As in this post, define the ff:





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,

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


  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.)

  • $\begingroup$ Please use \left and \right rather than \bigg, especially in titles. $\endgroup$ – Did Feb 8 at 15:13
  • $\begingroup$ 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 $\endgroup$ – Yuriy S May 22 at 12:33
  • $\begingroup$ @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. $\endgroup$ – Tito Piezas III May 24 at 8:37
  • $\begingroup$ Mathematica gives $I_3=6.53686311168760876289835638374$ with WorkingPrecision->30 $$ $$ It has no problem evaluating the integral, only takes a few seconds $\endgroup$ – Yuriy S May 25 at 12:06
  • 1
    $\begingroup$ A related case with a closed form in this question: math.stackexchange.com/q/3325678/269624 $\endgroup$ – Yuriy S Aug 17 at 9:56

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