# Is it true that $\int_0^1 \big(K(k^{1/2})\big)^2\,dk = \frac{7}2\zeta(3)$?

Define the complete elliptic integral of the first kind as,

$$K(k) = \tfrac{\pi}{2}\,_2F_1\left(\tfrac12,\tfrac12,1,\,k^2\right)$$

Part I. From the link above, we find some of the evaluations below,

\begin{aligned} \int_0^1 K(k^{1/1})\,dk &= 2C\\ \int_0^1 K(k^{1/2})\,dk &= 2\\ \int_0^1 K(k^{1/3})\,dk &= \frac34(2C+1) \\ \int_0^1 K(k^{1/4})\,dk &= \frac{20}9 \\ \int_0^1 K(k^{1/5})\,dk &= \frac5{64}(18C+13) \end{aligned}

and so on (?) where $$C$$ is Catalan's constant.

Part II. On a hunch, I decided to check 2nd powers. It turns out that,

\begin{aligned} \int_0^1 \big(K(k^{1/2})\big)^2\,dk &= \frac{7}2\zeta(3)\\ \int_0^1 \big(K(k^{1/4})\big)^2\,dk &= \frac{7}2\zeta(3)+1\\ \int_0^1 \big(K(k^{1/6})\big)^2\,dk &= \frac{231}{64}\zeta(3)+\frac{51}{32}\\ \int_0^1 \big(K(k^{1/8})\big)^2\,dk &= \frac{238}{64}\zeta(3)+\frac{881}{432}\\ \end{aligned}

and so on (?) where $$\zeta(3)$$ is Apery's constant.

Question: Does the pattern of Part I (involving Catalan's constant) and Part II (involving Apery's constant) really go on forever? What is the closed-form?

• It seems there is for 3rd powers as well. See this post. Commented Feb 14, 2019 at 6:14
• Interesting. Using Mathematica I have found other conjectural examples of integrals and families of series where a seemingly infinite sequence of similar results occur. When higher $\zeta(2n+1)$ and $\beta(2n)$ constants appear (perhaps also in this case for cubes and higher) they normally appear in repeating patterns along side all their smaller siblings. Commented Feb 14, 2019 at 11:24
• @TitoPiezasIII Did you mean $\int_{0}^{1} (K(k^{1/2}))^{2} = \frac{7}{2}\zeta(3) + 1$ ? This seems to check out numerically Commented Aug 10, 2022 at 18:36
• @MaxMuller In the OP, $k$ is the modulus of the elliptic function defined by the hypergeometric function (first equation). You probably used a program which takes the parameter $m=k^2$ as the argument of the function (as in the Wolfram Language, for example). Then, your numerical result corresponds to the second equation of Part II. Commented Aug 22, 2022 at 9:42

Set $$\theta_2(q):=\sum^{\infty}_{n=-\infty}q^{(n+1/2)^2}\textrm{, }\theta_3(q):=\sum^{\infty}_{n=-\infty}q^{n^2}\textrm{, }\theta_4(q):=\sum^{\infty}_{n=-\infty}(-1)^nq^{n^2}.$$ Then $$\theta_2(q)^2=\frac{2kK}{\pi}\textrm{, }\theta_3(q)^2=\frac{2K}{\pi}\textrm{, }\theta_4(q)^2=\frac{2k'K}{\pi}$$ and $$\frac{dk}{dr}=-\frac{k(k')^2K(k)^2}{\pi\sqrt{r}}.$$ Thus $$I=\int^{1}_{0}K\left(\sqrt{k}\right)^2dk=2\int^{1}_{0}K(k)^2kdk=-2\int^{0}_{\infty}K(k)^2k\frac{k(k')^2K(k)^2}{\pi\sqrt{r}}dr=$$ $$=2\int^{\infty}_{0}\frac{(kk')^2K(k)^4}{\pi\sqrt{r}}dr=2\int^{\infty}_{0}\frac{\pi^2\theta_2(q)^4}{4K^2}\frac{\pi^2\theta_4(q)^4}{4K^2}\frac{K^4}{\pi\sqrt{r}}dr=$$ $$=\frac{\pi^3}{8}\int^{\infty}_{0}\frac{\theta_2(q)^4\theta_4(q)^4}{\sqrt{r}}dr$$ But $$q=e^{-\pi\sqrt{r}}$$. Hence $$dq=\frac{-\pi q}{2\sqrt{r}}dr$$. Hence $$I=\frac{-\pi^3}{8}\int^{0}_{1}\theta_2(q)^4\theta_4(q)^4\frac{1}{\sqrt{r}}\frac{2\sqrt{r}}{\pi q}dq=\frac{\pi^2}{4}\int^{1}_{0}\theta_2(q)^4\theta_4(q)^4\frac{dq}{q}.$$ From the above integral we conclude easily that $$I=\frac{\pi^3}{2}\int^{\infty}_{0}\theta_2\left(e^{-2\pi t}\right)^4\theta_4\left(e^{-2\pi t}\right)^4dt.$$

