I want to verify the following equation:
$$E[(xe^{aY-\frac{1}{2}a^2}-b)^+]=x\Phi(l_1)-b\Phi(l_2)$$
where $Y\sim \mathcal{N}(0,1)$, $\Phi$ the distribution function of a standard normal distribution, constants $a,b,x$ and with $l_1=\frac{\log{\frac{x}{b}}+\frac{1}{2}a^2}{a}$, $l_2=\frac{\log{\frac{x}{b}}-\frac{1}{2}a^2}{a}$
This is what I've done: Let $A:=xe^{aY-\frac{1}{2}a^2}> b$, then
$$E[(xe^{aY-\frac{1}{2}a^2}-b)^+]=E[xe^{aY-\frac{1}{2}a^2}\mathbf1_A]-bP[A]$$
For the second term I got:
$$bP[A]=bP\left[Y>\frac{\log{\frac{b}{x}}+\frac{1}{2}a^2}{a}\right]=b\Phi\left(-\frac{\log{\frac{b}{x}}+\frac{1}{2}a^2}{a}\right)=b\Phi(l_2)$$
For the first term:
$$xE[e^{aY-\frac{1}{2}a^2}\mathbf1_A]=x\frac{1}{\sqrt{2\pi}a}\int_{\frac{b}{x}}^\infty e^{-\frac{(\log{y}+\frac{1}{2}a^2)^2}{2a^2}}\frac{1}{y}dy$$
I put $u:=\frac{\log{y}+\frac{1}{2}a^2}{a}$, hence $\frac{du}{dy}=\frac{1}{ay}$, or $dy=ay du$, therefore
$$x\frac{1}{\sqrt{2\pi}a}\int_{p}^\infty e^{-\frac{(\log{y}+\frac{1}{2}a^2)^2}{2a^2}}\frac{1}{y}dy=x\frac{1}{\sqrt{2\pi}a}\int_{p}^\infty e^{-\frac{u^2}{2}}\frac{1}{y}aydu=x\frac{1}{\sqrt{2\pi}}\int_{p}^\infty e^{-\frac{u^2}{2}}du$$
where $p=\frac{\log{\frac{b}{x}}+\frac{1}{2}a^2}{a}$, but then
$$x\frac{1}{\sqrt{2\pi}}\int_{p}^\infty e^{-\frac{u^2}{2}}du=x\Phi(-p),$$ which is not equal to $x\Phi(l_1)$. So where is my mistake?