Suppose I have the following equivalence for $x < y$: $$\int_{x+1}^{y+1} \exp\left(\frac{-u^2}{2}\right)du = \int_{-y+1}^{-x+1} \exp\left(\frac{-u^2}{2}\right)du$$
I know (from graphing these) that the only solution is $y = -x \quad (1)$, but I cannot rigorously prove this using the symmetry of the Gaussian integral. Why can we make the equality $(1)$? Thanks for any help you can provide!
By substituting $z = -u$ and using linearity of the integral we get:
$$\begin{align*} \int_{x+1}^{y+1} \exp\left(\frac{-u^2}{2}\right)du &= \int_{-y-1}^{-x-1} \exp\left(\frac{-u^2}{2}\right)du \\ &= \int_{-y-1}^{-y+1} \exp\left(\frac{-u^2}{2}\right)du &+ \int_{-y+1}^{-x+1} \exp\left(\frac{-u^2}{2}\right)du \\&&+ \int_{-x+1}^{-x-1} \exp\left(\frac{-u^2}{2}\right)du\end{align*}$$
By your equivalence it follows:
$$\int_{-x-1}^{-x+1} \exp\left(\frac{-u^2}{2}\right)du = \int_{-y-1}^{-y+1} \exp\left(\frac{-u^2}{2}\right)du$$
Now define a function $$F(x) = \int_{-x-1}^{-x+1} \exp\left(\frac{-u^2}{2}\right)du$$
and you will see that this function is increasing on $(-\infty,0]$ and decreasing on $[0,\infty)$, so the equation $$F(x) = F(y)$$ can have at most two solutions. One is $x=y$ the other one $y = -x$ and we are done.