# Limit of integral times power function (Barrier call option calculation)

I'm studying Shreve's Stochastic Stochastic Calculus for Finance_II.
In chapter 7, there's equation for Up-and-Out Call option that deals with CDF of normal distribution. It goes like

$$v(t,x)=x\left[N(\delta_+\left(\tau,\frac{x}{K}\right)-\delta_+\left(\tau,\frac{x}{B}\right)\right]\\ -e^{-r\tau}K\left[N(\delta_-\left(\tau,\frac{x}{K}\right)-\delta_-\left(\tau,\frac{x}{B}\right)\right]\\ -B\left(\frac{x}{B}\right)^{-\frac{2r}{\sigma}}\left[N(\delta_+\left(\tau,\frac{B^2}{Kx}\right)-\delta_+\left(\tau,\frac{B}{x}\right)\right]\\ +e^{-r\tau}K\left[N(\delta_-\left(\tau,\frac{B^2}{Kx}\right)-\delta_-\left(\tau,\frac{B}{x}\right)\right], \\ \quad 0\leq t where

$$\tau = T-t, \quad \delta_{\pm}(\tau,s)=\frac{1}{\sigma\sqrt{\tau}}\left[\log{s}+\left(r\pm\frac{1}{2}\sigma^2\right)\right],\quad N(y)=\frac{1}{\sqrt{2\pi}}\int_{-y}^\infty e^{-\frac{z^2}{2}}dz.$$ and $$T, t, x, \sigma, r$$ are constants.

I want to know how the above equation changes when $$B\rightarrow \infty$$.

What I've tried

From my intuition, when $$B\rightarrow \infty$$, it should match with plain European call regardless of $$r$$ and $$\sigma$$. I know that plain European Call option pricing formula is $$c(t,x)=xN\left(\delta_+\left(\tau,\frac{x}{K}\right)\right)-Ke^{-r\tau}N\left(\delta_-\left(\tau,\frac{x}{K}\right)\right),\\ \quad 0\leq t 0.$$

and the first two terms in the first formula is the same with it, so I think the third and fourth terms sould go to 0 but, it is the multiple of $$\infty$$ and 0. So, I don't know how to handle it. What I've tried is the change the third term into $$-\frac{x^{\frac{-2r}{\sigma}}}{\sqrt{2\pi}}\cdot B^{\frac{2r}{\sigma^2}+1}\int_{-\frac{1}{\sigma\sqrt{\tau}}\left(\log\left(\frac{B^2}{Kx}\right)+\left(r+\frac{1}{2}\sigma^2\right)\tau\right)}^{-\frac{1}{\sigma\sqrt{\tau}}\left(\log\left(\frac{B}{x}\right)+\left(r+\frac{1}{2}\sigma^2\right)\tau\right)}e^{-\frac{z^2}{2}}dz.$$ But, from here, I can't get a clue how to go further.