I would like to calculate the leading term of an asymptotic expansion of $$I(r)=\int_{-k}^{k} \frac{\sin{(\alpha R)}}{\alpha \sqrt{k^2-\alpha^2}}e^{ir(\sqrt{k^2-\alpha^2}\sin{\theta}+\alpha\cos{\theta})}d\alpha$$
as $r\to\infty$, where $R>0,k>0,\theta \in [0,\pi]$ is a constant.
My attempt
It looks like I can apply the method of stationary phase, so I tried to apply it.
Let $h(\alpha)=\sqrt{k^2-\alpha^2}\sin{\theta}+\alpha \cos{\theta}$. Then,
\begin{align} h'(\alpha) &= \frac{-\alpha \sin{\theta}}{\sqrt{k^2-\alpha^2}}+\cos{\theta}\\ h''(\alpha)&=\frac{-k^2 \sin{\theta}}{(k^2-\alpha^2)^{3/2}} \end{align}
We can easily verify that a solution to $h'(\alpha)=0$ is given by $\alpha = k\cos{\theta}$. It is the only solution since $h''(\alpha)<0$ which means $h'(\alpha)$ is strictly decreasing.
If $\alpha=k\cos{\theta}$, we have $\sqrt{k^2-\alpha^2}=k\sin{\theta}$, so
\begin{align} h(k\cos{\theta})&=k\sin^2{\theta}+k\cos^2{\theta}=k \\h''(k\cos{\theta})&=-\frac{k^2\sin{\theta}}{k^3\sin^3{\theta}}=-\frac{1}{k\sin^2{\theta}}\end{align}
Therefore, by the method of stationary phase,
\begin{align} I(r) &\sim \frac{\sin{(Rk\cos{\theta})}}{k^2\sin{\theta}\cos{\theta}}\sqrt{\frac{2\pi k\sin^2{\theta}}{r}}e^{i(kr-\frac{\pi}{4})} \\&= \frac{\sin{(Rk\cos{\theta})}}{k^{3/2}\cos{\theta}}\sqrt{\frac{2\pi}{r}}e^{i(kr-\frac{\pi}{4})} \end{align}
Let the asymptotic be $A(r)$.
I thought this was correct, but when I tried to verify numerically, when I had $k=1,\theta=1,R=0.1$ and $r=970$, I had $I(r)\simeq-0.0293163-0.01500839i$ while $A(r)\simeq-0.0002675+0.0080399i$ which is pretty far off. Is it something I can ignore "because it is asymptotic", or is there a fundamental mistake?