We set now $$P(z):=\theta_2(q)^4\theta_4(q)^4\textrm{, }q=e^{2\pi i z}\textrm{, }Im(z)>0$$ The function $$P(z)$$ is a weight 4 modular form in $$\Gamma_1(4)$$. The space $$M_4(\Gamma_1(4))$$ has dimension 3, with no cusp forms i.e. $$dim(S_4(\Gamma_1(4))=0$$ and $$dim(E_4(\Gamma_1(4))=3$$.

Consider now the functions $$E_{2k}(q):=2\zeta(2k)\left(1+\frac{2}{\zeta(1-2k)}\sum^{\infty}_{n=1}\sigma_{2k-1}(n)q^n\right),$$ where $$\sigma_{\nu}(n)=\sum_{d|n}d^{\nu}$$, $$\zeta(s)$$ being the Riemann zeta function. The functions $$E_{2k}(q)$$ are the classical Eisenstein series of weight $$2k$$, $$k-$$positive integer. For the present case we get $$k=2$$ and we will use the property $$E_{2k}(q)-lE_{2k}(q^l)$$ is a base element of $$M_{2k}(\Gamma_1(N))$$, when $$l|N$$.

Also in [1] I have proven that if $$q=e^{2\pi i z}$$, $$Im(z)>0$$, then $$H_k(q):=\frac{\pi^k}{k!}\left(\left(2-2^k\right)|B_{k}|+4ki^kF_{k}(q)\right),$$ $$F_k(q):=\sum^{\infty}_{n=1}\sigma^{*}_{k-1}(n)q^n,$$ where $$\sigma^{*}_{\nu}(n):=\sum_{d|n,d-odd}d^{\nu}$$, $$B_{k}$$ are the Bernoulli numbers, $$k-$$even positive integher, are modular forms of the space $$M_k\left(\Gamma_1(2)\right)$$, where $$\Gamma_1(N):=\left\{\left[ \begin{array}{cc} a\textrm{ }b\\ c\textrm{ }d \end{array}\right]:a,b,c,d\in\textbf{Z}\textrm{, }ab-cd=1\textrm{, }a,d\equiv1(N)\textrm{ and }b,c\equiv 0(N) \right\}.$$ By this way and from the fact that $$P(z)$$ is in dimension 3 space, comparing coefficients, we have $$P(z)=C_1\left(E_4(q)-4E_4(q^4)\right)+C_2H_4(q)+C_3H_4(-q),$$ where $$C_1=-\frac{14}{5\pi^4}\textrm{, }C_2=\frac{28}{\pi^4}\textrm{, }C_3=-\frac{92}{5\pi^4}.$$

Hence writing $$P(z)=a_P(0)+\sum^{\infty}_{n=1}a_P(n)q^n,$$ we get $$a_P(0)=0$$ and for $$n=1,2,\ldots$$, we get $$a_P(n)=-\frac{224}{15}\sigma_3(n)+\frac{896}{15}\sigma_3\left(\frac{n}{4}\right)+\frac{56}{3}\sigma^{*}_3(n)-\frac{184}{15}(-1)^n\sigma^{*}_3(n)$$ The Dirichlet series $$L(s)$$ coresponding to $$a_P(n)$$ are $$L(s)=\sum^{\infty}_{n=1}\frac{a_P(n)}{n^s}$$ and the function $$\Lambda_P(s):=\left(\frac{2}{i}\right)^4\int^{+\infty}_{0}P(it)t^{s-1}dt=G(s)\left(\frac{2}{i}\right)^4\sum^{\infty}_{n=1}\frac{a_P(n)}{n^s},$$ where $$G(s)=(2\pi)^{-s}\Gamma(s)$$ (here $$\Gamma$$ means the Euler's Gamma function), have the property (analytic continuation) via the functional equation $$\Lambda_P(s)=4^{2-s}\Lambda_P(4-s)$$ Hence we want to find $$\Lambda_P(1)=4\Lambda_P(3)$$. But $$\Lambda_P(s)=2^4(2\pi)^{-s}\Gamma(s)[-\frac{224}{15}\sum^{\infty}_{n=1}\frac{\sigma_3(n)}{n^s}+\frac{896}{15}4^{-s}\sum^{\infty}_{n=1}\frac{\sigma_3(n)}{n^{s}}+$$ $$+\frac{56}{3}\sum^{\infty}_{n=1}\frac{\sigma^{*}_3(n)}{n^s}-\frac{184}{15}\sum^{\infty}_{n=1}\frac{(-1)^n\sigma^{*}_3(n)}{n^s}]=$$ $$=2^4(2\pi)^{-s} \Gamma(s)[-\frac{224}{15}\zeta(s-3)\zeta(s)+\frac{896}{15}4^{-s}\zeta(s-3)\zeta(s)+$$ $$+\frac{56}{3}2^{-s}(-8+2^s)\zeta(s-3)\zeta(s) -\frac{184}{15}2^{-s}\left(2^{1-s}-1\right)(-8+2^s)\zeta(s-3)\zeta(s)].$$ Hence $$\Lambda_P(3)=\lim_{s\rightarrow 3}\Lambda_P(s)=2^4 (2\pi)^{-3} \Gamma(3)7\zeta(3).$$ Hence $$\Lambda_P(1)=4\Lambda_P(3)=\frac{28\zeta(3)}{\pi^3}=2\cdot 2^{4}I \pi^{-3}$$ and consequently $$I=\frac{7\zeta(3)}{2}.$$ QED

