I am trying to find a "simple" (for example a function $f$ that has a nice antiderivative) function $f$ such that : $$f \sim \frac{e^u}{u} \text{as } u \to +\infty$$
I have rewritten $e^{u-\ln(u)}$ yet it doesn't help, because finding $g$ such that : $g \sim \ln(u)$ is hard, because the series expansion of $\ln(u)$ has term of different sign...