Asymptotic expansion of $\int_{0}^{\infty}(1+\frac{u}{k})^{-k}e^{-u}du$ as $k\to \infty$ (Ablowitz) The following is from Complex Variables by Ablowitz & Fokas (P438).

Show that $\displaystyle \int_{0}^{\infty}\left(1+\frac{u}{k}\right)^{-k}e^{-u}du \sim \frac{1}{2}+\frac{1}{8k}$ as $k\to \infty$.

The original source had the third term, but it was omitted by the instructor of my course.
My attempt
Since the section is about Laplace's method, I tried to use it.
The integral can be written as
$\displaystyle \int_{0}^{\infty}e^{-k\log{(1+\frac{u}{k})}}e^{-u}du \tag*{}$
To use Laplace's method, we let $u/k=s$. Then we have $u=sk$ and $du=kds$. so the integral can be written as
$\displaystyle \int_{0}^{\infty}e^{-k\log{(1+s)}}e^{-sk}kds \tag*{}$
or
$\displaystyle k\int_{0}^{\infty}e^{-k(\log{(1+s)+s)}}ds \tag*{}$
Now we use Laplace's method. Since $h(s)=\log{(1+s)}+s$ attains minimum in $[0,\infty)$ at $s=0$, we have
$\displaystyle k\int_{0}^{\infty}e^{-k(\log{(1+s)+s)}}ds \sim \frac{1}{2}ke^{-kh(0)}\sqrt{\frac{2\pi}{k|h''(0)|}} = \sqrt{\frac{\pi k}{2}} \tag*{}$
...But this is obviously not what we wanted. I am stuck here.
 A: I am not sure if it is correct, so feel free to correct it. Denote the integral by $I(k)$.
We have $h'(s)=1+\frac{1}{1+s} = \frac{2+s}{1+s}$. Since $h'(s)>0$ on $[0,\infty)$, $h$ is a strictly increasing function. Thus, most contribution to the integral comes from the neighborhood of $s=0$, where $h(s)$ attains minimum (on $[0,\infty)$). Moreover, since it is a strictly increasing function, it has an inverse $h^{-1}$.
Now, we substitute $t=h(s)$. Then we have $s=h^{-1}(t)$. We have $t=0$ when $s=0$, and $t\to\infty$ when $s\to \infty$. Moreover, we have
$\displaystyle ds = \frac{dt}{h'(h^{-1}(t))} = \frac{1+h^{-1}(t)}{2+h^{-1}(t)}dt \tag*{}$
Therefore,
$\displaystyle I(k) = k \int_{0}^{\infty} \frac{1+h^{-1}(t)}{2+h^{-1}(t)}e^{-kt}dt \tag*{}$
We note that the contribution to the integral comes from the neighborhood of $t=0$. So, we find the asymptotic behavior of $\frac{1+h^{-1}(t)}{2+h^{-1}(t)}$ as $t\to 0$.
Since $t=h(s)=s +\log{(1+s)}$, we have
$\displaystyle t=s+s+O(s^2) =2s+O(s^2)\tag*{}$
Thus, $s=h^{-1}(t)=\frac{t}{2}+O(t^2)$.
Therefore, as $t\to 0$,
$\displaystyle \frac{1+h^{-1}(t)}{2+h^{-1}(t)} \sim \frac{1+\frac{t}{2}}{2+\frac{t}{2}} 
 \\ \displaystyle =\frac{1}{2} \frac{1+\frac{t}{2}}{1-(-\frac{t}{4})} \\ \displaystyle  = \frac{1}{2}(1+\frac{t}{2})(1-\frac{t}{4}+O(t^2)) \\ \displaystyle = \frac{1}{2}+\frac{t}{8} +O(t^2)$
Here, we used
$\frac{1}{1-x} = 1+x+O(x^2) \textrm{ as } x\to 0\tag*{}$
from the second line to the third line for $x=-\frac{t}{4}$.
Thus,
$\displaystyle I(k)\sim k\left(\frac{1}{2}\int_{0}^{\infty}e^{-kt}dt+\frac{1}{8}\int_{0}^{\infty}te^{-kt}dt\right) \\ \displaystyle = k\left(\frac{1}{2k}+\frac{1}{8k^2} \right) \\ = \displaystyle \frac{1}{2}+\frac{1}{8k}$
