I want to prove the equation:
$$ \int_0^\infty \sin x^2dx=\frac{1}{2}\sqrt\frac{\pi}2{} $$ This was my calculating process: $$ ||e^{iz^2}||=||e^{iR^2e^{i2\theta}}||=e^{-R^2\sin 2\theta} $$ Using the Jordan Inequality $$ \int_0^\frac{\pi}{2}e^{-Rsin\theta}d\theta\le \frac{\pi}{2R}(R>0) $$ I got $$ \int_0^\frac{\pi}{4}e^{-R^2\sin 2\theta}d\theta \le \frac{\pi}{4R^2} $$ Note that the symmetry in $\sin x$
I got $$ \int_0^\frac{\pi}{2}e^{-R^2\sin 2\theta}d\theta \le \frac{\pi}{2R^2} $$ Using the Cauchy-gursat theroem,I had $$ \int_0^Re^{ix^2}dx+\int_{C_R}e^{iz^2}dz+\int_R^0e^{-iy^2}dy=0 $$
The shape of the integral path was like this
Then $$ |\int_{C_R}e^{iz^2}dz| \le \int_0^{\frac{\pi}{2}}|e^{-R^2\sin 2\theta}Rie^{i\theta}|d\theta\le \frac{\pi}{2R} $$ When $R \rightarrow \infty$，I got $$ \int_{C_R}e^{iz^2}dz=0 $$ Then $$ \int_0^\infty e^{ix^2}dx+\int_\infty^0e^{-iy^2}dy=0 $$ Then $$ \int_0^\infty(\cos x^2+i\sin x^2) dx-\int_0^\infty(\cos y^2-i\sin y^2)dy=0 $$ Then $$ 2\int_0^\infty i\sin x^2 dx=0 $$ So $$ \int_0^\infty \sin x^2dx=0 $$ What's wrong with that?Where was my mistake?
I would appreciate it if you could help me.Thank you!
Your mistake is the missing imaginary unit in this equation: $$ \int_0^Re^{ix^2}dx+\int_{C_R}e^{iz^2}dz+\color{red}i\int_R^0e^{-iy^2}dy=0. $$ This amounts to the equation: $$ \int_0^\infty\cos(x^2)+i\sin(x^2)dx=\int_0^\infty\sin(y^2)+i\cos(y^2)dy $$ or $$ \int_0^\infty\cos(x^2)dx=\int_0^\infty\sin(x^2)dx, $$ which is though correct but does not help to evaluate the expression.
To avoid this stop the circle path at $\theta=\frac\pi4$, return to $0$ along the line $\theta=\frac\pi4$, and use the well-known result $\int_0^\infty e^{-x^2}dx=\sqrt\frac\pi2$.