How to prove approximation for fresnel integral converges

Question

I was looking at the fresnel integral $S(x)=\int^x_0\sin(s^2)ds$. From reading I learned that this integral approaches $\frac{1}{2} \sqrt{\frac{\pi}{2}}$ as $x \rightarrow \infty$. Through messing around on desmos, I found an excellent estimate: $S_2(x)=\frac{\sin(x^2)+2x^2\cos(x^2)}{-4x^3}+\frac{1}{2}\sqrt{\frac{\pi}{2}}$, which is not a good approximation near $x=0$, but quickly converges to $S(x)$ as $x\rightarrow \infty$. For my purposes, combining this with the taylor approximation for $S$ centered at $x=0$ yeilds sufficiently accurate results. While it is clear through solving the limit that $S_2$ will arrive at the same equilibrium value as $S$ at infinity, I neither know how to prove, or how to describe the sense in which it converges to the integral curve's shape over that range. I would appreciate if someone could shed some light on this.

Thanks

Answer 1

If you want to build an asymptotics, write $$S(x)=\int^x_0\sin(s^2)\,ds=\frac{1}{2}\sqrt{\frac{\pi }{2}}-\int_x^\infty\sin(s^2)\,ds.$$ Let $s=\sqrt x$ $$\int_x^\infty\sin(s^2)ds=\frac 12\int_{x^2}^\infty\frac{\sin (t)}{\sqrt{t}}\,dt.$$ Integrate by parts a few times and using the bounds, you will obtain \begin{gather*} \int_{x^2}^\infty\frac{\sin (t)}{\sqrt{t}}\,dt=\frac{\cos \left(x^2\right)}{x}+\frac{\sin \left(x^2\right)}{2 x^3}-\frac{3 \cos \left(x^2\right)}{4 x^5}+\dotsb \\ S(x)=\frac{1}{2}\sqrt{\frac{\pi }{2}}-\frac{\cos \left(x^2\right)}{2x}-\frac{\sin \left(x^2\right)}{4 x^3}+\frac{3 \cos \left(x^2\right)}{8 x^5}+\dotsb. \end{gather*}

Use $x=12.34$ : the above gives $0.622816621$ to be compared to the value of $0.622816642$.

Edit

Making the problem more general, we can write \begin{multline*} \color{blue}{S(x)=\frac{1}{2}\sqrt{\frac{\pi }{2}}-\frac{\cos \left(x^2\right)}{2x}\Bigl[1+\sum_{n=1}^\infty (-1)^n \frac {(4n)!}{(2n)!\, (16x^4)^n}\Bigr]-{}} \\ \color{blue}{\frac{\sin \left(x^2\right)}{4x^3}\Bigl[1+\frac 12\sum_{n=1}^\infty (-1)^n \frac{(4 n+2)!}{ (2 n+1)!\, (16x^4)^n}\Bigr]} \end{multline*} which is extremely good even for rather small values of $x$.