Let $X\sim N(-\frac{1}{2}\delta^2.\delta^2)$ What is $P(X\geq 0)$.
$$P(X\geq 0)=\frac{1}{\sqrt{2\delta^2\pi}}\int^\infty_0\exp\{\frac{-(X+\frac{1}{2}\delta^2)^2}{2\delta^2}\}dx$$
I use the substitution $Y=(X+1/2\delta^2)/\delta\Rightarrow dx=dy$
We have:
$$P(X\geq 0)={\sqrt{2\delta^2\pi}}\int^\infty_{1/2\delta}\exp\{-Y^2/2\}dy$$
$$=\frac{1}{\delta}{\frac{1}{\sqrt{2\pi}}}\int^\infty_{1/2\delta}\exp\{-Y^2/2\}dy$$
$$=\frac{1}{\delta}\Phi(-1/2\delta)$$
But my notes say it should be $\Phi(-1/2\delta)$
Notice that $$y=\frac{x+\frac{1}{2}\delta^2}{\delta}\implies \mathrm d x=\delta \cdot \mathrm d y$$