An intriguing pattern in Ramanujan's theory of elliptic functions that stops?

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

1. 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.)
2. If so, is it connected to why there are only 4 signatures of alternative bases?
• $H(1)=\frac{\pi}2.$ Jan 29, 2019 at 19:58
• Fascinating! I wish I could help :) Jan 30, 2019 at 2:16
• The numerators of the $2G$ series is given here oeis.org/A038534 In Mathermatica given by $$\binom{2 n}{n}^2 2^{-2\, \text{DigitCount}[n,2,1]}$$ The DigitCount function giving the number of 1's in a base 2 number $n$. Wild! Jan 30, 2019 at 13:08
• Nice integral representations for alternative $K$s. +1 Jan 31, 2019 at 5:19
• @ParamanandSingh: I believe it's the general form. The signature $2$ numerator only looks different since $a-b = \frac12-\frac12 = 0$ and $\cos(0\,\alpha) = 1$. Jan 31, 2019 at 6:07

Some speculative clues have appeared after a bit of detective work...

First the integral we are interested in with the associated hypergeometric function and infinite series.

$$I_n=\int_0^1 K_n(k)\, dk = {\tfrac{\pi}{2}\,_3F_2\left(\tfrac12,\tfrac{1}{n},\tfrac{n-1}{n};1,\tfrac32;1\right)}= \frac{ \pi}{2}\times\sum _{k=0}^{\infty } \frac{\prod _{j=0}^{k-1} \left(j+\frac{1}{n}\right) \prod _{j=0}^{k-1} \left(j+\frac{n-1}{n}\right)}{(2 k+1) (k!)^2}$$

Simplifying the infinite series a little I found that $$I_n=\frac{ \pi}{2}\,\sum _{k=0}^{\infty } \frac{\prod _{j=1}^k \left(j^2-\frac{1}{n^2}\right)}{(k n+1)(2 k+1) (k!)^2 }$$

Now some interesting links appear to your integral if we study the much simpler sum

$$S_n=\sum _{k=0}^{\infty } \frac{(-1)^k}{(k n+1)( 2k+1)}$$

we find from Mathematica that $$S_2=G$$ $$S_3=\pi \left(\frac{1}{\sqrt{3}}-\frac{1}{2}\right)+\log (2)$$ $$S_4=\frac{1}{4} \pi \left(\sqrt{2}-1\right)+\frac{\log \left(\sqrt{2}+1\right)}{\sqrt{2}}$$ $$S_6=\frac{1}{8} \left(\pi +2 \sqrt{3} \log \left(\sqrt{3}+2\right)\right)$$ $$S_8=\frac{1}{12} \pi \left(\sqrt{2}+1\right)+\frac{\log (2)}{3}+\frac{\log \left(\sqrt{2}+1\right)}{3 \sqrt{2}}$$

These are all the shortest and simplest closed forms between $$n=2$$ and $$n=12$$.

For $$I_2$$, $$I_3$$, $$I_4$$ and $$I_6$$ that you found closed forms for, the respective sums have one term with the same principal constant and have a maximum of 3 terms. The next simplest sum I found is $$S_8$$ with four terms.

Have fun.

• Interesting. I tried using an integer relations program on the 6 vectors $I_8, \sqrt2,\ln2,\sqrt2\ln2,\ln(1+\sqrt2),\sqrt2\ln(1+\sqrt2)$ assuming $I_8$ is a rational sum involving those. Unfortunately, it couldn't find a simple relation. Jan 31, 2019 at 3:07
• By the way, I fixed a small typo in your simplified series for $I_n$. There should be a $\pi/2$. Jan 31, 2019 at 3:20
• Thanks for fixing that. Jan 31, 2019 at 8:40

One can use the decomposition of these hypergeometric functions on the Legendre polynomials (see for example here): $${}_2F_1\left(a,b;1;x\right)=\frac{\Gamma \left(2-a -b \right)}{\Gamma \left(a \right) \Gamma \left(b \right)}\sum_{p=0}^\infty (2 p+1)\frac{ \Gamma \left(a +p \right) \Gamma \left(b +p \right) } { \Gamma \left(2 -a +p\right) \Gamma \left(2-b +p \right)}P_p(2x-1)$$ to express \begin{align} \mathbf K_{1/a}\left(x \right)&=\frac\pi2{}_2F_1\left( a,1-a;1;x^2 \right)\\ &=\frac{\sin\pi a}2\sum_{p=0}^\infty \frac{(-1)^p(2p+1)}{(p+a)(p+1-a)}P_p(1-2x^2) \end{align} Then, as (G&R 7.225.3), \begin{align} \int_0^1P_p(1-2x^2)\,dx&=2^{-3/2}\int_{-1}^1\frac{P_p(z)\,dz}{\sqrt{1-z}}\\ &=\frac{1}{2p+1} \end{align} one obtains $$$$\int_0^1 \mathbf K_{1/a}\left(x \right)\,dx=\frac{\sin\pi a}2\sum_{p=0}^\infty \frac{(-1)^p}{(p+a)(p+1-a)}$$$$ For $$a=1/2$$, as expected, $$$$\int_0^1 \mathbf K_2\left(x \right)\,dx=2G$$$$ For $$a\ne1/2$$, $$$$\int_0^1 \mathbf K_{1/a}\left(x \right)\,dx=\frac{1}{2(2a-1)}\left[\pi+\sin(\pi a)\left( \psi\left( \frac{a}{2} \right)-\psi\left( \frac{a+1}{2} \right) \right)\right]$$$$ where $$\psi$$ is the digamma function.

Gauss's_digamma_theorem shows that digamma function has values in closed form for positive rational numbers less than $$1$$, in terms of Euler's constant and a finite number of elementary functions: $$$$\psi\left({\frac{r}{m}}\right)=-\gamma-\ln(2m)-{\frac{\pi}{2}}\cot\left({\frac{r\pi}{m}}\right)+2\sum_{n=1}^{\lfloor{\frac{m-1}{2}}\rfloor}\cos\left({\frac{2\pi n r}{m}}\right)\ln\sin\left({\frac{\pi n}{m}}\right)$$$$ Then, after some simplifications $$$$\int_0^1 \mathbf K_{n/p}\left(x \right)\,dx=\frac{2 n}{2 p -n}\sin \left(\frac{p \pi}{n}\right)\sum^{\lfloor \frac{n}{2}\rfloor}_{s=1}\cos \left(\frac{\pi p \left(2 s -1\right)}{n}\right) \ln \left(\sin \left(\frac{\pi \left(2 s -1\right)}{2 n}\right)\right)$$$$ The proposed integrals, which correspond to $$p=1$$, can thus be expressed in closed form.