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Let $n>0$. How does one integrate $$\int _0^1\int _0^{\left(1-x^n\right)^{1/n}}\left(-x^n-y^n+1\right)^{1/n}dydx$$ ? This integral represents the volume enclosed by $$x>0,y>0,z>0,x^n+y^n+z^n<1$$. By substitution of $y=t\left(1-x^n\right)^{1/n}$ we get $$\int _0^1\left(1-x^n\right)^{1/n}\int _0^1\left(-x^n-\left(t\left(1-x^n\right)^{1/n}\right)^n+1\right)^{1/n}dtdx$$ $$=\int _0^1\left(1-x^n\right)^{1/n}\int _0^1\left((1-x^n)(1-t^n)\right)^{1/n}dtdx$$ $$=\int _0^1\left(1-x^n\right)^{2/n}dx\int _0^1\left(1-t^n\right)^{1/n}dt$$ Some help from Mathematica $$=\frac{\Gamma \left(1+\frac{1}{n}\right)^2}{\Gamma \left(\frac{n+2}{n}\right)}\int _0^1\left(1-x^n\right)^{2/n}dx$$ I am not sure how to do the rest. Judging by the pattern, it looks like the final answer is $$\frac{\Gamma \left(1+\frac{1}{n}\right)^2}{\Gamma \left(\frac{n+2}{n}\right)} \frac{2 \Gamma \left(1+\frac{1}{n}\right) \Gamma \left(\frac{2}{n}\right)}{3 \Gamma \left(\frac{3}{n}\right)}$$ but I don't see how to actually prove it.

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    $\begingroup$ For even natural $n$ this is a complex number. $\endgroup$ – user64494 Oct 13 '14 at 6:18
  • $\begingroup$ I asked the wrong question because I made a mistake. What I really wanted to know was not $$\int _0^1\int _0^1\left(-x^n-y^n+1\right)^{1/n}dydx$$. $\endgroup$ – Nazgand Oct 13 '14 at 6:45
  • $\begingroup$ Sorry; I did not notice the change and I was asking something which became irrelevant. $\endgroup$ – Claude Leibovici Oct 13 '14 at 6:47
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    $\begingroup$ This is a superellipsoid, as well as a superquadric. The integral in your last expression is the beta function. Just let $t=x^n$. $\endgroup$ – Lucian Oct 13 '14 at 7:37
  • $\begingroup$ Is there a reason why you are avoiding "=" signs? $\endgroup$ – andre Oct 13 '14 at 9:23
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Maple calclulates it for concrete rational values of the parameter $n$. For example,

VectorCalculus:-int((-x^3-y^3+1)^(1/3), [x, y] = Region(0 .. 1, 0 .. (-x^3+1)^(1/3)))

$$ {\frac {4\,\sqrt [3]{2}{\pi }^{5/2}}{81\,\Gamma \left( 2/3 \right) \Gamma \left( 5/6 \right) }},$$

VectorCalculus:-int((-x^4-y^4+1)^(1/4), [x, y] = Region(0 .. 1, 0 .. (-x^4+1)^(1/4)))

$$1/6\,{\frac {{\pi }^{3/2}{\it EllipticK} \left( i \right) }{ \left( \Gamma \left( 3/4 \right) \right) ^{2}}},$$ and

VectorCalculus:-int((-x^(3/2)-y^(3/2)+1)^(2/3), [x, y] = Region(0 .. 1, 0 .. (1-x^(3/2))^(2/3)))

$${\frac {4\,\sqrt {\pi }\sqrt {3}{2}^{2/3}\Gamma \left( 2/3 \right) \Gamma \left( 5/6 \right) }{81}}.$$ I don't see any general formula for the integral under consideration.

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  • $\begingroup$ I answered the question after the first its edit, but before the second one. $\endgroup$ – user64494 Oct 13 '14 at 7:35
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Final steps thanks to Lucian starting from $$\frac{\Gamma \left(1+\frac{1}{n}\right)^2}{\Gamma \left(\frac{n+2}{n}\right)}\int _0^1\left(1-x^n\right)^{2/n}dx$$ Let $x=s^{1/n}$ and we get $$\frac{\Gamma \left(1+\frac{1}{n}\right)^2}{n\Gamma \left(\frac{n+2}{n}\right)}\int _0^1\left(1-s\right)^{2/n}s^{1/n-1}ds$$ $$=\frac{\Gamma \left(1+\frac{1}{n}\right)^2}{n\Gamma \left(\frac{n+2}{n}\right)}\frac{\Gamma \left(\frac{1}{n}\right) \Gamma \left(\frac{n+2}{n}\right)}{\Gamma \left(\frac{n+3}{n}\right)}$$ $$=\frac{\Gamma \left(1+\frac{1}{n}\right)^3}{\Gamma \left(\frac{n+3}{n}\right)}$$

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