**I.** Define the ff integrals,

$$K(k)=K_2(k)=\int_0^{\pi/2}\frac{1}{\sqrt{1-k^2 \sin^2 x}}dx=\large{\tfrac{\pi}{2}\,_2F_1\left(\tfrac12,\tfrac12,1,\,k^2\right)}$$

$$K_3(k)=\int_0^{\pi/2}\frac{\cos\left(\frac13\,\arcsin\big(k\sin x\big)\right)}{\sqrt{1-k^2 \sin^2 x}}dx=\large{\tfrac{\pi}{2}\,_2F_1\left(\tfrac13,\tfrac23,1,\,k^2\right)}$$

$$K_4(k)=\int_0^{\pi/2}\frac{\cos\left(\frac12\,\arcsin\big(k\sin x\big)\right)}{\sqrt{1-k^2 \sin^2 x}}dx=\large{\tfrac{\pi}{2}\,_2F_1\left(\tfrac14,\tfrac34,1,\,k^2\right)}$$

$$K_6(k)=\int_0^{\pi/2}\frac{\cos\left(\frac23\,\arcsin\big(k\sin x\big)\right)}{\sqrt{1-k^2 \sin^2 x}}dx=\large{\tfrac{\pi}{2}\,_2F_1\left(\tfrac16,\tfrac56,1,\,k^2\right)}$$

These are Ramanujan's theory of elliptic functions for alternative bases of signature $2,3,4,6$, respectively. There are only 4 signatures.

**II.** Then, using Wolfram, I observed the closed-forms of the ff definite integrals,

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

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

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

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

where $G$ is *Catalan's constant*. (Curiously, other than the first, Wolfram didn't recognize the closed-form of those hypergeometrics. I had to use the Inverse Symbolic Calculator.)

**III.** Questions

- Does the generalized hypergeometric function, $$H(n)=\,_3F_2\left(\tfrac12,\tfrac1n,\tfrac{n-1}{n};1,\tfrac32;1\right)$$ have a closed form only for $n=2,3,4,6$? (I tried $n=5,7,8$, etc, and it doesn't seem to have a "neat" form using elementary functions.)
- If so, is it connected to why there are only 4 signatures of alternative bases?