Using Euler's formula to evaluate $\int_0^{\infty} \sin(x^2)dx$

Question

We let $$ I = \int_0^{\infty} \sin(x^2)dx $$ In other terms $$ I = \int_0^{\infty} \Im\left(e^{ix^2}\right)dx = \Im \left(\int_0^{\infty} e^{ix^2} dx\right) $$ We let $$ J = \int_0^{\infty} e^{ix^2}dx $$ This looks like the gaussian integral so using the standard technique of conversion to polar coordinates we have $$ J^2 = \left( \int_0^{\infty} e^{ix^2}dx \right)\cdot\left( \int_0^{\infty} e^{iy^2}dy \right) = \iint_{\mathbb{R}^2} e^{i(x^2+y^2)}dA $$ Converting: $$ \begin{cases} r^2 = x^2 + y^2 \\ dA = r \, dr d\theta \end{cases} $$ we obtain $$ J^2 = \int_0^{2\pi} \int_0^\infty re^{ir^2} \, dr d\theta $$ Evaluating and simplifying: $$ J^2 = \frac{\pi}{i} \left[e^{ir^2} \right]_0^\infty $$ Does this diverge? What can i do here, because I am pretty sure that $I$ converges? Thanks

Answer 1

$$\cos \left(x^2\right)+i \sin \left(x^2\right)=e^{ix^2}$$ $$\int_0^{\infty}e^{ix^2}\,dx=\left[-\frac{(1+i) \sqrt{\pi } \text{erf}\left(-\frac{(1-i) x}{\sqrt{2}}\right)}{2 \sqrt{2}}\right]_0^{\infty}=\left(\frac{1}{2}+\frac{i}{2}\right) \sqrt{\frac{\pi }{2}}$$ $$\int_0^{\infty}\sin(x^2)\,dx=\text{Im}\left(\int_0^{\infty}e^{ix^2}\,dx\right)=\text{Im}\left[\left(\frac{1}{2}+\frac{i}{2}\right) \sqrt{\frac{\pi }{2}}\right]=\frac{1}{2}\sqrt{\frac{\pi }{2}}$$