Infinite Series of the asymptotic expansion of Fresnel Integrals I need to find the infinite series for the asymptotic expansions of the fresnel integrals as $x\rightarrow \infty$ and $x\rightarrow 0$. 
Now I have computed the asyptotic expansions to be as follows
$$C(x)=\sqrt{\frac{\pi}{8}}+ \frac{\sin(x^2)}{2x} - \frac{\cos(x^2)}{4x^3} - \frac{3\sin(t^2)}{8t^5} + \mathcal{O}(x^{-7}) $$ and $$S(x)=\sqrt{\frac{\pi}{8}}- \frac{\cos(x^2)}{2x} - \frac{\sin(x^2)}{4x^3} + \frac{3\cos(t^2)}{8t^5} + \mathcal{O}(x^{-7}).$$
Not sure where to go from here, any ideas?
 A: I'll only cover the expansion as $x \to +\infty$ here, the expansion as $x \to 0$ is trivial. It will be easier to present the expansion if we group the real and imaginary parts together.
$$C(x) + iS(x) \asymp \sqrt{\frac{\pi}{8}}(1+i) + \frac{e^{ix^2}}{2}\sum_{n=0}^\infty \frac{(-1)^{n+1} \left(\frac12\right)_n}{x^{2n+1}}\tag{*1}$$
where $\left(\lambda\right)_n = \lambda(\lambda+1)\cdots(\lambda+n-1)$ is the $n^{th}$ rising Pochhammer symbol.
It is known that $\displaystyle\;C(+\infty) = S(+\infty) = \sqrt{\frac{\pi}{8}},\;$ we have
$$\sqrt{\frac{\pi}{8}}(1+i) - (C(x) + iS(x)) = \int_x^\infty e^{i z^2} dz$$
We can view the last integral as a contour integral from $x$ to $+\infty$.
Since the $e^{it^2}$ factor decays to zero rapidly as $|z| \to \infty$ in the $1^{st}$ quadrant, we can deform the contour to one from $x$ to $e^{i\pi/4} \infty$ without changing its value.
Introduce parametrization $z = x \sqrt{1 + it}$, we find
$$\int_x^\infty e^{iz^2} dz = \frac{i x e^{ix^2}}{2} \int_0^\infty e^{-x^2 t} \frac{dt}{\sqrt{1+it}}
= \frac{i x e^{ix^2}}{2} \int_0^\infty e^{-x^2 t}\underbrace{\sum_{n=0}^\infty \frac{(-it)^n \left(\frac12\right)_n}{n!}}_{\text{expansion of }1/\sqrt{1+it}} dt$$
Even though the expansion inside the rightmost integral is only valid for $t < 1$,
the whole thing is in a form which we can apply Watson's Lemma. We can integrate the expansion term by term and deduce the asymptotic expansion in $(*1)$.