REFERENCES

[1]: N.D. Bagis. ''Evaluations of certain theta functions in Ramanujan theory of alternative modular bases''. arXiv:1511.03716v2 [math.GM] 6 Dec 2017.

• A basis of the weight $4$ Eisenstein series for $\Gamma_1(N)$ are the $E_4(\alpha_j(z))$ with $\alpha_j \in GL_2(\mathbb{Q})$ such that $\Gamma_1(N) \subset \alpha_j^{-1} SL_2(\mathbb{Z}) \alpha_j$ ie. $SL_2(\mathbb{Z}) = \bigcup_{j=1}^J \alpha_j \Gamma_1(N) \alpha_j^{-1}$ which contains things like $E_4(z+1/N)$. To evaluate $\int_0^\infty y^{s-1} E_4(iy+1/N)dy$ we need to look at $L(s,\chi)L(s-3,\chi) = \sum_{n=1}^\infty \sigma_3(n) \chi(n) n^{-s}$ Commented Feb 21, 2019 at 1:07
• I don't know much about modular forms, but I think that the only shadowed point is the choise of $E_{4}(q)-4E_4(q^4)$ as a base element of $M_4(\Gamma_1(4))$. The $H_4(q)$ and $H_4(-q)$ are explained in [1]. Also if the dimension of the space is 3 and possess no cusp forms. A base ellement of $M_4(\Gamma)$ is also a base element of $M_4(\Gamma_1(4))$. Commented Feb 21, 2019 at 1:30

In part I, the integrals can be written as $$$$I_n=n\int_0^1x^{n-1}K(x)\,dx$$$$ The moments of the elliptic integrals seem to be well described in the literature (see here for example) Denoting $$K_n$$ and $$E_n$$ the moments of order $$n$$ of $$K$$ and $$E$$, $$$$K_n=\int_0^1x^nK(x)\,dx; \quad E_n=\int_0^1x^nE(x)\,dx$$$$ Then $$I_n=nK_{n-1}$$ the following recursions are derived: $$$$K_{n+2}=\frac{nK_n+E_n}{n+2};\quad E_n=\frac{K_n+1}{n+2}$$$$ with $$$$K_0=2C;\quad E_0=C+\frac{1}{2};\quad K_1=1;\quad E_1=\frac{2}{3}$$$$ which explains the observed pattern.

In part II, integrals can be written as $$$$J_{2p}=2p\int_0^1 x^{2p-1}K^2(x)\,dx$$$$ Denoting the moment of order $$n$$ of $$K^2$$, $$$${}_2K_n=\int_0^1x^nK^2(x)\,dx$$$$ we have $$$$J_{2p}=2p\,_2K_{2p-1}$$$$ In Moments of products of elliptic integrals by J.G. Wan, Theorem 2 expresses that, when $$p$$ is odd, the p-th moments of $$K'^2, E'^2, K'E', K^2, E^2$$ and $$KE$$ are expressible as $$a+b\zeta(3)$$. Moreover, the moments of $$K^2$$ satisfy a recursion $$$$(n+1)^3 {}_2K_{n+2}-2n\left( n^2+1 \right) {}_2K_n+(n-1)^3 {}_2K_{n-2}=2$$$$ and thus $$$$J_{2p+4}=\frac{p+2}{2\left( p+1 \right)^3}+\frac{\left( p+2 \right)\left( 2p+1 \right)\left( 2p^2+2p+1 \right)}{2\left( p+1 \right)^4}J_{2p+2}-\frac{p^2\left( p+2 \right)}{(p+1)^3}J_{2p}$$$$ In the linked paper, a method is given to obtain $${}_2K_1$$ and $${}_2K_3$$ using a theorem by Zudilin which express them in terms of a $${}_7F_6$$ hypergeometric function.