TITLE: Asymptotic expansion of $\int_0^{+\infty} \frac{ne^{-\sqrt{t}}}{1+n^2t^2}\,dt$
QUESTION [13 upvotes]: 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 ?

REPLY [4 votes]: Using a CAS.
$$I = \int_0^{+\infty} \frac{ne^{-\sqrt{t}}}{1+n^2t^2} \,dt=\frac{64 \sqrt{2} }{\pi ^{3/2}}n^2\, G_{1,5}^{5,1}\left(\frac{1}{256 n^2}|
\begin{array}{c}
 \frac{3}{2} \\
 1,\frac{5}{4},\frac{3}{2},\frac{3}{2},\frac{7}{4}
\end{array}
\right)$$ where appears the Meijer G function.
Expanding for large values of $n$
$$I=\frac{\pi }{2}-\frac{\pi }{\sqrt{2}\,n^{1/2}}+\frac{\log
   \left({n}\right)-2 \gamma +3}{2\, n}+\frac{\pi 
  }{6 \sqrt{2}\,n^{3/2}}-\frac{\pi }{48\,
   n^2}+O\left(\frac{1}{n^{5/2}}\right)$$ For $n=10$, the "exact" result would be $\approx 1.086826$ while the above gives $\approx 1.086776$.
No problem to get more terms (if you want).