Expectation of exponential function

Question

Let $X \sim \mathcal N(\mu, \sigma^2)$ which denotes some random costs. By making the investment $k > 0$ one can change costs to $Y(k) \sim \mathcal N(e^{-k}\mu,e^{-2k}\sigma^2)$ such that costs changes from $x$ to $y(k) + k$. I have from a paper the following relationship, but don't know how to verify it. Can someone hint me to what happens in between? \begin{align} \mathbb E\left[e^{Y(k) + k}\right] = \exp\left(e^{-k}\mu + \frac{1}{2}e^{-2k}\sigma^2 + k\right) \end{align}

Answer 1

\begin{align} & \operatorname{E} (e^{Y(k) +k }) = (2\pi )^{-\frac{1}{2}}e^k \sigma^{-1}\int_{\mathbb{R}}e^{x+k}e^{\frac{(x-e^{-k}\mu)^2}{2e^{-2k}\sigma^2}} \, dx \\[10pt] = {} & (2\pi )^{-\frac{1}{2}}e^{2k} \sigma^{-1}\int_{\mathbb{R}}e^{-e^k \mu +\frac{1}{2}e^{-2k}\sigma^2}e^{\frac{(x-(e^{-k}\mu -e^{-2k}\sigma^2))^2 }{2e^{-2k}\sigma^2}} \, dx \\[10pt] = {} & e^k \cdot e^{-e^k \mu +\frac{1}{2}e^{-2k}\sigma^2}=e^{-e^k \mu +\frac{1}{2} e^{-2k} \sigma^2 +k} \end{align}