Moments of Alternate Elliptic Integrals

Question

Let $K_s$ denote Ramanujan's Alternate Elliptic Integrals as follows:

$$K_s:=K_s(k)=\frac{\cos(\pi s)}{2}\int_0^1\frac{t^{s-1/2}}{(1-t)^{1/2+s}(1-k^2t)^{1/2-s}}dt$$

Where in Hypergeometric Form it is: $$K_s=\frac{\pi}{2}{}_2F_1\left[\frac{1}{2}+s,\frac{1}{2}-s,1,k^2\right]$$

The following signatures are important in this theory: $$s=\left\{0,\frac{1}{3},\frac{1}{4},\frac{1}{6}\right\}$$

I have found a General Theorem to calculate the following Integral:

$$\int_0^1kK_uK_vdk$$

As examples I will present the corollaries that follow from the Theorem while considering the combination of $4$ signatures.

This gives us the following 10 Integrals:

Corollaries: $$\int_0^1kK^2dk=\frac{7\zeta(3)}{4}$$

$$\int_0^1kK^2_{1/6}dk=\frac{9\sqrt{3}}{8}\operatorname{Cl_2}(\pi/3)$$

$$\int_0^1kK^2_{1/3}dk=\frac{15\sqrt{3}}{16}\operatorname{Cl_2}(\pi/3)$$

$$\int_0^1kK^2_{1/4}dk=2\beta(2)$$

$$\int_0^1kKK_{1/6}dk=\frac{27\sqrt{3}}{4}\ln 3-9\sqrt{3}\ln 2$$

$$\int_0^1kKK_{1/3}dk=\frac{27}{16}\ln 3$$

$$\int_0^1kKK_{1/4}dk=2\sqrt{2}\ln 2$$

$$\int_0^1kK_{1/3}K_{1/6}dk=\frac{3\sqrt{3}}{2}\ln 2$$

$$\int_0^1kK_{1/4}K_{1/6}dk=\frac{27\sqrt{6}}{5}\ln 2-\frac{27\sqrt{6}}{10}\ln3$$

$$\int_0^1kK_{1/4}K_{1/3}dk=\frac{27\sqrt{2}}{14}\ln 3-\frac{9\sqrt{2}}{7}\ln 2$$

The very first Integral is known in this list as $K_0$ is just the Complete Elliptic Integral of the First Kind $K$.

I have not seen the General Theorem and Other Integrals as well anywhere in Literature.

So I was wondering whether these results are new or already known?

This is only a Small List, one may produce as many Integrals as they want from the General Theorem.

Like for example after Differentiating:

$$\int_0^1kK\left[\frac{d}{dv}K_v\right]_{v=1/4}dk=2\sqrt{2}\left(8\beta(2)-\pi \ln2-8\ln 2\right)$$

This opens up new possibilities too.

The following is for example already known:

$$\int_0^1k^{2n+1}K^2dk=a_n\zeta(3)+b_n$$

where $a_n,b_n\in\mathbb{Q}$.

This is found using "FindIntegerNullVector":

Conjecture:

$$\int_0^1k^{2n+1}K^2_{1/6}dk=a_n\sqrt{3}\operatorname{Cl_2}(\pi/3)+b_n$$

$$\int_0^1k^{2n+1}K^2_{1/3}dk=a_n\sqrt{3}\operatorname{Cl_2}(\pi/3)+b_n$$

$$\int_0^1k^{2n+1}K^2_{1/4}dk=a_n\beta(2)+b_n$$

$$\int_0^1k^{2n+1}KK_{1/3}dk=a_n\ln 3+b_n$$

where $a_n,b_n\in\mathbb{Q}$.

The rest of the cases doesn't seem to continue the pattern.

I wonder if there is any greater meaning behind this:

$$\left[ \begin{array}{c} \text{Integral}&\text{Basis} \\\\ \displaystyle\int_0^1k^{2n+1}K_0^2dk&\zeta(3)\\\\ \displaystyle\int_0^1k^{2n+1}K_{1/6}^2dk&\sqrt{3}\operatorname{Cl_2}(\pi/3)\\\\ \displaystyle\int_0^1k^{2n+1}K_{1/3}^2dk&\sqrt{3}\operatorname{Cl_2}(\pi/3)\\\\ \displaystyle\int_0^1k^{2n+1}K_{1/4}^2dk&\beta(2)\\\\ \displaystyle\int_0^1k^{2n+1}K_0K_{1/3}dk&\ln3\\\\ \end{array} \right]$$

Answer 1

As in this answer , these integrals can be evaluated using the decomposition of the hypergeometric functions on the Legendre polynomials (see for example here): \begin{equation} {}_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) \end{equation} which gives \begin{align} \mathbf K_{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} From the orthogonality of the Legendre polynomials, one obtains \begin{align} \int_0^1xP_p(1-2x^2)P_q(1-2x^2)\,dx&=\frac14\int_{-1}^1P_p(u)P_q(u)\,du\\ &=\frac{1}{2p+1}\delta_{p,q} \end{align} and thus \begin{equation} \int_0^1x\mathbf K_{a}\left(x \right)\mathbf K_{b}\left(x \right)\,dx=\frac{\sin\pi a\sin\pi b}{8}\sum_{p=0}^\infty \frac{2p+1}{(p+a)(p+1-a)(p+b)(p+1-b)} \end{equation} In many cases, these series with rational terms can be evaluated in closed form which are simple, as those given in the OP.

