You seek the quantity
$$\delta_G(x,\lambda)=\frac{1}{x}\int_0^x \sqrt{1+f'(y)^2}\,dy,\;\;\text{with}\;\;f(x)=\lambda e^{-x^2/2},$$
where I have rescaled $x\mapsto x/\sigma$ to remove the parameter $\sigma$. There is no closed form answer for this integral, but for large $x$ it decays as
$$\delta_G(x,\lambda)\approx 1+\lambda^2\frac{\sqrt{\pi}}{8x},\;\;x\gg 1.$$
For small $x$ it grows as 
$$\delta_G(x,\lambda)\approx 1+\lambda^2\frac{x^2}{6},\;\;x\ll 1.$$
Here is a plot of $\delta_G(x,\lambda)$ for $\lambda=1$ (green) and the small-$x$ and large-$x$ asymptotes (blue and orange).

<IMG SRC="https://ilorentz.org/beenakker/MO/dissidence.png" WIDTH="400"/>