If I want to prove that \begin{equation*} \int_{0}^{\infty}\sin(x^2)dx=\int_{0}^{\infty}\cos(x^2)dx=\frac{\sqrt{2\pi}}{4} \end{equation*}
First method: It is possible to approach it by the method in which we consider a closed curve, then: Let gamma be one eighth of a circle ($\theta\in[0,\pi/4]$), then \begin{eqnarray*} 0 = \int_{\gamma} e^{iz^{2}} dz & = & \int_{\gamma_{1}} e^{iz^{2}} dz + \int_{\gamma_{2}} e^{iz^{2}} dz + \int_{\gamma_{3}} e^{iz^{2}} dz \\ & = & \int_{0}^{R} e^{iz^{2}(r)} dz(r) + \int_{0}^{\pi/4} e^{iz^{2}(\theta)} dz(\theta) + \int_{R}^{0} e^{iz^{2}(r)} dz(r) \\ & = & \int_{0}^{R} \cos(x^{2}) + i\sin(x^{2}) dx + \int_{0}^{\pi/4} e^{iz^{2}(\theta)} dz(\theta) + \int_{R}^{0} e^{iz^{2}(r)} dz(r) \\ \end{eqnarray*}
Then \begin{equation*} -\int_{0}^{R} \cos(x^{2}) + i\sin(x^{2}) dx = \int_{0}^{\pi/4} e^{iz^{2}(\theta)} dz(\theta) + \int_{R}^{0} e^{iz^{2}(r)} dz(r) \end{equation*} where \begin{eqnarray*} \int_{R}^{0}\exp\left(iz^{2}(r)\right)dz(r) & = & \int_{R}^{0}\exp\left(i(re^{i\pi/4})^{2}\right)e^{i\pi/4}dr \\ & = & \int_{R}^{0}\exp\left(ir^{2}e^{i\pi/2}\right) \cdot e^{i\pi/4}dr \\ & = & -e^{i\pi/4} \int_{0}^{R}\exp\left(ir^{2}e^{i\pi/2}\right)dr \\ & = & -e^{i\pi/4} \int_{0}^{R}\exp\left(ir^{2}[\cos(\pi/2)+i\sin(\pi/2)]\right)dr \\ & = & -e^{i\pi/4} \int_{0}^{R}\exp\left(-r^{2}\right)dr \\ \end{eqnarray*} and we know that \begin{eqnarray*} \lim_{R\to\infty} \int_{0}^{R}\exp\left(-r^{2}\right)dr = \frac{\sqrt{\pi}}{2} \end{eqnarray*} Then \begin{equation*} \begin{split} & \lim_{R\to\infty}\int_{R}^{0}\exp\left(iz^{2}\right)dz = -\left( \frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2} \right) \frac{\sqrt{\pi}}{2}\\ \Rightarrow & \lim_{R\to\infty}\int_{R}^{0}\exp\left(iz^{2}\right)dz = -\frac{\sqrt{2\pi}}{4}-i\frac{\sqrt{2\pi}}{4} \end{split} \end{equation*} And I wish that $\left|\int_{\gamma_{2}} e^{iz^{2}}dz\right|=|\int_{0}^{\pi/4} e^{iz^{2}(\theta)} dz(\theta)|\to 0$ when $R\to\infty$. How can I argue this in detail?
since using this I would have to \begin{equation*} \lim_{R\to\infty}\int_{0}^{R} \cos(x^{2}) + i\sin(x^{2}) dx = \frac{\sqrt{2\pi}}{4} + i\frac{\sqrt{2\pi}}{4} \end{equation*} then \begin{equation*} \int_{0}^{\infty}\sin(x^{2})dx = \int_{0}^{\infty}\cos(x^{2})dx = \frac{\sqrt{2\pi}}{4} \end{equation*} that concludes the desired.
Second method: On the other hand, Also study the possibility of doing this problem using power series. We know that \begin{equation*} \sin(z) = \sum_{i=0}^{\infty}\frac{(-1)^{n}z^{2n+1}}{(2n+1)!} \Rightarrow \sin(z^{2}) = \sum_{i=0}^{\infty}\frac{(-1)^{n}(z^{2})^{2n+1}}{(2n+1)!} \end{equation*} Then \begin{eqnarray*} \int_{0}^{r}\sin(x^{2})dx & = & \int_{0}^{r} \sum_{n=0}^{\infty}\frac{(-1)^{n}(x^{2})^{2n+1}}{(2n+1)!} dx\\ & = & \int_{0}^{r} \sum_{n=0}^{\infty}\frac{(-1)^{n}x^{4n+2}}{(2n+1)!} dx\\ & = & \sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)!}\int_{0}^{r} x^{4n+2} dx\\ & = & \sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)!}\left. \cdot\frac{x^{4n+3}}{4n+3} \right|_{0}^{r}\\ & = & \sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)!} \cdot \frac{r^{4n+3}}{4n+3} \\ \end{eqnarray*}
But I do not know how to reduce or work with this expression, if someone could help me I would be very grateful.
PD: If there is another method, perhaps by Fourier analysis it is also welcome, although my main interest is to exercise with the theory of the complex variable.
The integral which you mention is $$\int_{0}^{\frac{\pi}{4}} \exp(i (R cos(\theta) + i R sin(\theta))^2) i R \exp( i \theta) d\theta$$
Taking the modulus of this thing, it is less than $$\int_{0}^{\frac{\pi}{4}} R \exp(-R^2 sin(2 \theta)) d \theta = \frac{R}{2} \int_{0}^{\frac{\pi}{2}} \exp(-R^2 sin(\theta)) d\theta$$
on $[0,\frac{\pi}{2}]$ by concavity of sine, we have $\sin(\theta) \geq \frac{2 \theta}{\pi}$
so $$\frac{R}{2} \int_{0}^{\frac{\pi}{2}} \exp(-R^2 sin(\theta)) d\theta \leq \frac{R}{2} \int_{0}^{\frac{\pi}{2}} \exp(-R^2 \frac{2 \theta}{\pi}) d \theta = \frac{\pi(1 - \exp(-R^2))}{4 R}$$ and $$\lim_{R \rightarrow \infty} \frac{\pi(1 - \exp(-R^2))}{4 R} = 0$$ so we get the result you desired.