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 ?
Best Answer
Changing $u=z^2/n$ and $\zeta=n^{-1/2}$, the integral can be expressed as \begin{align} I(\zeta) &= \int_0^{+\infty} \frac{ne^{-\sqrt{t}}}{1+n^2t^2} \,dt\\ &=\int_0^{+\infty} \frac{2u}{1+u^4}e^{-\zeta u}\,du \end{align} We use the Mellin transform method for Laplace transforms with small parameters (DLMF). Here, \begin{align} h(u)&=\frac{2u}{1+u^4}\\ &\sim 2\sum_{s=0}^\infty (-1)^s u^{-3-4s} \end{align} for $u\to\infty$ and $h(u)=O\left( u \right)$ for $u\to 0$. The Mellin transform \begin{equation} \mathfrak M\left[ h(u)\right](z)=\frac{\pi}{2\sin\left( \frac{\pi}{4}(z+1) \right)} \end{equation} for $-1<z<3$. The asymptotic expansion can then be expressed as \begin{equation} I(\zeta)= \mathfrak M\left[ h(u)\right](1)+\sum_{k=2}^l \operatorname{Res}\left[-\zeta^{z-1}\Gamma(1-z)\frac{\pi}{2\sin\left( \frac{\pi}{4}(z+1) \right)};z=k \right]+O\left( \zeta^{l-\delta-1} \right) \end{equation} for $l>2$ and $\delta$ is a arbitrary small positive constant. Here $\psi(z)=\Gamma'(z)/\Gamma(z)$. We find, the values of the residues $R_k$ \begin{align} R_k&=\frac{2(-1)^{\frac{k+1}{4}}\zeta^{k-1}}{(k-1)!}\left[ \ln \zeta-\psi(k-2)-\frac{2k-3}{(k-1)(k-2)}\right]\text{ for } k= 3+4p \qquad (p=0,1,2\ldots)\\ &=\frac{(-1)^{k-1}\pi\zeta^{k-1}}{2(k-1)!\sin\left( \frac{\pi}{4}(k+1) \right)}\text{ for } k\ge 2 \text{ and } k\ne 3+4p\qquad (p=0,1,2\ldots) \end{align} By keeping the first terms ($k=2,3$) one obtains \begin{equation} I\sim\frac{\pi}{2}-\frac{\pi}{\sqrt{2n} }-\frac{1}{n}\left[ \ln \zeta+\gamma-\frac{3}{2}\right] \end{equation}