I thought I'd be able to do this but evidently not.
Let $S_t=S_0e^{(r-\frac{\sigma^2}{2})t+\sigma W_t}$ for all $t$. $W_t$ is a standard brownian motion. We have the following function for payoff at time $T$,
$h:=\frac{S_T}{K_1}$ if $S_T \in [K_1,K_2]$ where $K_1<K_2$ are positive constants and $h=0$ otherwise.
Find $V_0=e^{-rT}\mathbb{E}[h]$ to conclude the pricing of the option is $V_0=\frac{S_0}{K_1}(N(d_1)-N(d_2))$ where $N(x)$ is the CDF of a standard normal random variable. $d_1=\frac{\text{ln} \frac{S_0}{K_1}+(r+\frac{\sigma^2}{2})T}{\sigma \sqrt{T}}$, $d_1=\frac{\text{ln} \frac{S_0}{K_2}+(r+\frac{\sigma^2}{2})T}{\sigma \sqrt{T}}$.
I thought,
$\mathbb{E}[h]=\mathbb{E}[S_T \mathbb{1}_{[K_1,K_2]}S_T]=\mathbb{P}[K_2>S_T>K_1]$. But I immediately get stuck here; How do I find such $\mathbb{P}$?
So I tried a more naive "integration" attempt by definition of a continuous variable expected value, but no this gets overly complicated it seems. I believe I need to set $S_T$ as my $x$ to integrate but then I have $W_t$ in my integrand and it just messes with me.
How do I go about this? Please explain as you go along so I can follow, thank you so much as usual for all the help
Update
I managed to find $\mathbb{E}[h]$ using $\mathbb{P}[a>x>b]=\mathbb{P}[x<a]-\mathbb{P}[x<b]$. However, I still don't see how $e^{-RT}\mathbb{E}[h]$ would lead me to the price. I don't get the $\frac{S_0}{K_1}$. Where does this come from?
The Black-Scholes model under the real-world measure is described by \begin{align}dS_t &= \mu S_t dt + \sigma S_t dW_t \\ dB_t &= rB_t dt\end{align}
The fundamental theorem of asset pricing says the market above is free of arbitrage if and only if for every choice of numeraire ("currency"), $N_t$, there is a probability measure equivalent to the real-world measure such that under this measure the relative price of any asset, $\frac{Y_t}{N_t}$, is a martingale. It is this principle (combined with completeness of course) that makes pricing of contingent claims possible.
The expression you have for $S_t$ is valid under the risk-neutral measure, i.e. when you take the bond as the numeraire ($N_t = B_t$). In that case, the asset prices follow the process \begin{align}dS_t &= \color{red}r S_t dt + \sigma S_t dW^{\color{red}M}_t \\ dB_t &= rB_t dt\end{align} You can check for yourself that $\frac{S_t}{B_t}$ is now a martingale.
But $B_t$ is not the only possible choice for a numeraire. You can also choose the stock price, $S_t$ as the numeraire. In that case you would require $\frac{S_t}{S_t}$ and $\frac{B_t}{S_t}$ to be martingales. The former is always a martingale regardless of the measure. For the latter to be a martingale it must hold that
\begin{align}dS_t &= \color{red}{(r+\sigma^2)} S_t dt + \sigma S_t dW^{\color{red}S}_t \\ dB_t &= rB_t dt\end{align}
You can then say $$S_T = S_0 e^{\left(r \color{red}+ \frac{1}{2}\sigma^2\right)T + \sigma W_T^S} $$
So now for the value of this contract you can write (due to the martingale property) $$\frac{C_0}{S_0} = E\left[\frac{C_T}{S_T}\right]$$
We have $$\frac{C_T}{S_T} = \frac{1}{S_T}\frac{S_T}{K_1}\mathbb{1}_{K_1 \leq S_T \leq K_2} = \frac{1}{K_1}\mathbb{1}_{K_1 \leq S_T \leq K_2}$$
Hence $$E\left[\frac{C_T}{S_T}\right] = \frac{1}{K_1}P\{K_1 \leq S_T \leq K_2\}$$
Then $$C_0 = \frac{S_0}{K_1}P\{K_1 \leq S_T \leq K_2\}$$
$$C_0 = \frac{S_0}{K_1}P\{\log{K_1} \leq \log{S_T} \leq \log{K_2}\}$$ $$C_0 = \frac{S_0}{K_1}P\{\log{K_1} \leq \log{S_0}+ \left(r + \frac{1}{2}\sigma^2\right)T + \sigma W_T^S \leq \log{K_2}\}$$
$$C_0 = \frac{S_0}{K_1}P\{\log{\frac{K_1}{S_0}}-\left(r + \frac{1}{2}\sigma^2\right)T \leq \sigma W_T^S \leq \log{\frac{K_2}{S_0}}-\left(r + \frac{1}{2}\sigma^2\right)T \}$$
$$C_0 = \frac{S_0}{K_1}P\{\frac{\log{\frac{K_1}{S_0}}-\left(r + \frac{1}{2}\sigma^2\right)T}{\sigma \sqrt{T}} \leq Z \leq \frac{\log{\frac{K_2}{S_0}}-\left(r + \frac{1}{2}\sigma^2\right)T}{\sigma \sqrt{T}} \}$$ In the last equality $Z$ denotes a standard normal random variable. If you then define $$d_i = \frac{\log{\frac{K_i}{S_0}}-\left(r + \frac{1}{2}\sigma^2\right)T}{\sigma \sqrt{T}}$$ for $i = 1,2$, you get the formula $$C_0 = \frac{S_0}{K_1} \left(\Phi(d_2) - \Phi(d_1) \right)$$ where $\Phi(x)$ is the cumulative distribution function of the standard normal random variable.