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Using the $L^p$ norm for $p>1$, the unit circle is the first quadrant is the curve $y=\sqrt[p]{1-x^p}$. The length of a curve is found using $\int_{a}^{b} \sqrt[p]{\left|{\partial x}\right|^p+\left|{\partial y}\right|^p}$ Let $\tau_p$ be the circumference of an $L^p$ unit circle. Then: $$\tau_p= 4\int_{0}^{1} \sqrt[p]{1+{\left({1-x^p}\right)^{1-p}x^{p^2-p}}} \,dx$$ $\tau_1=8,\tau_2=2\pi$ are trivial. Substitute $x^p=z$ $$\tau_p= \frac{4}{p}\int_{0}^{1} \frac{\sqrt[p]{z\left(1-z\right)\left({\left(1-z\right)^{p-1}+z^{p-1}}\right)}}{\left(1-z\right)z} \,dz$$ Substitute $\frac{1}{2}+x=z$, use symmetry $$\tau_p= \frac{8}{p}\int_{0}^{\frac{1}{2}} \frac{\sqrt[p]{\left(\frac{1}{2}+x\right)\left(\frac{1}{2}-x\right)\left({\left(\frac{1}{2}-x\right)^{p-1}+\left(\frac{1}{2}+x\right)^{p-1}}\right)}}{\left(\frac{1}{2}-x\right)\left(\frac{1}{2}+x\right)} \,dx$$ Somehow Wolfram Mathematica can evaluate this last form for $p=3$. $$\tau_3=\frac{2}{3} \Gamma \left(\frac{1}{3}\right) \left(\frac{\Gamma \left(\frac{1}{4}\right)}{\Gamma \left(\frac{7}{12}\right)}+\frac{\Gamma \left(\frac{3}{4}\right)}{\Gamma \left(\frac{13}{12}\right)}\right) =\frac{2}{3} \left(B\left(\frac{1}{3},\frac{1}{4}\right)+B\left(\frac{1}{3},\frac{3}{4}\right)\right)$$ where $B$ is the Beta function.

I am interested in the evaluation of the integral. Preferable for reals, then rationals, then integers. Failing that, the chosen answer will be given for new specific exact values (e.g. $\tau_4$). Failing that, a proof of the exact value $\tau_3$ will be the chosen answer.

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  • $\begingroup$ Note: math.stackexchange.com/questions/2044223/… is a related question, yet does not give non-integral exact values. $\endgroup$
    – Nazgand
    May 11 '18 at 17:59
  • $\begingroup$ Mathematica shows $\tau_4=$ -((4 Sqrt[[Pi]] Gamma[1/4] Hypergeometric2F1[-(1/4), 1/2, 3/4, -3])/ Gamma[-(1/4)]) using t3[p_]:=8/p*Integrate[((1/2+x)^(1-p)(1/2-x)^(1-p)((1/2-x)^(p-1)+(1/2+x)^(p-1)))^(1/p),{x,0,1/2}] and t3[4]. $\endgroup$
    – Nazgand
    Sep 12 '20 at 16:49

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