are the way to go. They are incorporated in modern software, such as Matlab or Mathematica, so evaluating them is faster than numerically evaluating the original integral.