Improper Integral: Fourier Transform? Or Complex Numbers?

We all know that $\int^{\infty}_0 e^{-x^2}dx=\frac{\sqrt\pi}{2}$ (Result of Gaussian Integral, the integrand in it is an even function). Also, $\int^\infty_0e^{ix^2}dx=\frac{e^{i\frac\pi4}}{2}\sqrt\pi$ Which can easily be proved by the previous claim.

Now comes the drum roll! $e^{ix^2}=\cos(x^2)+i\sin(x^2)$ By the second claim, we get $\int^\infty_0\cos(x^2)dx+i\int^\infty_0\sin(x^2)dx=\frac{\sqrt\pi}{2}\left(\frac{1}{\sqrt2}+i\frac1{\sqrt2}\right)$ Comparing the imaginary parts, we get $\int^{\infty}_0\sin(x^2)dx=\sqrt{\frac\pi8}\approx0.62$

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(This solution is not completely mine, my idea was to use the unnormalized sinc function and take the inverse Fourier Transform, but this also is good)

We know that $\int_0^\infty\sin(x^2)dx=\int_0^\infty\cos(x^2)dx$

$\hat{f}(\xi)=\int_{-\infty}^\infty dx \frac{1}{\sqrt{|x|}}e^{-2\pi\mathrm{i}x\xi}=2\mathrm{Re}\int_0^\infty dx \frac{1}{\sqrt{x}}e^{-2\pi\mathrm{i}x\xi}=\frac{1}{\sqrt{|\xi|}}\$ $I=\int_0^\infty dx\cos(x^2)=\mathrm{Re}\int_0^\infty dx\ e^{\mathrm{i}x^2}=\frac{1}{2}\mathrm{Re}\int_0^\infty dy\ \frac{e^{\mathrm{i}y}}{\sqrt{y}}=\frac{1}{4}\hat{f}\left(-\frac{1}{2\pi}\right)=\sqrt{\frac{\pi}{8}}\ .$

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Finally, the $\text{\sinc}$ function solution!

Fourier Transform of the sinc function is defined as: $\hat{g}(\xi)=\int_{-\infty}^\infty \frac{\sin(x^2)}{x^2}e^{-2 \pi\mathrm{i}\xi x}dx$ $\frac{\partial^2}{\partial\xi^2}\hat{g}(\xi)=-4\pi^2\int_{-\infty}^\infty dx\sin(x^2)e^{-2 \pi\mathrm{i}\xi x}$ Evaluating at $\xi=0$ , $\int_0^\infty dx\sin(x^2)=-\frac{1}{8\pi^2}\frac{\partial^2}{\partial\xi^2}\hat{g}(\xi)\Big|_{‌​\xi\to 0}$ Which gives the required answer.