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I am trying to find the path length of a Gaussian $f(x)=e^{-x^2/a^2}$ from $x=0$ to some positive point $x_0$. I've tried this by integrating the differential length, $ds^2=dx^2+dy^2$, but getting stuck at the integral.

$$f'(x)\equiv\frac{dy}{dx}=\frac{-2x}{a^2}e^{-x^2/a^2}$$

\begin{align} L &= \int_0^{x_0}ds \\ &=\int_0^{x_0}\sqrt{dx^2+dy^2} \\ &=\int_0^{x_0}\sqrt{1+\left(f'\right)^2}dx\\ &= \int_0^{x_0}\sqrt{1+\frac{4}{a^4}x^2e^{-2x^2/a^2}}dx, \end{align}

which neither I nor Wolfram are able to do, so maybe I am barking up the wrong tree.

Using this method or any other, can anyone help me find the path length of a Gaussian?

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    $\begingroup$ Arc length integrals are rarely expressible in closed form in terms of elementary functions. Even the simple ellipse leads to elliptic integrals. $\endgroup$ Jul 22, 2016 at 18:39
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    $\begingroup$ The tree is correct, barking up bit tough. $\endgroup$
    – Narasimham
    Jul 22, 2016 at 19:46

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Elementary functions whose path-length integrals can be done in closed form are quite rare, and this is not one of them.

You might, however, write the square root as a series in powers of $(4 x^2/a^4) e^{-2x^2/a^2}$. This leads to

$$L = x_0 + \sum_{k=1}^\infty \dfrac{(-1)^{k+1} a^{-2k} (2k)!}{(2k-1)(k!)^2} \int_0^{x_0/a} t^{2k} e^{-2k t^2}\; dt $$

where each of these integrals is of the form

$$\eqalign{&(\text{odd polynomial in $x_0/a$ of degree $2k-1$}) e^{-2k (x_0/a)^2}\cr &+ {\frac {{2}^{-3\,k-1} \left( 2\,k-1 \right) !}{ k ^{k +1/2} (k-1)!}} \sqrt{2\pi} \;\text{erf}\left(\sqrt{k x_0/a}\right)}$$

The series should converge for all real $x_0$ as long as $a > \sqrt{2/e}$ (which ensures that $(4 x^2/a^4) e^{-2x^2/a^2}$ stays in the region where the series for the square root converges).

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  • $\begingroup$ As Robert Israel and Mark Fischler said, very very few curves, even among the most prominent ones have an EXACT "analytical" expression for their lengths. A question and an an opinion. Question : what need do you have for the length of an arc of a gaussian, especialy an "exact" value of it ? Opinion : it is often, on SE, that people ask for an "exact/analytical" expression of such and such... ; I think that in a vast majority of cases it is a deadend, it is just a way to loose hours without profit ; it's better to concentrate on richer parts of mathematics and they are so numerous... $\endgroup$
    – Jean Marie
    Jul 22, 2016 at 22:09

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