Let
$$
S(t)=\int_{0}^{t}\sin(x^{2})\,dx\quad\text{and}\quad C(t)=\int_{0}^{t}\sin(x^{2})\,dx\,.
$$
Using the identities
$$
\sin(x^{2})=-\frac{1}{2x}\frac{d}{dx}\cos(x^{2})\quad\text{and}\quad\cos(x^{2})=\frac{1}{2x}\frac{d}{dx}\sin(x^{2})\,,
$$
integrating by parts over $[1,t]$, and passing to the limit as $t\to\infty$, one shows that there exist
$$
\lim_{t\to\infty}S(t)=S_{\infty}\in\mathbb{R}\quad\text{and}\quad\lim_{t\to\infty}C(t)=C_{\infty}\in\mathbb{R}\,.
$$
Moreover, since
$$
S_{\infty}=\int_{0}^{\infty}\frac{\sin y}{2\sqrt y}\,dy=\sum_{k=0}^{\infty}(-1)^{k}a_{k}\quad\text{where}\quad a_{k}=\int_{k\pi}^{(k+1)\pi}\frac{|\sin y|}{2\sqrt y}\,dy
$$
and $0<a_{k+1}<a_{k}$, we infer that $S_{\infty}>0$. To evaluate $S_{\infty}$ with the Feynman method, let us introduce the mappings
$$
f(t)=\left(\int_{0}^{t}\sin(x^{2})\,dx\right)^{2}+\left(\int_{0}^{t}\cos(x^{2})\,dx\right)^{2},\quad
g(t)=\int_{0}^{1}\frac{\sin(t^{2}(1-x^{2}))}{1-x^{2}}\,dx\,.
$$
We can check that $f'(t)=g'(t)$ and $f(0)=g(0)$, hence $f(t)=g(t)$ for every $t\ge 0$. Since
$$
g(t)=\int_{0}^{t^{2}}\frac{\sin(2x+x^{2}/t^{2})}{2x+x^{2}/t^{2}}\,dx\to\int_{0}^{\infty}\frac{\sin(2x)}{2x}\,dx=\frac{\pi}{4}\quad\text{as }t\to\infty\,,
$$
we obtain that
$$
S_{\infty}^{2}+C_{\infty}^{2}=\frac{\pi}{4}\,.
$$
Proving that
$$
(*)\qquad S_{\infty}^{2}=C_{\infty}^{2}\,,
$$
we conclude that $S_{\infty}=\sqrt{\pi/8}$. To show $(*)$ we introduce the mappings
$$
F(t)=\left(\int_{0}^{t}\cos(x^{2})\,dx\right)^{2}-\left(\int_{0}^{t}\sin(x^{2})\,dx\right)^{2},\quad
G(t)=\int_{0}^{1}\frac{\sin(t^{2}(1+x^{2}))}{1+x^{2}}\,dx\,.
$$
One can check that $F'(t)=G'(t)$ and $F(0)=G(0)$, hence $F(t)=G(t)$ for every $t\ge 0$ and in particular
$$
C_{\infty}^{2}-S_{\infty}^{2}=\lim_{t\to\infty}G(t)=\lim_{t\to\infty}\int_{0}^{t}\frac{t}{y^{2}+t^{2}}\,\sin(y^{2}+t^{2})\,dy\,.
$$
We split
$$
\int_{0}^{t}\frac{t}{y^{2}+t^{2}}\,\sin(y^{2}+t^{2})\,dy=\int_{0}^{1}\frac{t}{y^{2}+t^{2}}\,\sin(y^{2}+t^{2})\,dy+\int_{1}^{t}\frac{t}{y^{2}+t^{2}}\,\sin(y^{2}+t^{2})\,dy
$$
and we observe that
$$
\left|\int_{0}^{1}\frac{t}{y^{2}+t^{2}}\,\sin(y^{2}+t^{2})\,dy\right|\le\frac{1}{t}
$$
whereas
\begin{equation*}
\begin{split}
\int_{1}^{t}\frac{t}{y^{2}+t^{2}}\,\sin(y^{2}+t^{2})\,dy&=\left[-\frac{t\cos(y^{2}+t^{2})}{2y(y^{2}+t^{2})}\right]_{y=1}^{y=t}\\
&\qquad-\frac{1}{2t}\int_{1}^{t}\frac{3t^{2}y^{2}+t^{4}}{y^{2}(y^{4}+2t^{2}y^{2}+t^{4})}\,\cos(y^{2}+t^{2})\,dy
\end{split}
\end{equation*}
and one can easily see that each term tends to zero as $t\to\infty$. Hence $(*)$ is proved.
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.
Best Answer
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 $(*1)$.