Integrals of elliptic integrals.

Question

Being: $$R(k)=\left(-\frac{K(k)^2}{k}+K(k)E(k)\left(\frac{1}{k}-\frac{k}{2(k^2-1)}\right)\right)$$

We have these evaluations:

$$\int_{0}^{1/\sqrt{2}}R(k)dk=\frac{3\Gamma{(1/4)}^4}{128\pi}-\frac{\pi^2}{8},\tag{1}$$

$$\int_{0}^{2^{5/4}(\sqrt{2}-1)}R(k)dk=\frac{3\Gamma{(1/4)}^4}{64\pi}-\frac{\pi^2}{8},\tag{2}$$

$$\int_{0}^{\sqrt{2}-1}R(k)dk=\frac{\Gamma(1/8)^2\Gamma(3/8)^2\left(\frac{3\sqrt{2}}{512}+\frac{1}{256} \right)}{\pi}-\frac{\pi^2}{8},\tag{3}$$ $$\int_{0}^{\sqrt{2\sqrt{2}-2}}R(k)dk=\frac{\Gamma(1/8)^2\Gamma(3/8)^2}{64\pi}-\frac{\pi^2}{8}.\tag{4}$$ Where $K(k)$ and $E(k)$ are the complete elliptic integrals of the first and second kind respectively. Are particular cases of: $$\int_{0}^{l}R(k)dk=\frac{K(l)^2(2-l^2)}{4}-\frac{\pi^2}{8},\tag{5}$$ and: $$\int_{0}^{\frac{2\sqrt{l}}{1+l}}R(k)dk=\frac{K(l)^2(1+l^2)}{2}-\frac{\pi^2}{8}.\tag{6}$$ Questions: Are these integrals known? Can we improve this result with Clausen's transformation and derive some series?

Answer 1

After some work something has come. The answer to the question is yes, but the sums are not obvius and with not beatiful forms. We have: $$\frac{8K(k)E(k)}{\pi^2}=\frac{2-k^2}{\sqrt{1-k^2}}\sum_{n=0}^{\infty}\frac{(2n+1)(2n)!^3}{2^{8n}n!^6}{\left(\frac{k^4}{ k^2-1}\right)}^n,\tag{1}$$ with $-\frac{1}{\sqrt{2}}\leq k\leq\frac{1}{\sqrt{2}}.$ Where: $$K(k)=\int_{0}^{\pi/2}\frac{dx}{\sqrt{1-k^2\sin^2{(x)}}},\tag{2}$$ and $$E(k)=\int_{0}^{\pi/2}\sqrt{1-k^2\sin^2{(x)}}dx,\tag{3}$$ are the complete elliptic of the first and second kind respectively.

As suggested by Bob Dobbs and Paramanand Singh we provide a proof of $(1)$.

Since we have: $$K(k)(1-k^2)=E(k)-K'(k)(k-k^3),\tag{4}$$ where $K'(k)$ here is the derivative of $K(k)$. Multiplying both sides by $K(k)$: $$K^2(k)(1-k^2)+K(k)K'(k)(k-k^3)=K(k)E(k),\tag{5}$$ We recognize $K(k)K'(k)$ as $\left(\frac{K^2(k)}{2}\right)'$. Then: $$K^2(k)(1-k^2)+\left(\frac{K^2(k)}{2}\right)'(k-k^3)=K(k)E(k).\tag{6}$$ Now we invoke Landen's transformation: $$K(k)(1+k)=K(\frac{2\sqrt{k}}{1+k}),\tag{7}$$ and Clausen's formula (1828) (https://arxiv.org/pdf/1302.5984.pdf) formula as in the version of linked papers. $$_3F_2(1/2,1/2,1/2;1,1;4k^2(1-k^2))=\frac{4K^2(k)}{\pi^2}.\tag{8}$$ Some calculus shows that the combination of both gives: $$\frac{4K^{2}(k)}{\pi^2}=\frac{1}{\sqrt{1-k^2}}\sum_{n=0}^{\infty}\frac{(2n)!^3}{2^{8n}n!^6}\left(\frac{k^4}{k^2-1} \right)^{n},\tag{9}$$ with $|k|\leq \frac{1}{\sqrt{2}}$. Finally, derivation of $(9)$ and $(8)$ applied to $(6)$ gives: $$\frac{8K(k)E(k)}{\pi^2}=\frac{2-k^2}{\sqrt{1-k^2}}\sum_{n=0}^{\infty}\frac{(2n+1)(2n)!^3}{2^{8n}n!^6}{\left(\frac{k^4}{ k^2-1}\right)}^n.$$

