I am reading '3D imaging and mechanical modeling of helical buckling in Medicago truncatla plant roots' by Silverberg et al.
At a certain point there is a 3D curve (that is supposed to be like an 'irregular helix') parametrized by $$ (x(z),y(z),z) $$ and the authors claim that the average length, i.e. the average vertical extent of the helix can be calculated using $$ \frac{\left( \int r(z)^2 dz\right)^2}{\int r(z)^4 dz } $$ where $$ r(z)^2= x(z)^2+y(z)^2$$ is measured from the central axis of the helix, which is oriented to coincide with the z axis.
What's the mathematical derivation of this formula?
The expression they provide is just a useful measure, but it is not exact.
Let me first recast the problem in a simpler form, maybe easier to handle. The helical geometry is not essential, as the only relevant value is the radius $r$. So one can translate everything to the plane.
Given a continuous function $f(x)$ such that it is different from zero only on a certain interval, without loss of generality $[-\frac{L}{2};\frac{L}{2}]$, we can ask the question of looking for a quantity that could give us an approximation of the said interval length.
This is analogous to what they do in the paper you cite: they deal with “localised” helices, such that $r$ is different from $0$ only over an interval (this explains their comment on why the expression they mention is rather insensitive to integration limits).
So, looking for such quantity, first of all we would like it to be dimensionally right: and the expression you quote has the dimension of length (the denominator has dimension $L^6$ and the denominator $L^5$), so we indeed get a length out of it, dimensionally speaking.
Further, we can check how the provided expression works for “box” type functions, zero outside an interval (of length, say, $L$) and equal to a constant, which we call $a$, inside it. Then, the expression you quote simplifies to $$ \frac{(L a^2)^2}{L a^4} = \frac{L^2 a^4}{L a^4} = L $$ as desired.
For a general function $f$, one can use the mean value theorem, and say $$ \frac{(\int_{-\infty}^{\infty} f^2(x) \mathrm{d}x)^2} {\int_{-\infty}^{\infty} f^4(x)\mathrm{d}x} = \frac{(\int_{-L/2}^{L/2} f^2(x) \mathrm{d}x)^2} {\int_{-L/2}^{L/2} f^4(x)\mathrm{d}x} = \frac{(L f^2(\xi_1))^2}{L f^4(\xi_2)} = L \frac{f^4(\xi_1)}{f^4(\xi_2)} $$ And it all boils down to the fact $\xi_1$ and $\xi_2$ are “close”.
We can further check that the answer is not exact for less simple functions: one can try functions such as $ f(x) =\beta \sqrt{cos(x)}$ defined on $[-\pi/2;\pi/2]$ and zero everywhere else, or $ f(x) = \beta \sqrt{1-x^2}$ defined on $[-1;1]$ and zero everywhere else , and verify that the answer is only an approximation, albeit very useful.