black scholes - 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?

Answer

Let $\tau = T-t$. Then \begin{align*} S_T = S_t e^{(\mu - \frac{1}{2}\sigma^2) \tau + \sigma \sqrt{\tau}\, Z}, \end{align*} where $Z$ is a standard normal random variable, independent of $\mathcal{F}_t$. Moreover, \begin{align*} E\left(S_T 1_{\{S_T >K\}}\mid \mathcal{F}_t \right) &= E\left(S_t e^{(\mu - \frac{1}{2}\sigma^2) \tau + \sigma \sqrt{\tau}\, Z}\, 1_{\left\{S_t e^{(\mu - \frac{1}{2}\sigma^2) \tau + \sigma \sqrt{\tau}\, Z} >K\right\}}\mid \mathcal{F}_t \right)\\ &=\int_{-\infty}^{\infty}S_t e^{(\mu - \frac{1}{2}\sigma^2) \tau + \sigma \sqrt{\tau}\, z}\, 1_{\left\{S_t e^{(\mu - \frac{1}{2}\sigma^2) \tau + \sigma \sqrt{\tau}\, z} >K\right\}} \frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}} dz\\ &=\int_{z^*}^{\infty}S_t e^{(\mu - \frac{1}{2}\sigma^2) \tau + \sigma \sqrt{\tau}\, z}\frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}} dz\\ &=\int_{z^*}^{\infty}\frac{1}{\sqrt{2\pi}} S_t e^{(\mu - \frac{1}{2}\sigma^2) \tau + \sigma \sqrt{\tau}\, z - \frac{z^2}{2}}dz\\ &=\int_{z^*}^{\infty}\frac{1}{\sqrt{2\pi}} S_t e^{(\mu - \frac{1}{2}\sigma^2) \tau -\frac{1}{2}(z-\sigma \sqrt{\tau})^2 +\frac{1}{2}\sigma^2 \tau }dz\\ &=S_t e^{u\tau} \int_{z^* - \sigma \sqrt{\tau}}^{\infty} \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx\\ &=S_t e^{u\tau}N(\sigma \sqrt{\tau}-z^*)\\ &=S_t e^{u\tau}N(d_1^*), \end{align*} where \begin{align*} d_1^* &= \sigma \sqrt{\tau}-z^*\\ &=\frac{\ln\frac{S_t}{K}+(\mu+\sigma^2/2)\tau}{\sigma\sqrt{\tau}}. \end{align*}