Proof of the Black - Scholes pricing formula for European Call Option I want to prove the following
The price of a European call option with strike price $K$ and time of maturity $T$ is given by the formula $\Pi(t) = F(t,S(t))$, where 
$$F(t,s) = sN[d_1(t,s)]-e^{-r(T-t)}KN[d_2(t,s)]$$
$$d_1(t,s) = \frac{1}{\sigma\sqrt{T-t}}\left[\ln \frac{s}{K}+\left(r-\frac{\sigma^2}{2}\right)\left(T-t\right)\right]$$
$$d_2(t,s) = d_1(t,s) - \sigma\sqrt{T-t} $$
and $$N(x) = \int^{\infty}_{-\infty} e^{-\frac{x^2}{2}}\,dx,\,\,X\sim N(0,1)$$
I've gotten this far...
$$S(T) = s\exp\left[\left(r-\frac{\sigma^2}{2}\right)\left(T-t\right) + \sigma\left(W(T)-W(t)\right)\right]$$
and we define 
$$Z = \left(r-\frac{\sigma^2}{2}\right)\left(T-t\right) + \sigma\left(W(T)-W(t)\right) $$
with 
\begin{align*}
\mathbb{E}\left(Z\right) & = \left(r-\frac{\sigma^2}{2}\right)\left(T-t\right) \\
\mbox{Var}\left(Z\right) & = \sigma^2\left(T-t\right)
\end{align*}
and so 
\begin{align*}
Z\sim N\left(\left(r-\frac{\sigma^2}{2}\right)\left(T-t\right),\sigma^2\left(T-t\right)\right)
\end{align*}
then 
\begin{align*}
F(t,s) & = e^{-r(T-t)}\mathbb{E}^Q\left(\Phi\left(S(T)\right)\right) \\
\,\, & = e^{-r(T-t)} \int^\infty_{-\infty} \Phi\left(S(T)\right)f(z)\,dz \\
\,\; & = e^{-r(T-t)} \int^\infty_{-\infty} \Phi\left(se^z\right)f(z)\,dz 
\end{align*}
with the case of an European option we put $\Phi(x) = \max\left[x-K,0\right]$ and 
then 
\begin{align*}
F(t,s) & = e^{-r(T-t)}\int^\infty_{-\infty} \max\left[se^z-K,0\right]f(z)\,dz \\
\,\, & = e^{-r(T-t)}\left(\int^{\ln \frac{K}{s}}_{-\infty} 0\cdot f(z)\,dz + \int^{\infty}_{\ln\frac{K}{s}} \left(se^z-K\,f(z)\right)\,dz\right) \\
\,\, & = e^{-r(T-t)}\int^{\infty}_{\ln\frac{K}{s}}\left(se^z-K\right)\,f(z)\,dz \\
\,\, & = e^{-r(T-t)} \left(s\int^{\infty}_{\ln\frac{K}{s}} e^zf(z)\,dz -K\int^{\infty}_{\ln\frac{K}{s}}f(z)\,dz \right) \\
\,\, & = e^{-r(T-t)} \left(s\int^{\infty}_{\ln\frac{K}{s}} e^ze^{-\frac{z^2}{2}}\,dz -K\int^{\infty}_{\ln\frac{K}{s}}e^{-\frac{z^2}{2}}\,dz \right)  \\
\,\, & = e^{-r(T-t)} \left(s\int^{\infty}_{\ln\frac{K}{s}} e^ze^{-\frac{z^2}{2}}\,dz -KN\left(-\frac{\ln\frac{K}{s}- \left(r-\frac{\sigma^2}{2}\right)\left(T-t\right)}{\sigma\sqrt{T-t}}\right)\right) \\
\,\, & = e^{-r(T-t)} \left(s\int^{\infty}_{\ln\frac{K}{s}} e^{z-\frac{z^2}{2}}\,dz -KN\left(-\frac{\ln\frac{K}{s}- \left(r-\frac{\sigma^2}{2}\right)\left(T-t\right)}{\sigma\sqrt{T-t}}\right)\right) \\
\,\, & = e^{-r(T-t)}  \left(s\int^{\infty}_{\ln\frac{K}{s}} e^{-\frac{1}{2}\left(z-1\right)^2+\frac{1}{2}}\,dz-KN\left(\frac{\ln\frac{s}{K}+\left(r-\frac{\sigma^2}{2}\right)\left(T-t\right)}{\sigma\sqrt{T-t}}\right)\right) \\
\,\, & = e^{-r(T-t)}  \left(se^{\frac{1}{2}}\int^{\infty}_{\ln\frac{K}{s}} e^{-\frac{1}{2}\left(z-1\right)^2}\,dz-KN\left(\frac{\ln\frac{s}{K}+\left(r-\frac{\sigma^2}{2}\right)\left(T-t\right)}{\sigma\sqrt{T-t}}\right)\right) 
\end{align*}
Were do I go from here ?? I can't solve the remaining integral !!
