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



How to prove approximation for fresnel integral converges
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 $x$.


