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).

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).

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.