 A: Now lets solve the above $SDE$. This is just a GBM with solution 
$$S(T) = s\exp\left[\left(r-\frac{\sigma^2}{2}\right)\left(T-t\right) + \sigma\left(W(T)-W(t)\right)\right]$$
and we define 
$$Z = \left(r-\frac{\sigma^2}{2}\right)\left(T-t\right) + \sigma\left(W(T)-W(t)\right) $$
$$Z = \left(r-\frac{\sigma^2}{2}\right)\left(T-t\right) + \sigma\sqrt{T-t}Y,\,\,Y\sim N(0,1)$$
with 
\begin{align*}
\mathbb{E}\left(Z\right) & = \left(r-\frac{\sigma^2}{2}\right)\left(T-t\right) \\
\mbox{Var}\left(Z\right) & = \sigma^2\left(T-t\right)
\end{align*}
and so 
\begin{align*}
Z\sim N\left(\left(r-\frac{\sigma^2}{2}\right)\left(T-t\right),\sigma^2\left(T-t\right)\right)
\end{align*}
then 
\begin{align*}
F(t,s) & = e^{-r(T-t)}\mathbb{E}^Q\left(\Phi\left(S(T)\right)\right) \\
\,\, & = e^{-r(T-t)} \int^\infty_{-\infty} \Phi\left(S(T)\right)f(z)\,dz \\
\,\; & = e^{-r(T-t)} \int^\infty_{-\infty} \Phi\left(se^z\right)f(z)\,dz 
\end{align*}
with the case of an European option we get have $\Phi(x) = \max\left[x-K,0\right]$
hen 
\begin{align*}
F(t,s) & = e^{-r(T-t)}\int^\infty_{-\infty} \max\left[se^z-K,0\right]f(z)\,dz \\
\,\, & = e^{-r(T-t)}\left(\int^{\ln \frac{K}{s}}_{-\infty} 0\cdot f(z)\,dz + \int^{\infty}_{\ln\frac{K}{s}} \left(se^z-K\right)\,f(z)\,dz\right) \\
\,\, & = e^{-r(T-t)}\int^{\infty}_{\ln\frac{K}{s}}\left(se^z-K\right)\,f(z)\,dz \\
\,\, & = e^{-r(T-t)} \left(s\int^{\infty}_{\ln\frac{K}{s}} e^zf(z)\,dz -K\int^{\infty}_{\ln\frac{K}{s}}f(z)\,dz \right) \\
\,\, & = e^{-r(T-t)} \left(s\int^{\infty}_{\ln\frac{K}{s}} e^zf(z)\,dz -K\int^{\infty}_{\ln\frac{K}{s}}f(z)\,dz \right) \\
\,\, & = \frac{e^{-r(T-t)}}{\sqrt{2\pi}} \left(s\int^{\infty}_{\ln\frac{K}{s}}  e^{\left(r-\frac{\sigma^2}{2}\right)\left(T-t\right) + \sigma\sqrt{T-t}y}e^{-\frac{y^2}{2}}\,dy -K\int^{\infty}_{\ln\frac{K}{s}}e^{-\frac{z^2}{2}}\,dz \right)  \\
\,\, & = \frac{e^{-r(T-t)}}{\sqrt{2\pi}} \left(s\int^{\infty}_{\ln\frac{K}{s}} e^{\left(r-\frac{\sigma^2}{2}\right)\left(T-t\right) + \sigma\sqrt{T-t}y-\frac{y^2}{2}}\,dy\right) -Ke^{-r(T-t)}\Phi\left(-\frac{\ln\frac{K}{s}- \left(r-\frac{\sigma^2}{2}\right)\left(T-t\right)}{\sigma\sqrt{T-t}}\right) \\
