Derive partial expectation of a lognormal variable I have been trying to figure out how to derive the partial mean/expectation of a lognormally distributed variable. I have seen many pages providing the solution but I have not been able to find a proper derivation.
Hence, what I would like to show is that (from wikipedia):
$$
g(k)=\int_k^{\infty} x f_X(x \mid X>k) d x=e^{\mu+\frac{1}{2} \sigma^2} \Phi\left(\frac{\mu+\sigma^2-\ln k}{\sigma}\right)
$$
I am asking this because in the end, I would like to find an expression for:
$$
g(k)=\int_k^{\infty} x^\phi f_X(x \mid X>k) d x
$$
where $\phi$ is some constant parameter.
Thanks a lot in advance.
 A: The task is much easier than you think and also easier than what I see in that Wikipedia page:
The random variable $X$ is lognormal which means it is $X=e^{\mu+\sigma x}$ for some $x\sim N(0,1)\,.$ We know that the PDF of $x$ is
$$
p(x)=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}\,.
$$
Obviously, $X>k$ if and only if $x>\frac{\ln k-\mu}{\sigma}\,.$ Then your task is to calculate
$$
\mathbb E[X;X>k]=\int_{\frac{\ln k-\mu}{\sigma}}^\infty e^{\mu+\sigma x}\,p(x)\,dx\,.
$$
Hint: Write out $p(x)$ and merge it with $e^{\mu+\sigma x}\,.$ This will lead to a shifted normal distribution and a factor in front of the integral.
Spoiler I:

 \begin{align}&\int_{\frac{\log k-\mu}{\sigma}}^\infty e^{\mu+\sigma }\,\frac{1}{\sqrt{2\pi}}e^{-x^2/2}\,dx= \frac{1}{\sqrt{2\pi}}e^{\mu+\sigma^2/2}\int_{\frac{\ln >!k-\mu}{\sigma}}^\infty e^{-(x-\sigma)^2/2}\,dx \end{align}

Spoiler II:

 The RHS of the last equation is essentially the CDF of the normal  distribution evaluated at $$-\frac{\ln k-\mu}{\sigma}+\sigma=-\frac{\ln k-\mu-\sigma^2}{\sigma}=\frac{\mu+\sigma^2-\ln k}{\sigma}\,.$$

