Understanding the solution of this integral The following integral represents an expected value of a geometric brownian motion for $S_T>K$ (i.e. part of the Black-Scholes call option price):
$$\int_{z^*} (S_te^{\mu\tau-\frac{1}{2}\sigma^2\tau+\sigma\sqrt{\tau}z})\frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}}dz
=e^{\mu\tau}S_tN(d_1^*)$$
with $z^*=\frac{\ln\frac{K}{S_t}-(\mu-\sigma^2/2)\tau}{\sigma\sqrt{\tau}}$,
$d_1^*=\frac{\ln\frac{S_t}{K}+(\mu+\sigma^2/2)\tau}{\sigma\sqrt{\tau}}$, and $N(\cdot)$ cumulative Standardnormal distribution.
Can you please explain how this equality is obtained?
 A: Integrals are linear:
$$\int_{z^*}^\infty  \left(S_t\, e^{\mu\tau-\sigma^2\tau/2+\sigma\sqrt{\tau}z}\right)\frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}}dz
=\color{blue}{\left( S_t\, e^{\mu\tau-\sigma^2\tau/2}\right)}\frac{1}{\sqrt{2\pi}}\int_{z^*}^\infty e^{\sigma\sqrt{\tau}z-\frac{z^2}{2}}dz.$$
The exponent in the integrand looks very much like a Normal density, but shifted.  Let's force it to look like that:
$$\sigma\sqrt{\tau}z - z^2/2 = -\frac{1}{2}\left(z - \sigma\sqrt{\tau}\right)^2 + \sigma^2\tau/2.$$
Let us therefore change to the variable $y = z - \sigma\sqrt{\tau}$.  It presents no difficulty because $dy = dz$.  Once again, linearity allows us to factor out the part that does not depend on $z$:
$$\frac{1}{\sqrt{2\pi}}\int_{z^*}^\infty e^{\sigma\sqrt{\tau}z-\frac{z^2}{2}}dz
=\color{red}{e^{\sigma^2\tau/2}}\frac{1}{\sqrt{2\pi}}\int_{z^*-\sigma\sqrt{\tau}}^\infty e^{-\frac{y^2}{2}}dy.$$
The right hand side obviously is a multiple of the right tail of a Standard Normal distribution.  By virtue of its symmetry, this integral gives the same value as integrating from $-\infty$ to $-\left(z^{*}-\sigma\sqrt{\tau}\right)$, which is (by definition) given by the CDF $\Phi$:
$$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{-z^*+\sigma\sqrt{\tau}} e^{-\frac{y^2}{2}}dy = \Phi\left(-z^*+\sigma\sqrt{\tau}\right) = \Phi(d_1^*).$$
(The algebraic equivalence of $-z^*+\sigma\sqrt{\tau}$ and $d_1^*$ assumes $\sigma \gt 0$.)
Plugging everything back where it belongs shows us the original integral equals
$$\color{blue}{\left(S_t\, e^{\mu\tau-\sigma^2\tau/2}\right)} \color{red}{e^{\sigma^2\tau/2}} \Phi(d_1^*) = S_t\, e^{\mu\tau}\, \Phi(d_1^*).$$
