Integral $\int_0^{\infty} \sin(x^2)/x^2\,dx$

I'll extend @sos440's answer a bit. $$ I = \int_o^\infty \frac {\sin x^2}{x^2}dx = -\left. \frac{\sin x^2}x \right|_0^\infty + 2\int_0^\infty \cos x^2 dx \\ \left .\frac{\sin x^2}x \right |_0^\infty = \lim_{x \rightarrow \infty} \frac {\sin x^2}x - \lim_{x \rightarrow 0} \frac {\sin x^2}x = 0-\lim \frac 1x \left ( x^2 + O(x^4)\right ) = -\lim_{x \rightarrow 0}\left( x + O(x^3)\right ) = 0 \\ $$ To take Fresnel's integral $C(x) = \int_0^x \cos x^2 dx$ at $\infty$ just take a contour like here, and you're done.

PS: $I = 2\lim_{x \rightarrow \infty} C(x) = 2 \sqrt{\frac \pi 8} = \sqrt{\frac \pi 2}$

Another solution. (which does not use complex analysis)
Substitute $u=x^2$, then the integral becomes $$A:=\int_{0}^{\infty}\frac{\sin (x^2)}{x^2}dx=\frac{1}{2}\int_{0}^{\infty}u^{-3/2}\sin u du$$ Now we'll consider more general one; $$f(p):=\int_{0}^{\infty}\frac{\sin u}{u^p}du\phantom{a} (p\in (1, 2))$$ From the definition of the gamma function, we can easily deduce that $$\frac{\Gamma(p)}{u^p}=\int_{0}^{\infty}v^{p-1}e^{-uv}dv$$ This formula gives $$f(p)\Gamma(p)=\int_{0}^{\infty}\sin u\left(\int_{0}^{\infty}v^{p-1}e^{-uv} dv\right)du$$ By Fubini's theorem, $$f(p)\Gamma(p)=\int_{0}^{\infty}\int_{0}^{\infty}v^{p-1}e^{-uv}\sin u dvdu=\int_{0}^{\infty}\int_{0}^{\infty}v^{p-1}e^{-uv}\sin u du dv$$ So $$f(p)\Gamma(p)=\int_{0}^{\infty}v^{p-1}\left(\int_{0}^{\infty}e^{-uv}\sin udu\right)dv=\int_{0}^{\infty}\frac{v^{p-1}}{v^2+1}dv$$ Substitute $v=\tan \theta$, then $$\begin{aligned}f(p)\Gamma(p)&=\int_{0}^{\pi /2}\tan^{p-1}\theta d\theta=\int_{0}^{\pi /2}\cos^{2\cdot(1-p/2)-1}\theta\sin^{2\cdot p/2-1}\theta d\theta\\&=\frac{1}{2}\operatorname{B}\left(1-\frac{p}{2},\frac{p}{2}\right)=\frac{1}{2}\Gamma\left(1-\frac{p}{2}\right)\Gamma\left(\frac{p}{2}\right)\end{aligned}$$ ($\operatorname{B}(\phantom{x},\phantom{y})$ denotes Beta function.)
By Euler's reflection formula, $$f(p)\Gamma(p)=\frac{1}{2}\Gamma\left(1-\frac{p}{2}\right)\Gamma\left(\frac{p}{2}\right)=\frac{1}{2}\frac{\pi}{\sin(\pi p/2)}$$ Therefore $$f(p)=\int_{0}^{\infty}\frac{\sin u}{u^p}du=\frac{1}{2}\frac{\pi}{\Gamma(p) \sin(\pi p/2)}$$ Substituting $\displaystyle p=\frac{3}{2}$, we obtain $$A=\frac{1}{2}f\left(\frac{3}{2}\right)=\frac{1}{2}\cdot\frac{1}{2}\frac{\pi}{\sqrt{\pi}/2 \cdot \sqrt{2}/2}=\frac{\sqrt{\pi}}{\sqrt{2}}$$