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

Question

(Update in last section.)

I. Define the integral,

$$K_n(k)=\int_0^{\pi/2}\frac{\cos\left((1-\frac{2}n)\arcsin(k\sin x)\right)}{\sqrt{1-k^2 \sin^2 x}}dx=\tfrac{\pi}{2}\,_2F_1\left(\tfrac1{n},\tfrac{n-1}{n},1,\,k^2\right)$$

hence,

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

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

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

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

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 following definite integrals,

$$I_2 = \int_0^1 K_2(k)\, dk = {\tfrac{\pi}{2}\,_3F_2\left(\tfrac12,\tfrac12,\tfrac12;1,\tfrac32;1\right)}=2G$$

$$I_3 =\int_0^1 K_3(k)\, dk = {\tfrac{\pi}{2}\,_3F_2\left(\tfrac13,\tfrac23, \tfrac12;1,\tfrac32;1\right)}=\tfrac{3\sqrt3}2\, \ln2$$

$$I_4 =\int_0^1 K_4(k)\, dk = {\tfrac{\pi}{2}\,_3F_2\left(\tfrac14,\tfrac34,\tfrac12;1,\tfrac32;1\right)}=2\ln(1+\sqrt2)$$

$$I_6 =\int_0^1 K_6(k)\, dk = {\tfrac{\pi}{2}\,_3F_2\left(\tfrac16,\tfrac56,\tfrac12;1,\tfrac32;1\right)}=\tfrac{3\sqrt3}4\, \ln(2+\sqrt{3})$$

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.)

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III. Question: Does the integral, or equivalently the generalized hypergeometric function, $$I_n=\tfrac{\pi}{2}\,_3F_2\left(\tfrac1n,\tfrac{n-1}n, \tfrac12; 1,\tfrac32;1\right)$$ 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.)

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$\color{blue}{\text{IV. Update (Oct 2024)}}$

I somehow missed that Paul Enta last Jan 2024 answered this question and showed $I_n$ has a beautiful and general closed-form for all integer $n>1$. Thus, we finally have,

\begin{align} I_5 &=\int_0^1 K_5(k)\, dk = \tfrac{\pi}{2}\,_3F_2\left(\tfrac15,\tfrac45,\tfrac12;1,\tfrac32;1\right)\\ &=-\frac{10}3\,\sin\frac{\pi}5\times\left(\cos\frac{\pi}5\, \ln\left(\sin\frac{\pi}{10}\right)+\cos\frac{3\pi}5\, \ln\left(\sin\frac{3\pi}{10}\right)\right)\ \end{align}

and,

\begin{align} I_8 &=\int_0^1 K_8(k)\, dk = \tfrac{\pi}{2}\,_3F_2\left(\tfrac18,\tfrac78,\tfrac12;1,\tfrac32;1\right)\\ &=-\frac{8}3\,\sin\frac{\pi}8\times\left(\cos\frac{\pi}8\, \ln\left(\tan\frac{\pi}{16}\right)+\cos\frac{3\pi}8\, \ln\left(\tan\frac{3\pi}{16}\right)\right)\ \end{align}

and so on for other $n$. Or using radicals,

\begin{align} I_5 &= \frac{\,5^{5/4}}6\sqrt{\frac1{\phi}}\, \ln\left(2\,\phi^\sqrt5\right)\\[6pt] I_8 &= \frac{\,2^{1/2}}3\sqrt{\frac1{r}}\, \ln\left( \left(\frac{2^{3/4}+\sqrt{r}}{2^{3/4}-\sqrt{r}}\right)^\sqrt{r} \left(\frac{2^{3/4}+\sqrt{1/r}}{2^{3/4}-\sqrt{1/r}}\right)^\sqrt{1/r} \right) \end{align}

with golden ratio $\phi=\frac{1+\sqrt5}2$ and silver ratio $r = 1+\sqrt2$.

P.S. The transcendental constant $\phi^\sqrt5$ also appears as a limiting ratio in this post. And I have a feeling the $\ln(z)$ of $I_8$ can be simplified.

Answer 1

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.

Answer 2

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 \begin{equation} \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)} \end{equation} For $a=1/2$, as expected, \begin{equation} \int_0^1 \mathbf K_2\left(x \right)\,dx=2G \end{equation} For $a\ne1/2$, \begin{equation} \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] \end{equation} 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: \begin{equation} \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) \end{equation} Then, after some simplifications \begin{equation} \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) \end{equation} The proposed integrals, which correspond to $p=1$, can thus be expressed in closed form.