Superelliptic Area Of $x^5+y^5=r^5$

Question

$${\LARGE\int}_0^\tfrac\pi2\frac{dx}{\bigg(\sqrt[{\Large 5}]{\cos^5x+10\cos^3x\sin^2x+5\cos x\sin^4x}\bigg)^{\large 2}}~=~?$$

Its numerical value is about $1.40171345128228$. Maple, Mathematica, and the Inverse Symbolic Calculator are not able to return any closed form.

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Motivation:

The above integral is, up to a certain scaling factor, nothing else than the area between the graphic of $x^n+y^n=r^n$, and the second bisector, for $n=5$. For even values of the exponent, we have the area of $x^{2k}+y^{2k}=r^{2k}$ being equal to $A_{2k}=\displaystyle(2r)^2\cdot{2a\choose a}^{-1}$, where $a=\dfrac1{2k}$. For $n\!=\!3$ we have $A_3=\dfrac{r^2}{\sqrt[3]4}\cdot B\bigg(\dfrac12,\dfrac13\bigg)$. But how to compute its value for odd exponents greater than $3$ is beyond

me. In this case, $A_5=r^2\cdot\sqrt[5]8\cdot I$

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$\qquad\qquad$

$$x^5+y^5=1$$

Answer 1

Method due to Dirichlet, for your $r=1$ the area in the first quadrant is $$ \frac{\Gamma \left( \frac{6}{5} \right)^2}{\Gamma \left( \frac{7}{5} \right) } \approx 0.95015. $$

Since your number is larger than $1,$ i guess you are getting area for the whole thing, along the line $x+y = 0$ included. I suggest trying ratios of Gamma functions, see if you can get your $1.4017$

Answer 2

The general formula for odd $n=2k+1$ is $~A_{n}=\displaystyle{2a\choose a}^{-1}+{-a\choose a}^{-1}$, where $a=\dfrac1n$ , and

the first term represents the area inside the first sector or quadrant. Also, the integral could

simply have been written as $\displaystyle\int_{-\infty}^\infty\Big(x+\sqrt[\large^n]{1-x^n}\Big)dx$, but both Maple and Mathematica face

significant trouble when evaluating it on $(1,\infty)$, since they both identify the n-th root of a

number as the $($complex$)$ quantity with the smallest argument, i.e., whose angle formed with

the positive semi-axis is least, even when a real negative solution exists, in which case the

angle is $180^\circ>\dfrac{360^\circ}n$ for odd $n\ge3$.