# 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

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