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This question is a follow-up to MSE2735980, about the volume and surface area of unit balls in $\mathbb{R}^n$ for non-Euclidean norms. Focusing on the $4$-norm in $\mathbb{R}^2$, the enclosed area for the unit ball is $$ A=4\int_{0}^{1}(1-x^4)^{1/4}\,dx = \int_{0}^{1}x^{-3/4}(1-x)^{1/4}\,dx = \frac{\Gamma\left(\frac{1}{4}\right)^2}{2\sqrt{\pi}} \tag{Area}$$ which is strictly related to the lemniscate constant. Due to the relations with the complete elliptic integral of the first kind and the AGM mean, $$ A = \frac{\pi}{\text{AGM}\left(1,\frac{1}{\sqrt{2}}\right)}\tag{FastArea} $$ hence $A$ can be efficiently evaluated through a numerical procedure with quadratic convergence, and the trivial $\pi<A<4$ can be immediately improved into the much sharper $2\pi(2-\sqrt{2})<A<\pi 2^{1/4}.$

What about the perimeter? It is given by $$L= 4\int_{0}^{1}\sqrt{1+\frac{x^6}{(1-x^4)^{3/2}}}=\int_{0}^{1}\sqrt{\frac{1}{x^{3/2}}+\frac{1}{(1-x)^{3/2}}}\,dx=\int_{0}^{+\infty}\frac{\sqrt{1+x^{3/2}}\,dx}{x^{3/4}(1+x)^{5/4}} $$ which is a highly non-elementary integral related to the Meijer G-function.

Q1) Is there an efficient numerical procedure for the evaluation of $L$?
Q2) Even if the outcome of Q1 is uncertain, can we prove sharp inequalities for $L$ (like $L>7$) in a simple way? The trivial inequality here is $2\pi<L<8$.

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  • $\begingroup$ what about $L< 4 \frac{\Gamma(\tfrac{5}{4})\Gamma(\tfrac{1}{4})}{\sqrt{\pi}} \approx 7.4163$ $\endgroup$ – tired Apr 14 '18 at 13:05
  • $\begingroup$ @tired: well, that's a start. You got it through the Cauchy-Schwarz inequality or through Jensen's inequality? $\endgroup$ – Jack D'Aurizio Apr 14 '18 at 13:19
  • $\begingroup$ i was thinking of CS.. $\endgroup$ – tired Apr 14 '18 at 13:21
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    $\begingroup$ @tired: I got a sharper upper bound by interpolation methods. They can be combined with CS and integration by parts, so we probably have a similarly sharp lower bound, too. $\endgroup$ – Jack D'Aurizio Apr 14 '18 at 13:44
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Here it is an approach for Q2, waiting for further improvements. Over the interval $(0,1)$ we have $$ \sqrt{\frac{1}{x^{3/2}}+\frac{1}{(1-x)^{3/2}}}\leq \frac{1-4(1-2^{-1/4})x(1-x)}{x^{3/4}(1-x)^{3/4}} \tag{1}$$ hence by Euler's Beta function $$ L \leq \frac{(4+2^{3/4})}{6\sqrt{\pi}}\,\Gamma\left(\tfrac{1}{4}\right)^2\leq \color{green}{7.0}23. \tag{2}$$ $(1)$ holds in the opposite direction if $4(1-2^{-1/4})$ is replaced by $\frac{3}{4}$, and that leads to $$ L \geq \frac{15}{16\sqrt{\pi}}\,\Gamma\left(\tfrac{1}{4}\right)^2\geq 6.952.\tag{3} $$

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