Calculating $\int_{0}^{\infty}\sin(x^{2})dx$

Solution 1:

Your parametrisation of the third integral is rather complicated. Why not just write $\gamma_3:[0,R]\rightarrow \mathbb{C}$, $t\mapsto -e^{\pi i/4}t$. Then the integral becomes $$ \int_R^0 e^{-e^{\pi i/2}t^2}e^{\pi i/4}dt = \int_R^0 e^{-it^2}e^{\pi i/4}dt = e^{\pi i/4} \int_R^0 \cos t^2 - i \sin t^2 dt. $$ I am sure you can take it from there.

As for bounding the integral $\int_{\gamma_2}e^{-z^2}dz$, the length of the contour grows linearly with $R$. How fast does the maximum of the integrand decay? It's the standard approach, using the fact that $$ \left|\int_\gamma f(z) dz\right|\leq \sup\{|f(z)|: z \in \text{ image of }\gamma\}\cdot \text{length of }\gamma. $$

Solution 2:

$$\int_0^R e^{-t^2}dt= \frac{\sqrt{\pi}}{2}+o(1). \tag{1}$$

$$t\in\left[0,\frac{\pi}{4}\right] \implies \operatorname{Re}(Re^{it})\ge \frac{R}{\sqrt{2}} \implies \left|\int_{\gamma_2}e^{-z^2}dz\right|\le R\frac{\pi}{4}e^{-R^{\,2}/2}\to0 \tag{2}$$

$$\alpha=\int_0^M e^{it^2}dt \implies \int_0^M \sin(t^2)dt=\frac{\alpha+\overline{\alpha}}{2}. \tag{3}$$

Try using these deductions for $\gamma_1$, $\gamma_2$ and $\gamma_3$ respectively.

Solution 3:

An easy way to evaluate $\int_{0}^{\infty}\sin(x^{2})dx$

$$\int_0^{\infty}e^{-ax^2}dx=\frac{\sqrt{\pi}}{2\sqrt{a}}$$ Now replace $a\rightarrow ia$

$$ \int\limits_0^\infty \cos \left( {a{x^2}}\right)dx-i \int\limits_0^\infty \sin \left( {a{x^2}}\right)dx= \frac{\sqrt{\pi}}{2\sqrt{a}\sqrt{i}} $$ But

$$\frac{1}{\sqrt{i}}= \frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}} $$ So

$$ \int\limits_0^\infty \cos \left( {a{x^2}}\right)dx=\int\limits_0^\infty \sin \left( {a{x^2}}\right)dx= \frac{\sqrt{\pi}}{2\sqrt{2a}} $$