Asymptotic expansion of $\int_0^{+\infty} \frac{ne^{-\sqrt{t}}}{1+n^2t^2}\,dt$

I would like to get a general formula to get the asymptotic expansion at the order $n$ (so at whatever precision I want) of the following integral :

$$I = \int_0^{+\infty} \frac{ne^{-\sqrt{t}}}{1+n^2t^2} \mathrm{d}t$$

Some thoughts :

For now I am able to find an asymptotic expansion of order $1$. First intuitively we have :

$$I = O \left ( \frac{1}{n} \right )$$

From the expression :$$\frac{ne^{-\sqrt{t}}}{1+n^2t^2}$$

The change of variable : $\tan(\theta) = nt$ suggests itself and the problem boils down to calculating an asymptotic expansion of :

$$\int_0^{\pi/2} e^{-\sqrt{\frac{\tan(\theta)}{n}}} \mathrm{d}\theta$$

Now it is possible to use the dominated convergence theorem because we have :

$$e^{-\sqrt{\frac{\tan(\theta)}{n}}} \leq 1$$

Yet there is still a problem since $I$ is an improper and integral, so by using the dominated convergence theorem we only get an asymptotic expansion and not the exact value.

That’s why I get the following asymptotic (of order $1$):

$$\frac{\pi}{2} + O \left ( \frac{1}{n} \right )$$

Now is it possible to expand this in order to get a general formula for the asymptotic expansion of $I$ at whatever order we want ?