With $(1)$ and $(9)$ we can now rewrite the $R(k)$ in the question in terms of an $_3F_2$ function and perform integration term by term. The first case in question is $\int_{0}^{k_{1}}R(k)dk$, where $k_{1}=\frac{1}{\sqrt{2}}$ which maybe can be interesting for first instance but the resulting sum involves also Beta functions in the general terms.

Note : Putting $k_{1}=\frac{1}{\sqrt{2}}$ in $(9)$ and $(1)$ we get: $$\frac{4K^2(k_{1})}{\sqrt{2}\pi^2}=\sum_{n=0}^{\infty}\frac{(-1)^n(2n)!^3}{2^{9n}n!^6}=\frac{\Gamma{(1/4)}^4}{4\sqrt{2}\pi^3},\tag{10}$$ and $$\frac{8\sqrt{2}E(k_{1})K(k_{1})}{3\pi^2}\\=\sum_{n=0}^{\infty}\frac{(-1)^n(2n)!^3(2n+1)}{2^{9n}n!^6}=\frac{2\sqrt{2}}{3\pi}+\frac{\Gamma(\frac{1}{4})^4}{6\sqrt{2}\pi^3},\tag{11}$$ and we deduce: $$-\frac{8K^2(k_{1})}{\sqrt{2}\pi^2}+\frac{8\sqrt{2}E(k_{1})K(k_{1})}{\pi^2}\\=\sum_{n=0}^{\infty}\frac{(-1)^n(6n+1)(2n)!^3}{2^{9n}n!^6}=\frac{2\sqrt{2}}{\pi}.\tag{12}$$

Whis is one of famous Ramanujan's like series for $1/\pi$. And playing with some more singular moduli also others can be obtained.

Answer 2

For the first question (Are these integrals known?), Altough OP is not interested integrals now, turned his attention to series:

First we need to compute the derivative of $K=K(k)=\int_0^{\pi/2}\frac{1}{\Delta}d\phi$ where $\Delta=\sqrt{1-k^2\sin^2\phi}\,\,\,$: $$\frac{dK}{dk}=\int_0^{\pi/2}-\frac{\Delta'}{\Delta^2}d\phi=\int_0^{\pi/2}\frac{k\sin^2\phi}{\Delta^3}d\phi=\frac1k\int_0^{\pi/2}(\frac{1}{\Delta^3}-\frac1\Delta)\,d\phi=\frac{E-(1-k^2)K}{k(1-k^2)}$$ due to the reduction formula $$\int\frac1{\Delta^3}d\phi=-\frac{k^2\sin\phi\cos\phi}{(1-k^2)\Delta}+\frac{1}{(1-k^2)}\int\Delta\,d\phi.$$ Note that $E=E(k)=\int_0^{\pi/2}\Delta\,d\phi$.

Now we can check: $$\frac{d}{dk}\left(\frac{K^2(2-k^2)}{4}\right)=\frac12K\frac{dK}{dk}(2-k^2)-\frac12kK^2=\frac12K\frac{E-(1-k^2)K}{k(1-k^2)}(2-k^2)-\frac12kK^2\\ =-\frac{K^2}{k}+\frac{2-k^2}{2k(1-k^2)}EK=R(k).$$ Therefore, $\int_0^{l}R(k)dk=\frac{K(l)^2(2-l^2)}{4}-\frac{K(0)^2(2-0^2)}{4}=\frac{K(l)^2(2-l^2)}{4}-\frac{\pi^2}{8}$, since $K(0)=\frac\pi 2$.