Continuing the improper integrals!

$\begin{array}{rcl} \displaystyle \int_0^X \sin(x^2)\sin x\,dx & = &\displaystyle \tfrac12\int_0^X \big[\cos(x^2-x) - \cos(x^2+x)\big]\,dx \\ & = &\displaystyle \tfrac12\int_0^X \big[\cos\big((x-\tfrac12)^2 - \tfrac14\big) - \cos\big((x+\tfrac12)^2 - \tfrac14\big)\big]\,dx \\ & = & \displaystyle \tfrac12\left(\int_{-\frac12}^{X-\frac12} \cos(x^2-\tfrac14)\,dx - \int_{\frac12}^{X+\frac12} \cos(x^2-\frac14)\,dx\right) \\ & = & \displaystyle \tfrac12\left( \int_{-\frac12}^{\frac12} \cos(x^2-\tfrac14)\,dx - \int_{X-\frac12}^{X+\frac12} \cos(x^2 - \tfrac14)\,dx\right) \end{array}$ Letting $X \to \infty$ , we deduce that $\begin{array}{rcl} \displaystyle \int_0^\infty \sin(x^2) \sin x\,dx & = &\displaystyle \tfrac12\int_{-\frac12}^{\frac12} \cos(x^2-\tfrac14)\,dx \\ & = &\displaystyle \cos\tfrac14 \int_0^{\frac12} \cos (x^2)\,dx + \sin\tfrac14\int_0^{\frac12} \sin(x^2)\,dx \\ & = &\displaystyle \sqrt{\frac{\pi}{2}}\left[ \cos\tfrac14 C\big(\tfrac{1}{\sqrt{2\pi}}\big) + \sin\tfrac14S\big(\tfrac{1}{\sqrt{2\pi}}\big) \right] \end{array}$ where $C$ and $S$ are the Fresnel integrals. At this point, I reached for my copy of Mathematica to get $0.4917...$ .

This is not an appropriate solution, but I show you how I solve this using some tool.

First we divide into two parts, namely $\int_{0}^{1} \sin(x)\sin(x^2) \, dx \, + \, \int_{1}^{\infty} \sin(x)\sin(x^2) \, dx.$ For the right one, we change $x$ to be $\frac{1}{x}$ , and we finally express it as $\int_{0}^{1} \sin(x)\sin(x^2) \, dx \, + \, \int_{0}^{1} \frac{1}{x^2}\sin\left(\frac{1}{x}\right)\sin\left(\frac{1}{x^2}\right) \, dx.$ Therefore, what we need to find is $\int_{0}^{1} \sin(x)\sin(x^2) + \frac{1}{x^2}\sin\left(\frac{1}{x}\right)\sin\left(\frac{1}{x^2}\right) \, dx.$ I use Mathematica to numerically compute this integral. Because It is more precise when we use a fixed interval, here it is $(0,1)$ . The integral equals 0.492 approximately, so the answer is $\boxed{0.49}$ .

Some may wonder if we can calculate it using numerical methods. But take note that the original integral is over an infinite range, which is not applicable, of course. Even the form I derived, it is not bounded as $x \to 0^+$ , so I cannot evaluate by standard methods like composite integration rules, etc.