Show that $e^{-rt}\mathbb E[\Phi(S_T)]=S_0N(d_+)-Ke^{-rT}N_{d_-}$ where
$S_t=S_0e^{(r-\sigma ^2/2)t+\sigma W_t}$ for $t\in[0,T]$ , $W_t\sim \mathbb N(0,t)$ and $N$ is the cumulative density function of the standard normal distribution.
Also:
$$d_\pm=\frac {ln(S_0/K)(r \pm \sigma ^2/2)T}{\sigma\sqrt T}$$ and $\Phi(S_t)=(S_t-K)^+=max(S_t-K,0)$
What I tried: Let's do it for the 2nd part of the Integral - The K term
$e^{-rT}\mathbb E[\Phi(S_T)]=e^{-rT} \int_K ^\infty (S_T-K)\phi(z)dz+ e^{-rt} \int_{-\infty}^K 0$
So I am assuming I only need to find $$e^{-rT} \int_K ^\infty (S_T-K)\phi(z)dz$$ Using $S_t=S_0e^{(r-\sigma ^2/2)t+\sigma W_t}$ Since $W_t=\mu+\sqrt T Z$ where $Z\sim (0,1)$ I can rewrite this as a density function.
$\int_K ^\infty (S_0e^{(r-\sigma ^2/2 t + \sigma (\mu + \sqrt T Z))}-K) \times \phi(z) dz=$
But I'm having trouble from them on. I can't even equate the simple second term together. That is if I take $K$ out.
I should get $\int_K ^\infty -K \phi(z) dz=-KN_{d_-}$
What Am I doing wrong? $\int_K ^\infty -K \phi(z) dz=-K P(Z>K)=-KN$ where $N$ is the cumulative distribution. But why am I missing the $d_-$?