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).
