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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} ?

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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.

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    (+1) You might consider mentioning the reason that while $u$ is unbounded, the expansion herein remains valid. -Mark – 2017-02-27
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    when you do exponentiating, do we have e to the u^2/2k ? – 2017-02-27
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    how you get second equation from first equation? when you do exponential, the right hand side are not equal – 2017-02-27
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    anyone can show me how to get from the right hand side of first equation to the right hand side of second equation after taking exponential? – 2017-02-27
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    $$ \exp{\left(\frac{u^2}{2k}+ O\left(\frac{u^3}{k^2}\right)\right)} = 1+\frac{u^2}{2k}+O\left( \frac{u^3}{k^2} \right) $$ by using Taylor's Theorem again. – 2017-02-27