 Solution

Apply part a with $\alpha =r$. For $\tau >0$, the transition density is the normal distribution with mean $m=S{e}^{r\tau }$ and positive variance $v={\sigma }^2(e^{2r\tau }-1)/(2r)$. Integrating the financial payoff against this probability density function givesThus $d_{\ast }=(x-S{e}^{r\tau })/\sqrt{v}$, as required. The variance is positive even if $r<0$, because numerator and denominator have the same sign. For $r=0$, replace $m$ by $S$ and $v$ by ${\sigma }^2\tau$; at expiry use the terminal value directly. The integral presumes an integrable financial payoff.