Getting a finite value for curve Dissidence

Question

If I've got $f: x \to f(x)$, one may define the Arithmetic Dissidence $\delta_A[f(x)]$ as the real value of the difference between the length following the curve of $f$ and the length of the $x$ axis (we could define it on intervals or on the whole axis). You may also want to define the Geometric Dissidence $\delta_G[f(x)]$ as the ratio of these lengths.

For example, if can be easily shown that for a function $f: x \to |x|$, $\delta_G[f(x)]=\sqrt2$.
For a distribution $\mathcal{T}(1-|x|)_{[-1, 1]}$, $\delta_G[\mathcal{T}]$ can't be different from $1$ but $\delta_A[\mathcal{T}]=2(\sqrt2-1)$.

I need to find the value for $\delta_A[f(x)]$ if it's finite for $f=\lambda e^{\frac{-x^2}{2\sigma^2}}$. Otherwise, I'd need the value of $\delta_G[f(x)]$. I've still found no way of finding these values by integrating $d\ell=\sqrt{dx^2+df^2}$ but maybe you know one way.

Thanks for your help,

Answer 1

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).

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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.