<s>[INST] Integrating $\int_0^\infty \sin(1/x^2) \, \operatorname{d}\!x$ How would one compute the following improper integral: 
$$\int_0^\infty \sin\left(\frac{1}{x^2}\right) \, \operatorname{d}\!x$$ 
without any knowledge of Fresnel equations?
I was thinking of using the comparison theorem since $\sin x$ is never greater than 1. However, I can't find a function always greater than 1 between 0 and infinity such that its integral from 0 to infinity converges.
Thanks.
 [/INST] I prefer to evaluate the more general integral below
$$\displaystyle  I(a):=\int_{-\infty}^{\infty} \sin \left(\frac{a}{x^2} \right)dx, \tag*{}$$
where $a\geq 0.$
Letting $y=\frac{1}{x^2}$ transforms the integral into
$\displaystyle \begin{aligned} I(a)&=2 \int_{\infty}^0 \sin (a y)\left(-\frac{1}{2 y^{\frac{3}{2}}} d y\right) = \int_0^{\infty} \frac{\sin (a y)}{y^{\frac{3}{2}}} d y \end{aligned} \tag*{} $
Differentiating $I(a)$ w.r.t. $a$, we have
$\displaystyle \begin{aligned} I^{\prime}(a) &=\int_0^{\infty} \frac{\cos (a y)}{y^{\frac{1}{2}}} d y\stackrel{u^2=ay}{=}\frac{2}{\sqrt{a}} \int_0^{\infty} \cos u^2 d u \end{aligned} \tag*{} $
Using the result(refer to footnote for details):$ $
$$\displaystyle \int_0^{\infty} \cos x^2 d x=\frac{1}{2}\int_{-\infty}^{\infty} \cos x^2 d x =\sqrt{\frac{\pi}{8}}\tag*{} $$
we obtain
$$\displaystyle I^{\prime}(a)=\sqrt{\frac{\pi}{2}} a^{-\frac{1}{2}} \tag*{} $$
Integrating back yields
$$\boxed{\displaystyle \int_{-\infty}^{\infty} \sin \left(\frac{a}{x^2} \right)dx=\sqrt{\frac{\pi}{2}} \int_0^a t^{-\frac{1}{2}} d t=\sqrt{2 a \pi}} \tag*{} $$
Putting $a=1$ yields our integral
$\boxed{\displaystyle \int_0^{\infty} \sin \left(\frac{1}{x^2} \right)dx=\sqrt{2  \pi}} \tag*{} $

Furthermore, for $a<0$, we have
$\displaystyle \int_{-\infty}^{\infty} \sin \left(\frac{a}{x^2} \right)dx= -\sqrt{-2 a\pi}\tag*{}$

Footnote:
$\displaystyle \begin{aligned} \int_{-\infty}^{\infty} \cos x^2 d x-i \int_{-\infty}^{\infty} \sin x^2 d x =& \int_{-\infty}^{\infty} e^{-x^2 i} d x \\ =&\frac{\sqrt{2}}{1+i} \int_{-\infty}^{\infty} e^{-\left[\left(\frac{1+i}{\sqrt{2}}\right)x\right]^2} d\left(\frac{1+i}{\sqrt{2}} x\right) \\ =& (1-i)\sqrt{\frac{\pi}{2} }  \end{aligned} \tag*{} $
By comparing the real and imaginary parts, we have
$\displaystyle \int_{-\infty}^{\infty} \cos x^2 d x=\int_{-\infty}^{\infty} \sin x^2 d x=\sqrt{\frac{\pi}{2}} \tag*{} $
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