I want to apply the Watson's theorem to find an asymptotic expansion for the function $$f(z)=\int_{- \infty}^{\infty} e^{-z \frac{y^{2}}{2}} \sin(y^{2})dy$$
(Assume $z \rightarrow \infty, z>0$). But I need to find a domain which contains the real axis and the function:
$$g(z)=\sin(z^{2})$$
Is bounded on it? Is there any such domain? If $z=x+iy$ then: $$|g(z)| \leq \frac{e^{-2xy}+e^{2xy}}{2}$$ But the right hand side is not bounded for any domain $|y|<c, c>0$.
Observe that $$ f(z)=\int_{- \infty}^{\infty} e^{-z \frac{y^{2}}{2}} \sin(y^{2})\mathrm dy=2\int_{0}^{\infty} e^{-z \frac{y^{2}}{2}} \sin(y^{2})\mathrm dy $$ and putting $t=\frac{y^2}{2}$ we have $$ f(z)=\int_{0}^{\infty} e^{-z t} 2\frac{\sin (2t)}{\sqrt{2t}}\mathrm dt. $$ Observe that, $$ g(t)=2\frac{\sin (2t)}{\sqrt{2t}} $$ is continuous in $(0,\infty)$, and using the Taylor's series for $\sin(2t)$, $$ g(t)=2\frac{\sin (2t)}{\sqrt{2t}}\sim \frac{2}{\sqrt{2t}}\sum_{k=0}^{\infty} \frac{(-1)^k (2t)^{2 k+1}}{(2 k+1)!}=\sum_{k=0}^{\infty} \underbrace{\frac{(-1)^k 2^{2 k+\frac{3}{2}}}{(2 k+1)!}}_{a_k}t^{\overbrace{2 k+\frac{3}{2}}^{\lambda_k}-1}. $$ that is $g(t)\sim \sum_{k=0}^{\infty} a_k t^{{\lambda_k}-1}$ for $t\to 0$ with $0<\lambda_0<\lambda_1<\ldots$.
Moreover, for some fixed $c>0$, $g(t)=\mathcal{O}(e^{ct})$ for $t\to\infty$.
Thus, for the Watson's Lemma, $$ f(z)\sim \sum_{k=0}^{\infty} a_k \frac{\Gamma(\lambda_k)}{z^{\lambda_k}}=\sum_{k=0}^{\infty}\frac{(-1)^k 2^{2 k+\frac{3}{2}}}{(2 k+1)!}\frac{\Gamma\left(2 k+\frac{3}{2}\right)}{z^{2 k+\frac{3}{2}}} $$ for $|z|\to\infty$ and $|\mathrm{arg}(z)|<\frac{\pi}{2}-\delta<\frac{\pi}{2}$ for some $\delta$ such that $0<\delta<\frac{\pi}{2}$.