To obtain higher odd moments, \begin{equation} S_n=\int_0^1x^{2n+1}\mathbf K_{a}\left(x \right)\mathbf K_{b}\left(x \right)\,dx \end{equation} one can use the recurrence relation for the Legendre polynomials : \begin{equation} zP_p(z)=\frac{1}{2p+1}\left[(p+1)P_{p+1}(z)+pP_{p-1}(z)\right] \end{equation} with $z=1-2x^2$, one has \begin{equation} (1-2x^2)P_p(1-2x^2)=\frac{1}{2p+1}\left[(p+1)P_{p+1}(1-2x^2)+pP_{p-1}(1-2x^2)\right] \end{equation} and thus \begin{equation} x^2P_p(1-2x^2)=\frac12\left[P_p(1-2x^2)-\frac{p+1}{2p+1}P_{p+1}(1-2x^2)-\frac{p}{2p+1}P_{p-1}(1-2x^2)\right] \end{equation} For the third moments, the integrals to be evaluated read \begin{align} \int_0^1x^3P_p(1-2x^2)P_q(1-2x^2)\,dx&=\frac12I_{pq}-\frac{p+1}{2(2p+1)}I_{p+1,q}-\frac{p}{2(2p+1)}I_{p-1,q}\\ I_{n,q}&= \int_0^1xP_n(1-2x^2)P_q(1-2x^2)\,dx\\ &=\frac{1}{2q+1}\delta_{n,q} \end{align} The third moment can then be written as \begin{align} S_3&=\int_0^1x^3\mathbf K_{a}\left(x \right)\mathbf K_{b}\left(x \right)\,dx\\ &=\frac{S1}{2}-\frac{\sin\pi a\sin\pi b}{16}\sum_{p,q=0}^\infty \frac{(-1)^{p+q}(2p+1)(2q+1)}{(p+a)(p+1-a)(q+b)(q+1-b)}\\ &\hspace{3cm}\times\frac{(p+1)\delta_{p+1,q}+p\delta_{p-1,q}}{(2p+1)(2q+1)}\\ &=\frac{S1}{2}+\frac{\sin\pi a\sin\pi b}{16}\sum_{p=0}^\infty \frac{p+1}{(p+a)(p+1-a)(p+1+b)(p+2-b)}\\ &+\frac{\sin\pi a\sin\pi b}{16}\sum_{p=0}^\infty \frac{p}{(p+a)(p+1-a)(p-1+b)(p-b)} \end{align} As above, these series can often be evaluated under closed forms. For example, \begin{equation} \int_0^1x^3\mathbf K\left(x \right)\mathbf K_{1/6}\left(x \right)\,dx=\frac{1755}{2048}\log3+\frac{81}{512} \end{equation} while \begin{equation} \int_0^1x\mathbf K\left(x \right)\mathbf K_{1/6}\left(x \right)\,dx=\frac{27}{16}\log 3 \end{equation} (I think that there is a typo in the OP, $K_{1/3}$ and $K_{1/6}$ are exchanged. Present results were numerically checked).

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Edit 19/01/2024

To express the moment of order $1$, one can use the summation method by the polygamma function express the series as a weighted sum of $\psi(a),\psi(1-a),\psi(b),\psi(1-b)$ with rational coefficients. Indeed, partial decomposition of the terms of the series gives \begin{equation} \frac{2p+1}{(p+a)(p+1-a)(p+b)(p+1-b)}= \frac{1}{(b-a)(1-a-b)}\left[ \frac{1}{p+a}+\frac{1}{p+1-a} -\frac{1}{p+b}-\frac{1}{p+1-b}\right] \end{equation} Then, assuming $a\ne b$, $a\ne1/2,b\ne1/2$ and non-integer, \begin{equation} \int_0^1x\mathbf K_{a}\left(x \right)\mathbf K_{b}\left(x \right)\,dx=\frac{\sin\pi a\sin\pi b}{8(a-b)(1-a-b)}\left[ \psi(a)+\psi(1-a)-\psi(b)-\psi(1-b)\right] \end{equation} With the reflection formula $\psi\left(z\right)-\psi\left(1-z\right)=-\pi/\tan\left(\pi z\right)$, \begin{align} S_1=\frac{\sin\pi a\sin\pi b}{8(a-b)(1-a-b)}\left[ 2\psi(a)-2\psi(b)+\pi\cot\pi a-\pi\cot\pi b\right] \end{align} If $a=1/2$ and/or $b=1/2$ or $a=b$ the same method gives the result in terms of polygamma functions.

To obtain higher order moments, one can apply the recurrence method described above. It can be seen that for a moment $2n+1$, Legendre polynomials of order $p$ are converted into a weighted sum of polynomials of order $p+n,p+n-2,\ldots,p-n$. In the corresponding series, the factors in the denominators $p+b$ become $p+k+b$ and the factors $p+1-b$ become $p+1-k-b$, with $k=-n,-n+2,\ldots,n$.

Assuming $a,b\ne1/2,a\ne b$ and non-integer, the corresponding poles are all different. The summation method expresses the series as weighted sums of $\psi(a),\psi(1-a),\psi(k+b),\psi(1-k-b)$ with rational coefficients.

The recurrence relation $\psi\left(z+1\right)=\psi\left(z\right)+\frac{1}{z}$ and the reflection formula $\psi\left(z\right)-\psi\left(1-z\right)=-\pi/\tan\left(\pi z\right)$ show that the resulting series are expressible with weighted terms of $\psi(b)$ and $\cot \pi b$ plus rational terms. The former terms are already present in the evaluation $S_1$. Thus, \begin{equation} S_{2n+1}=\sin\pi a\sin\pi b\left[\alpha_n\psi(b)+\beta_n\psi(b)+\gamma_n\pi\cot\pi a+\delta_n\pi\cot\pi b+a_n\right] \end{equation} where all the coefficient are rational.

Here again, the cases $a=1/2$ and/or $b=1/2$ or $a=b$ the same method can be used to express the results. These expressions could be used to validate the conjecture in the given specific cases.