\,\, & = \frac{e^{-r(T-t)}}{\sqrt{2\pi}}  e^{\left(r-\frac{\sigma^2}{2}\right)\left(T-t\right)}\left(s\int^{\infty}_{\ln\frac{K}{s}} e^{\sigma\sqrt{T-t}y-\frac{y^2}{2}}\,dy\right) -Ke^{-r(T-t)}\Phi\left(-\frac{\ln\frac{K}{s}- \left(r-\frac{\sigma^2}{2}\right)\left(T-t\right)}{\sigma\sqrt{T-t}}\right) \\
\,\, & =    \frac{e^{-\frac{\sigma^2}{2}\left(T-t\right)}}{\sqrt{2\pi}}  \left(s\int^{\infty}_{\ln\frac{K}{s}} e^{-\frac{1}{2}\left(y^2-2\sigma\sqrt{T-t} y+\sigma^2\left(T-t\right)\right)}e^{\frac{1}{2}\sigma^2\left(T-t\right)}\,dy\right)-Ke^{-r(T-t)}\Phi\left(\frac{\ln\frac{s}{K}+\left(r-\frac{\sigma^2}{2}\right)\left(T-t\right)}{\sigma\sqrt{T-t}}\right) \\
\,\, & = \frac{e^{-\frac{\sigma^2}{2}\left(T-t\right)}}{\sqrt{2\pi}}   \left(s\int^{\infty}_{\ln\frac{K}{s}} e^{-\frac{1}{2}\left(y-\sigma\sqrt{T-t}\right)^2+\frac{1}{2}\sigma^2\left(T-t\right)}\,dz\right)-Ke^{-r(T-t)}\Phi\left(\frac{\ln\frac{s}{K}+\left(r-\frac{\sigma^2}{2}\right)\left(T-t\right)}{\sigma\sqrt{T-t}}\right) \\
\,\, & = \frac{e^{-\frac{\sigma^2}{2}\left(T-t\right)}e^{\frac{\sigma^2}{2}\left(T-t\right)}}{\sqrt{2\pi}}   \left(s\int^{\infty}_{\ln\frac{K}{s}} e^{-\frac{1}{2}\left(y-\sigma\sqrt{T-t}\right)^2}\,dz\right)-Ke^{-r(T-t)}\Phi\left(\frac{\ln\frac{s}{K}+\left(r-\frac{\sigma^2}{2}\right)\left(T-t\right)}{\sigma\sqrt{T-t}}\right) \\
\,\ & = \frac{1}{\sqrt{2\pi}}   \left(s\int^{\infty}_{\ln\frac{K}{s}} e^{-\frac{1}{2}\left(y-\sigma\sqrt{T-t}\right)^2}\,dz\right)-Ke^{-r(T-t)}\Phi\left(\frac{\ln\frac{s}{K}+\left(r-\frac{\sigma^2}{2}\right)\left(T-t\right)}{\sigma\sqrt{T-t}}\right) \\
\,\, & = s\Phi\left(-\frac{\ln\frac{K}{s}-\left(r-\frac{\sigma^2}{2}\right)\left(T-t\right)}{\sigma\sqrt{T-t}}+\sigma\sqrt{T-t}\right)-Ke^{-r(T-t)}\Phi\left(\frac{\ln\frac{s}{K}+\left(r-\frac{\sigma^2}{2}\right)\left(T-t\right)}{\sigma\sqrt{T-t}}\right) \\
\,\, & = s\Phi\left(\frac{\ln\frac{s}{K}+\left(r+\frac{\sigma^2}{2}\right)\left(T-t\right)}{\sigma\sqrt{T-t}}\right)-Ke^{-r(T-t)}\Phi\left(\frac{\ln\frac{s}{K}+\left(r-\frac{\sigma^2}{2}\right)\left(T-t\right)}{\sigma\sqrt{T-t}}\right)
\end{align*}
and thats it !
A: I think the American case is out of scope for the book I'm using. One has to know how to solve optimization problems to derive the pricing formula for American options, This i will be able to do after i've read a course on in next semester. :)
