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As in this post, define the ff:

$$K_2(k)={\tfrac{\pi}{2}\,_2F_1\left(\tfrac12,\tfrac12,1,\,k^2\right)}$$

$$K_3(k)={\tfrac{\pi}{2}\,_2F_1\left(\tfrac13,\tfrac23,1,\,k^2\right)}$$

$$K_4(k)={\tfrac{\pi}{2}\,_2F_1\left(\tfrac14,\tfrac34,1,\,k^2\right)}$$

$$K_6(k)={\tfrac{\pi}{2}\,_2F_1\left(\tfrac16,\tfrac56,1,\,k^2\right)}$$

We find that,

$$\int_0^1 K_2(k)\, dk = {\tfrac{\pi}{2}\,_3F_2\left(\tfrac12,\tfrac12,\tfrac12;1,\tfrac32;1\right)}=2G$$

$$\int_0^1 K_3(k)\, dk = {\tfrac{\pi}{2}\,_3F_2\left(\tfrac12,\tfrac13,\tfrac23;1,\tfrac32;1\right)}=\tfrac{3\sqrt3}2\, \ln2$$

$$\int_0^1 K_4(k)\, dk = {\tfrac{\pi}{2}\,_3F_2\left(\tfrac12,\tfrac14,\tfrac34;1,\tfrac32;1\right)}=2\ln(1+\sqrt2)$$

$$\int_0^1 K_6(k)\, dk = {\tfrac{\pi}{2}\,_3F_2\left(\tfrac12,\tfrac16,\tfrac56;1,\tfrac32;1\right)}=\tfrac{3\sqrt3}4\, \ln(2+\sqrt{3})$$

where $G$ is Catalan's constant. However, this post gives a third power,

$$\int_0^1\big(K_2(k)\big)^3 dk=\frac35 \bigg(\tfrac{\pi}2\,_2F_1\big(\tfrac12,\tfrac12,1,\tfrac12\big)\bigg)^4=\frac35 \big(K(k_1)\big)^4 = \frac{3\,\Gamma (\frac14)^8}{1280\,\pi^2} \approx 7.0902$$

where $K(k_d)$ is an elliptic integral singular value while YuriyS in his comment below gives,

$$\int_0^1\big(K_3(k)\big)^3 dk \approx 6.53686311168760876289835638374 $$


Questions:

  1. What are the closed-forms of: $$\int_0^1\big(K_n(k)\big)^m dk=\,?$$ for power $m=2$ or $m=3$?
  2. Or at least their numerical evaluation up to 20 digits?

P.S. My old version of Mathematica can't evaluate it with sufficient precision, nor does WolframAlpha. (Enough digits may make it amenable to the Inverse Symbolic Calculator.)

Tito Piezas III
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    Please use `\left` and `\right` rather than `\bigg`, especially in titles. – Did Feb 08 '19 at 15:13
  • Could you clarify: does your 2nd question ask for numerical evaluation of the titular integral? Or the general integral from the 1st question? Above you have already provided the closed form for the titular integral, so I'm confused – Yuriy S May 22 '19 at 12:33
  • @YuriyS: The closed-form is ***not*** the titular integral $I_3$. Note the closed-form involves $\frac12,\tfrac12$, while the titular integral $I_3$ involves $\frac13,\frac23$. P.S. If you can provide a numerical evaluation of $I_3$ for 20 digits or more, that would be nice. – Tito Piezas III May 24 '19 at 08:37
  • Mathematica gives $I_3=6.53686311168760876289835638374$ with `WorkingPrecision->30` $$ $$ It has no problem evaluating the integral, only takes a few seconds – Yuriy S May 25 '19 at 12:06
  • @YuriyS: I revised the post to give a clearer version of the problem. – Tito Piezas III May 30 '19 at 03:40
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    A related case with a closed form in this question: https://math.stackexchange.com/q/3325678/269624 – Yuriy S Aug 17 '19 at 09:56
  • At one point, I convinced myself that https://dlmf.nist.gov/15.5 (at least a couple of them) worked for fractional derivatives. If so, then the order might be reduced To 1F1(a,b;k^2), which might be easier to evaluate (and invert). If you're still interested, I might take a stab at that. No warranties implied :) – rrogers Jan 17 '22 at 15:28

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