TITLE: Integral involving an error function
QUESTION [7 upvotes]: For $\sigma>0$, how can we prove that
$$\frac{1}{2}\int_{-1}^1  \text{erf}\left(\frac{\sigma}{\sqrt{2}}+\text{erf}^{-1}(x)\right) \, \mathrm{d}x= \text{erf}\left(\frac{\sigma}{2}\right)$$
where erf is the error function, $$\text{erf}(x)=\frac{2}{\sqrt{\pi}}\int_{0}^x e^{-t^2}dt.$$
The result was obtained by tinkering, and I was wondering if there a concise derivation.

REPLY [4 votes]: I did it fast without checking it too much. Might give inspiration.$$\int erf( \frac{\sigma}{\sqrt{2}} + erf^{-1}(x)) dx$$ To solve, we first do the variable exchange $u = \frac{\sigma}{\sqrt{2}} - erf^{-1}(x)$;
By the inverse function derivative we find that $du = \frac{\sqrt\pi}{2}e^{erf^{-1}(x)} dx$. Solving for x from above equation and plugging it in yields us with $du =\frac{\sqrt{\pi}}{2}e^{u-\frac{\sigma}{2}}$ Therefore the integral is transformed to:
$$\int erf(u)*e^{-{u-\frac{\sigma}{\sqrt{2}}}} du$$
Using partial differentiation with $ s = erf (u)$ and $ dt = e^{-({u-\frac{\sigma}{2}})}$ should solve quite easily.