The exercise consists in pricing in the Black & Scholes model a contract whose payoff at the maturity is given by: \begin{equation} F(n,S_T)=\textbf{1}_{n<S_T<n+1}=\textbf{1}_{S_T>n}-\textbf{1}_{S_T>n+1} \end{equation} Observing that the payoff at the maturity corresponds to the one of a long position in a digital option with strike $n$ and a short position on a digital option with strike $n+1$ applying the B&S formula I end up with the following: \begin{equation} price_t(F(n,S_T))=e^{-R(T-t)}\Phi(d_2^n)-e^{-R(T-t)}\Phi(d_2^{n+1}) \end{equation} where: \begin{equation} d_2^n=\dfrac{\ln(\frac{S_t}{n})+(r-\frac{\sigma^2}{2})(T-t)}{\sigma\sqrt{T-t}}\\ d_2^{n+1}=\dfrac{\ln(\frac{S_t}{n+1})+(r-\frac{\sigma^2}{2})(T-t)}{\sigma\sqrt{T-t}} \end{equation} and $\Phi(x)=P(X\leq x)$ with $X\sim N(0,1)$. Now the exercise asks me to compute the limit as $n\to\infty$ of the price. My idea is to say that since both $d_2^{n}$ and $d_2^{n+1}$ goes to $-\infty$ as $n\to\infty$ but I don't know how to rigorously compute: \begin{equation} \lim_{n\to\infty}e^{-R(T-t)}\int_{-\infty}^{d_2^n}\dfrac{1}{2\pi}e^{-z^2/2}dz \end{equation} Is it just sufficient to bring the limit to the upper extremal of integration or have I to remove the $n$ terms in the upper extremal and bring them inside the integral with a change of variable?
EDIT
consider the following change of variable: $y=z+\frac{\ln(n)}{\sigma\sqrt{T-t}}$. thus by the very definition of $d_2^n$ we have that the new upper extremal of integration becomes: \begin{equation} \hat{y}=\dfrac{\ln(\frac{S_t}{n})+(r-\frac{\sigma^2}{2})(T-t)}{\sigma\sqrt{T-t}}+\frac{\ln(n)}{\sigma\sqrt{T-t}}=\dfrac{\ln(S_t)+(r-\frac{\sigma^2}{2})(T-t)}{\sigma\sqrt{T-t}} \end{equation} then the global integral becomes: \begin{equation} \int_{-\infty}^{\hat{y}}\dfrac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}(y-\frac{\ln(n)}{\sigma\sqrt{T-t}})^2}dy \end{equation} But now I'm not able to find an integrable function which is larger in absolute value in order to apply dominated convergence theorem