I hope my question is well posed in the math section. I am pondering over the following question: Find $$ M=\mathrm{E}\left(\max{\left(e^X-e^Y,0\right)}\right)$$ where $X$ and $Y$ are bivariate normally distributed random numbers. The textbook states (without proof) that the solution computes to $$M=\mathrm{E}\left(e^X\right)\Phi(d)-\mathrm{E}\left(e^Y\right)\Phi(d-s),$$ where $s^2 = Var(X-Y)$ and $$d=\frac{1}{s}\ln{\left(\frac{\mathrm{E}\left(e^X\right)}{\mathrm{E}\left(e^Y\right)}+\frac{1}{2}s^2\right)}$$ and where $\Phi(z)$ denotes the cdf of the standard normal distribution. I wanted to come up with a proof, but all my approaches failed so far. I stuck with the following ansatz: $$\int\limits_{x=-\infty\;}^{\infty}\int\limits_{y=-\infty}^{x}\left(e^x-e^y\right)f(x,y)dydx=\int\limits_{x=-\infty}^{\infty}\left(e^x\int\limits_{y=-\infty}^{x}f(y\mid x)dy-\int\limits_{y=-\infty}^{x}e^yf(y\mid x)dy\right)f(x)dx$$ Now, let $\mu_c=\mathrm{E}(Y\mid X=x)$ and $\sigma_c^2=Var(Y\mid X=x)$, then the integrals can be stated as $$M=\int\limits_{x=-\infty}^{\infty}\left[e^x\Phi\left(\frac{x-\mu_c}{\sigma_c}\right)-e^{\mu_c+\frac{1}{2}\sigma_c^2}\Phi\left(\frac{x-\mu_c+\frac{1}{2}\sigma_c^2}{\sigma_c}\right)\right]f(x)dx$$
From this point onwards, I find no way of showing that the integrals would solve to the stated solution.
Any hint / help is highly appreciated. Br