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 $(*1)$.
Hint: $\cos(\tau)$ is the real part of $\exp(i\tau)$, which you combine with the $\exp(-p \tau)$ in one integral, and use the definition of the Gamma function.
Best Answer
I suppose you want an asymptotic expansion as $x\to \infty$. We start with
$$S(x) = \int_0^x \sin \left(\frac{\pi\cdot t^2}{2}\right)\,dt = \frac{1}{2} - \int_x^\infty \sin \left(\frac{\pi\cdot t^2}{2}\right)\,dt.$$
Now, to get a handle on that integral, we substitute $u = \frac{\pi t^2}{2}$ and obtain
$$\int_x^\infty \sin \left(\frac{\pi\cdot t^2}{2}\right)\,dt = \frac{1}{\sqrt{2\pi}} \int_{\pi x^2/2}^\infty \frac{\sin u}{\sqrt{u}}\,du.$$
Then we want an asymptotic expansion of
$$\int_y^\infty \frac{\sin u}{\sqrt{u}}\,du$$
which we get via integration by parts:
\begin{align} \int_y^\infty \frac{\sin u}{\sqrt{u}}\,du &= \left[-\frac{\cos u}{\sqrt{u}}\right]_y^\infty - \frac{1}{2}\int_y^\infty \frac{\cos u}{u^{3/2}}\,du\\ &= \frac{\cos y}{\sqrt{y}} - \frac{1}{2}\int_y^\infty \frac{\cos u}{u^{3/2}}\,du\\ &= \frac{\cos y}{\sqrt{y}} - \frac{1}{2}\left(\left[\frac{\sin u}{u^{3/2}}\right]_y^\infty + \frac{3}{2}\int_y^\infty \frac{\sin u}{u^{5/2}}\,du\right)\\ &= \frac{\cos y}{\sqrt{y}} + \frac{\sin y}{2y^{3/2}} - \frac{3}{4}\int_y^\infty \frac{\sin u}{u^{5/2}}\\ &= \frac{\cos y}{\sqrt{y}} + \frac{\sin y}{2y^{3/2}} - \frac{3\cos y}{4y^{5/2}} - \frac{15}{8} \int_y^\infty \frac{\cos u}{u^{7/2}}\,du. \end{align}
An elementary estimate shows that the last integral is $O(y^{-5/2})$ [actually, it is $O(y^{-7/2})$, as one sees when thinking about what a further integration by parts yields], so we get the asymptotic expansion
$$\int_y^\infty \frac{\sin u}{\sqrt{u}}\,du = \frac{\cos y}{\sqrt{y}} + \frac{\sin y}{2y^{3/2}} + O(y^{-5/2}),$$
and hence, inserting $y = \frac{\pi x^2}{2}$,
$$S(x) = \frac{1}{2} - \frac{\cos \frac{\pi x^2}{2}}{\pi x} - \frac{\sin \frac{\pi x^2}{2}}{\pi^2 x^3} + O(x^{-5}).$$
To get higher-order asymptotic expansions, integrate by parts more often.