I have this integral for which I need to calculate the leading asymptotic behavior: $$ I\left(x\right)=\int_0^{\frac{\pi}{2}}e^{-x\tan\left(t\right)}dt,\quad x\rightarrow \infty$$ $\phi\left(t\right)=\tan t \rightarrow t_0=0$
I should get this solution but I'm not quite sure how I get there correctly: $$I\left(x,\varepsilon\right)=\int_0^{\varepsilon}e^{-x\tan\left(t\right)}dt\rightarrow I\left(x\right)\sim\frac{1}{x}$$
On
Without Laplace mathod.
Using the exponential integral function $$\int e^{-x\tan\left(t\right)}\,dt=\frac i 2\left(e^{i x} \text{Ei}(-x (\tan (t)+i))-e^{-i x} \text{Ei}(-x (\tan (t)-i))\right)$$ Using the bounds (assuming $x>0$) $$I(x)=\int_0^{\frac{\pi}{2}}e^{-x\tan\left(t\right)}\,dt=\text{Ci}(x) \sin (x)+\frac{1}{2} (\pi -2 \,\text{Si}(x)) \cos (x)$$ Using asymptotics $$I(x)=\frac 1 x-\frac 2 {x^3}+\cdots$$ which is a good approximation as soon as $x \geq 2\pi$.
For this specific value $$I(2\pi)=\frac{\pi }{2}-\text{Si}(2 \pi )=0.152645$$ while the approximation gives $$\frac{1}{2 \pi }-\frac{1}{4 \pi ^3}=0.151092$$
We don't use Laplace's Method here. Instead, enforce the substitution $t\mapsto \arctan(t/x)$. Then, we have
$$\begin{align} \int_0^{\pi/2} e^{-x\tan(t)}\,dt&=\frac1x\int_0^\infty \frac{e^{-t}}{1+t^2/x^2}\,dt \end{align}$$
Finally, note that the Dominated Convergence Theorem guarantees that $\lim_{x\to\infty}\int_0^\infty \frac{e^{-t}}{1+t^2/x^2}\,dt= 1$. And we are done!