Integral involving an error function 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.
 A: 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.
A: Let $u=\text{erf}^{-1}(x)$ so that $x=\text{erf}\left(u\right)$, and $dx=\frac{2}{\sqrt{\pi}}e^{-u^{2}}du.$ Then $$I(\sigma)=\frac{1}{2}\int_{-1}^{1}\text{erf}\left(\frac{\sigma}{\sqrt{2}}+\text{erf}^{-1}(x)\right)dx=\frac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-u^{2}}\text{erf}\left(\frac{\sigma}{\sqrt{2}}+u\right)du.$$ Differentiating with respect to $\sigma$ we find that $$\frac{\partial I(\sigma)}{\partial\sigma}=\frac{\sqrt{2}}{\pi}\int_{-\infty}^{\infty}e^{-u^{2}}e^{-(u+\sigma/\sqrt{2})^{2}}du,$$ and so substituting $u=y/\sqrt{2}$ we have $$\frac{\partial I(\sigma)}{\partial\sigma}=\frac{1}{\pi}\int_{-\infty}^{\infty}e^{-y^{2}+y\sigma-\sigma^{2}/2}dy=\frac{1}{\pi}\int_{-\infty}^{\infty}e^{-(y+\sigma/2)^{2}-\sigma^{2}/4}dy=\frac{1}{\sqrt{\pi}}e^{-\sigma^{2}/4}.$$ Hence $$I(\sigma)=\frac{1}{\sqrt{\pi}}\int_{0}^{\sigma}e^{-t^{2}/4}dt=\frac{2}{\sqrt{\pi}}\int_0^{\sigma/2}e^{-t^2}dt=\text{erf}\left(\sigma/2\right),$$ as desired. 
