I'm staring at $\int_k^\infty e^x g(x) dx$, where $g(x)$ is the density of a normal distribution.
Now, I know that for log normal variables, the partial expectation of $\int_k^\infty x f(x) dx$, with $f(x)$ the density of a log normal, this partial expectation actually has an explicit solution.
My hunch is that I should be able to use the relationship between the normal and the log-normal distribution to find this explicit solution also for my initial expression, but I'm unable to... is there a straight-forward way to show this or am I barking up the wrong tree?
If you substitute $u=e^x$ in the integral you get $$\int_k^\infty e^x g(x) \: dx = \int_{e^k}^\infty ug(\ln(u))\frac{1}{u}\:du = \int_{e^k}^\infty uf(u) \: du,$$ which gives the desired relation between the two partial expectations.