Jerack curve - $\int_{-1}^{1}\frac{\sqrt{4w^{4}-20w^{3}+24w^{2}-20w+13}}{(w-2)^{2}\sqrt{1-w^{2}}}dw$

Question

I wanted to calculate the length of the Jerabek curve which has equation:

$$(x^2+y^2)(x-1)^2=4(x^2+y^2-x)^2$$

My work

1. I used the coordinates to transform it into parametric form and arrived at: $$\left(\frac{2\cos(t)-1}{2-\cos(t)}\cos(t),\frac{2\cos(t)-1}{2-\cos(t)}\sin(t)\right)$$
2. I used the formula to calculate the length of parametric curves and arrived at this integral: $$\int_{0}^{2\pi}\frac{\sqrt{4\cos(t)^4-20\cos(t)^3+24\cos(t)^2-20\cos(t)+13}}{\left(\cos(t)-2\right)^{2}}\mathrm{d}t$$ $$2\int_{-1}^{1}\frac{\sqrt{4w^{4}-20w^{3}+24w^{2}-20w+13}}{(w-2)^{2}\sqrt{1-w^{2}}}\mathrm{d}w$$

But I have no idea how to continue, can anyone help me?

Answer 1

Personally I don't think this integral has a closed form in terms of known functions or constants (it's hard to work with that fourth degree polynomial)

Allow me to suggest a numerical solution, you can use Chebyshev–Gauss quadrature since you have an integral between $-1$ and $1$ with $\sqrt{1-x^2}$ in the denominator (which of all is the one with the easiest weight since it doesn't depend on $j$):

Let: $$f(x):=\frac{\sqrt{4x^{4}-20x^{3}+24x^{2}-20x+13}}{(x-2)^{2}}\quad \text{and}\quad x_j=\cos\left(\frac{2j-1}{n}\pi\right)$$

Note that $f(x)$ is the same integrand as the problem but without $\sqrt{1-x^2}$ to the denominator

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Your integral can be calculated through this series:

$$2\int_{-1}^{1}\frac{\sqrt{4w^{4}-20w^{3}+24w^{2}-20w+13}}{(w-2)^{2}\sqrt{1-w^{2}}}\mathrm{d}w\approx\frac{2\pi}{n}\sum_{j=1}^{n}\frac{\sqrt{4x_{j}^{4}-20x_{j}^{3}+24x_{j}^{2}-20x_{j}+13}}{\left(x_{j}-2\right)^{2}}$$

Where $n$ it is a number that you can choose as large as you want (the higher it is, the more precise the integral will be). To have equality you must have:

$$\begin{matrix}\displaystyle2\int_{-1}^{1}\frac{\sqrt{4w^{4}-20w^{3}+24w^{2}-20w+13}}{(w-2)^{2}\sqrt{1-w^{2}}}\mathrm{d}w=\lim_{n\to\infty}\frac{2\pi}{n}\sum_{j=1}^{n}\frac{\sqrt{4x_{j}^{4}-20x_{j}^{3}+24x_{j}^{2}-20x_{j}+13}}{\left(x_{j}-2\right)^{2}}\end{matrix}$$ To avoid having to calculate the cosine often, you can also write it by putting everything under the root and using partial fractions

Let $y_j:=2-\cos\left(\frac{2j-1}{2n}\pi\right)$: $$\color{blue}{2\int_{-1}^{1}\frac{\sqrt{4w^{4}-20w^{3}+24w^{2}-20w+13}}{(w-2)^{2}\sqrt{1-w^{2}}}\mathrm{d}w\approx\frac{2\pi}{n}\sum_{j=1}^{n}\sqrt{4-\frac{12}{y_{j}}+\frac{36}{y_j^3}-\frac{27}{y_j^4}}}$$

Even a term like this is simplified for you, you have one less

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The integral is approssimativament equal to $6.562068241855779$ (calculated with Wolfram)

• For $n=5$ you have: $\color{green}{6.5}\color{red}{763761665}$
• For $n=10$ you have: $\color{green}{6.562}\color{red}{19948899}$
• For $n=15$ you have: $\color{green}{6.5620}\color{red}{7041605}$
• For $n=20$ you have: $\color{green}{6.5620682}\color{red}{8605}$
• For $n=25$ you have: $\color{green}{6.56206824}\color{red}{286}$
• For $n=30$ you have: $\color{green}{6.5620682418}\color{red}{8}$