How to use Laplace method to show $$\int_0^\infty\left(1+\frac uk\right)^{-k} e^{-u}du \sim \frac12 + \frac 1{8k}$$ as $k \to \infty$
I know $\left(1+\frac uk\right)^{-k} \to e^{-u}$ as $k \to \infty$, so the integration go to 1/2, but how to show the 1/2+1/{8k} ?
We want to find a better approximation to $(1+u/k)^{-k}$ than $e^{-u}$, so let's take logs. Then Taylor's theorem with integral remainder gives $$ -k\log{\left( 1 + \frac{u}{k} \right)} = -u + \frac{u^2}{2k} + \int_0^u \frac{(u-t)^2}{2} \frac{-2}{k^2(1+t/k)^3} \, dt, $$ and the latter term is clearly $O(u^3/k^2)$. Exponentiating, we find $$ \left(1+\frac{u}{k}\right)^{-k} = e^{-u} \left( 1 + \frac{u^2}{2k} + O(u^3/k^2) \right), $$ and then we can use $\int_0^{\infty} u^n e^{-2u} \, du = n!/2^{n+1}$ to arrive at the answer.