Integrating $\int_0^\infty \sin(1/x^2) \, \operatorname{d}\!x$

By "Fresnel equations," I assume you mean stuff like

$$\mathcal{F}(z) = \int_0^z dt \: e^{i \frac{\pi}{2} t^2}$$

Have no fear: I will produce everything from scratch without reference to anyone named Fresnel. That said, I will produce integrals with quadratic phase, but I will show you how to deal with them.

Begin with a substitution $u=1/x^2$:

$$\begin{align}\int_0^{\infty} dx \: \sin{\left ( \frac{1}{x^2}\right)} &= \frac{1}{2} \underbrace{\int_0^{\infty} du \: u^{-3/2} \, \sin{u}}_{\text{integrate by parts}} \\ &= \frac{1}{2} \underbrace{\left [ -2 u^{-1/2} \sin{u} \right]_0^{\infty}}_{\text{this is zero}} + \underbrace{\int_0^{\infty} du \: u^{-1/2} \, \cos{u}}_{u=v^2}\\ &= 2 \int_{0}^{\infty} dv \: \cos{v^2} \\ &= \Re{\left[\int_{-\infty}^{\infty} dv \: e^{i v^2} \right ]} \end{align}$$

Now, let's focus on this last integral. It is far from obvious that this even converges; I will show below that it does. I hope you are familiar with Cauchy's integral theorem, though.

Consider the integral in the complex plane

$$\oint_C dz \: e^{-z^2}$$

where $C$ is the following contour:

That is, a $45^{\circ}$ wedge of radius $R$, where we will set $R \rightarrow \infty$. Because there are no poles inside the contour, the above integral is zero. We then write the integral around this contour as

$$\oint_C dz \: e^{-z^2}=\int_0^R dx \: e^{-x^2} + R \int_0^{-\pi/4} d\theta \: e^{-R^2 \cos{2 \theta}} e^{-i R^2 \sin{2 \theta}} + e^{-i \pi/4} \int_R^0 dv \: e^{i v^2}=0$$

I hope you see that the second integral, that about the arc, vanishes as $R \rightarrow \infty$. It vanishes because of the exponential term in the integrand which vanishes very quickly in this limit. Because this integral vanishes, we are then left with the following relation:

$$\int_0^{\infty} dx \: e^{-x^2} - e^{-i \pi/4} \int_0^{\infty} dv \: e^{i v^2} = 0$$

Thus

$$\int_0^{\infty} dv \: e^{i v^2} = e^{i \pi/4} \int_0^{\infty} dx \: e^{-x^2} $$

Using the fact that the integral on the RHS is equal to $\sqrt{\pi}$ and that we are only interested in the real part of the integral on the LHS, we finally get

$$\int_0^{\infty} dx \: \sin{\left ( \frac{1}{x^2}\right)} = \sqrt{\frac{\pi}{2}}$$

A related problem. Recalling the Mellin transform

$$ F(s) = \int_{0}^{\infty} x^{s-1} f(x) dx. $$ We are going to relate our integral to the above integral. Using the change of variables $x=\frac{1}{y}$ to the integral in consideration gives

$$ \int_0^\infty \sin\left(\frac{1}{x^2}\right) \, \operatorname{d}\!x = \int_0^\infty \frac{\sin\left(y^2\right) }{y^2}\, \operatorname{d}\!y. $$

Following it with another change of variables $y=\sqrt{u}$ yields

$$ \int_0^\infty \frac{\sin\left(y^2\right) }{y^2}\, \operatorname{d}\!y = \frac{1}{2}\int_0^\infty u^{-\frac{3}{2}}{\sin\left(u\right) }\, \operatorname{d}\!x. $$

Now, the last integral is nothing but the Mellin transform of $\sin(x)$ with $s=-\frac{1}{2}$, that is

$$\frac{1}{2}\int_0^\infty u^{-\frac{3}{2}}{\sin\left(u\right) }\, \operatorname{d}\!x= \frac{1}{2}\lim_{s \to -\frac{1}{2} }\Gamma\left( s \right) \sin\left( \frac{\pi s}{2} \right)= \frac{\sqrt {\pi }}{\sqrt{2}},$$

where $\Gamma(s)$ is the gamma function

$$ \Gamma(s)= \int_{0}^{\infty} x^{s-1} e^{-x} dx, \quad Re(s)>0 .$$

Note:

$$ \Gamma\left( -\frac{1}{2} \right) = -2\Gamma\left( \frac{1}{2} \right)=-2 \sqrt{\pi}. $$