<s>[INST] Asymptotic expansion of integral Find the first two terms in asymptotic expansion (as $x \rightarrow \infty$) of the following integral
$$\int_{0}^{\frac{\pi}{2}}e^{-x \sin^{2} t}dt.$$
 [/INST] As Lucian pointed out in a comment $$f(x)=\int_{0}^{\frac{\pi}{2}}e^{-x \sin^{2} t}dt=\frac{ \pi}{2}  e^{-x/2}\, I_0\left(\frac{x}{2}\right)$$ At this point, one could use the asymptotic expansion given by Hankel for the modified Bessel function of the first kind $$I_{\alpha}(z)=\frac{e^z}{\sqrt{2\pi z}}\Big(1-\frac {4\alpha^2-1}{8z}+\frac {(4\alpha^2-1)(4\alpha^2-9)}{2! (8z)^2}-\frac {(4\alpha^2-1)(4\alpha^2-9)(4\alpha^2-25)}{3! (8z)^3}+\cdots\Big)$$ Applied to $\alpha=0$ and $z=\frac x2$, this gives $$f(x)\approx\frac{\sqrt{\pi } }{2
   \sqrt{x}}\left(1+\frac{1}{4 x}+\frac{9}{32 x^2}+\frac{75}{128 x^3}+\cdots\right)$$
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