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I have to find the area enclosed by $x^{2n}+y^{2n}=1, n \in \mathbb{Z}$ in terms of $F(\alpha)=\int_0^{\pi}\frac{dt}{(\sin{t})^{\alpha}}$

I tried the substitution $x=r\cos^{1/n}{\theta}, y=r\sin^{1/n}{\theta}$, but the Jacobian determinant is really horribly ugly, and I don't think this works. I ended up with $\frac{1}{2^{1/n}n} \int_0^{2\pi}\sin^{1/n}({2\theta})(\tan{\theta}+\cot{\theta})d\theta$. I don't see any other reasonable substitutions.

vukov
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2 Answers2

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The area $A$ of the supercircle is

$$A=4\int_{0}^1 (1-x^{2n})^{1/2n}\,dx \tag 1$$

Enforcing the substitution $x^{2n}\to x$ in $(1)$ reveals

$$\begin{align} A&=\frac2n \int_0^1 (1-x)^{1/2n} x^{1/2n - 1}\,dx\\\\ &=\frac2n B\left(\frac{1}{2n},1+\frac{1}{2n}\right) \tag 2\\\\ &=\frac2n \frac{\Gamma\left(1+\frac{1}{2n}\right)\Gamma\left(\frac{1}{2n}\right)}{\Gamma\left(1+\frac{1}{n}\right)} \tag 3\\\\ &=\frac2n \frac{1}{2n}\frac{\Gamma^2\left(\frac{1}{2n}\right)}{\frac1n \Gamma\left(\frac1n\right)} \tag 4\\\\ &= \frac{\Gamma^2\left(\frac{1}{2n}\right)}{n \Gamma\left(\frac1n\right)} \tag 5\\\\ &=\frac{2^{1-1/n}\sqrt{\pi}\,\Gamma\left(\frac{1}{2n}\right)}{n\Gamma\left(\frac12+\frac{1}{2n}\right)}\tag 6 \end{align}$$

In arriving at $(1)$, we used the standard integral representation of the Beta function, $B(x,y) = \int_0^1 t^{x-1}(1-t)^{y-1} \,dt$.

In going from $(2)$ to $(3)$ we used the relationship between the Beta and Gamma functions, $B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$.

In going from $(3)$ to $(4)$, we used the functional equation for the Gamma function, $\Gamma(1+x)=x\Gamma(x)$.

In going from $(4)$ to $(5)$, we simplified the terms in $(4)$.

In going from $(5)$ to $(6)$, we used the duplication formula, $\Gamma(z)\Gamma(z+1/2)=2^{1-2z}\sqrt{\pi}\,\Gamma(2z)$

Mark Viola
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  • I am hoping for a solution without special functions. – vukov Apr 20 '16 at 16:59
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    I don't believe that this can be reduced to a form that uses elementary functions only. In the OP, the request was to have a form in terms of an integral involving only powers of the sine function. We can cast it in terms of an integral involving powers of sine and cosine by letting $x=\sin^2(\theta)$ or $x=\cos^2(\theta)$ in the equation preceding $(2)$. This is one of the forms of the Beta function, given in $(2)$. Does that help? – Mark Viola Apr 20 '16 at 17:02
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Here is another idea.

You can parametrize the curve $x^{2n}+y^{2n}=1$ as follows (if $n\neq 0$): $$ \begin{cases} x(t)=(\cos t)^\frac{1}{n}\\ y(t)=(\sin t)^\frac{1}{n} \end{cases} \quad t\in [0,2\pi] $$ And then use Green's theorem to compute the area inside the curve: $$ A=\oint_Cx\,dy = \int_0^{2\pi}x(t)y'(t)\,dt=\frac{1}{n}\int_0^{2\pi} \frac{(\cos(t))^{1+\frac{1}{n}}} {(\sin(t))^{1-\frac{1}{n}}} \,dt $$

This being said, maybe there is a way to make $F(\alpha)$ appear in this expression?

Kuifje
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