change of variable to the arclength for $y=\sec^2(x)$

Question

I have a curve defined by $y=\sec^2(x)$ for $-\frac{\pi}{2} < x < \frac{\pi}{2}$. I'm looking to do a change of variable from $x$ to $l$, where $l(x)$ is defined as the signed arclength of this curve starting at $x=0$, that is $l(0)=0$. Then we will have $l(-\frac{\pi}{2})=-\infty$ and $l(\frac{\pi}{2}) = +\infty$. I would like to find a closed form for $y$ as a function of $l$.

Here is what I have tried:

I only need to do it for the positive side, the other side follows by symmetry.

First invert the equation to find $$x(y) = \cos^{-1}\left(\frac{1}{\sqrt{y}}\right)$$ then calculate the arclength as a function of $y$ $$ l(y) = \int_1^y \sqrt{1 + \left(\left.\frac{dx}{dy}\right|_{y'}\right)^2}dy' = \int_1^y \sqrt{1+\frac{1}{4y'^2(y'-1)}}dy' $$ and if I could explicitly compute this integral, then I could invert it, but I can't.

So, the question is: (1) is there a closed form for this integral, or since I don't need the integral itself, I just need its inverse function, (2) is there another way to directly calculate $y(l)$?

Answer 1

Here is a series solution from:

On the generalization of the Lambert W function

$$x=iz:i(2\tanh(z)-z)=2i-\frac{4i}{e^{2z}+1}-iz=y$$

Rearrange:

$$\frac4{e^{2z}+1}=iy+2-z\implies e^{2z}=\frac{4}{iy+2-z}-1$$

Now use a change of variable:

$$e^{2z}=\frac{2-iy+z}{iy+2-z}\mathop\implies^{-2z=w}e^{-w}=\frac{2-iy-\frac w2}{iy+2+\frac w2}=\frac{-w-2iy+4}{w+2iy+4}$$

Now we solve:

$$-1=e^w\frac{w-(4-2iy)}{w-(-4-2iy)}$$

To get an inverse using the Generalized Laguerre function $\text L_n^a(x)$ and page $5$ of the source:

$$2\tan(x)-x=z\implies x=-2i-z+\frac i2\sum_{n=1}^\infty \sum_{k=1}^n\frac{\left(-e^{2iz-4}\right)^n}{n^2}\binom nk\frac{(-8n)^k}{(k-1)!}=-z-2i-4i\sum_{n=1}^\infty\frac{\text L_{n-1}^1(8n)\left(-e^{2iz-4}\right)^n}n$$

Which works when comparing both links’ results.

For fun, let’s use the Laguerre function integral representation solution which may not work in $|x|<\frac\pi2$

$$2\tan(x)-x=z\implies x=-z-2i+4i \sum_{n=1}^\infty\frac{\left.\frac{d}{dx}\right|_{x=8n} \frac1{2\pi}\int_0^{2\pi}e^{-e^{it}x}\left(e^{-it}+1\right)^ndt\left(-e^{2iz-4}\right)^n}n =-z-2i+\frac{2i}{\pi}\int_0^{2\pi}e^{it}\ln\left((e^{it}+1)e^{2iz-8e^{it}-it-4}+1\right)dt;\left|e^{it}+1\right|e^{-8\cos(t)-2\text{Im}(z)-4}<1$$

Shown here roughly. A closed form for the series and integral comes from the block quoted source, but these “generalized Lambert W” functions are not available in WolframAlpha, Maple, etc. unless you use the series and integral solutions in this answer.