Check out this answer for a real method to evaluate this integral.
First note that $ \int_0^\infty e^{-x^2}\,\mathrm{d}x=\frac{\sqrt\pi}2\tag{1} $
The integral $ \int_\gamma e^{-z^2}\,\mathrm{d}z=0\tag{2} $ over the curve $\gamma$ consisting of the line from $0$ to $R$ then counterclockwise along the circular arc from $R$ to $Re^{i\pi/4}$ then back along the line from $Re^{i\pi/4}$ to $0$ must be $0$ since $e^{-z^2}$ has no singularities inside $\gamma$.
Next note that $\int_\gamma e^{-z^2}\,\mathrm{d}z$ along the line from $0$ to $R$ as $R\to\infty$ tends to $(1)$.
Next note that $\int_\gamma e^{-z^2}\,\mathrm{d}z$ along the arc of the circle of radius $R$ from $R$ to $Re^{i\pi/4}$ as $R\to\infty$ goes to zero.
Finally, note that $\int_\gamma e^{-z^2}\,\mathrm{d}z$ along the line from $Re^{i\pi/4}$ to $0$ is $ -e^{i\pi/4}\int_0^\infty e^{-ix^2}\,\mathrm{d}x\tag{3} $ Therefore, $(1)$ plus $(3)$ is $0$, so we get $ \int_0^\infty e^{-ix^2}\,\mathrm{d}x=\frac{1-i}{\sqrt2}\frac{\sqrt\pi}2\tag{4} $ Taking the imaginary part of $(4)$ yields that $ \int_0^\infty\sin(x^2)\,\mathrm{d}x=\sqrt{\frac\pi8}\tag{5} $ or $ \int_{-\infty}^\infty\sin(x^2)\,\mathrm{d}x=\sqrt{\frac\pi2}\tag{6} $