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
$$\delta_G(x,\lambda)\approx 1+\lambda^2\frac{\sqrt{\pi}}{8x},\;\;x\gg 1,\;\;\lambda\ll 1.$$
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"/>

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The alternative quantity
$$\delta_A(x,\lambda)=\int_0^x \sqrt{1+f'(y)^2}\,dy-x$$
approaches a constant in the limit $x\rightarrow\infty$. For $\lambda\ll 1$ I find
$$\lim_{x\rightarrow\infty}\delta_A(x,\lambda)=\frac{1}{8}\lambda^2\sqrt{\pi}.$$
Here is a plot of this large-$x$ limit as a function of $\lambda$. The orange curve is the exact result, the blue curve the small-$\lambda$ approximation.

<IMG SRC="https://ilorentz.org/beenakker/MO/dissidence_2.png" WIDTH="400"/>