# 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!

• wolframalpha.com/input/… – Michael May 21 '19 at 0:00
• @Michael Thanks, but I think the Fresnel C integral is not an elementary function. Is it possible to turn it into a form such as $\frac{\sqrt \pi}{2}$? – Zero May 21 '19 at 0:03
• Yes: Multiply the integral by 0 and then add $\frac{\sqrt{\pi}}{2}$. – Michael May 21 '19 at 3:14
• @Michael Are we seriously going to do that? Or Was that sarcasm? – Soumalya Pramanik May 21 '19 at 4:07
• @SoumalyaPramanik :both. – Michael May 21 '19 at 14:23

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.