Elliptic integrals and $\zeta(5)$.

Question

Who can evaluate this one? $$ \int_0^1 \frac{K'(k)^4}{K(k)^2} \;k\;dk = \frac{31}{8} \zeta(5) . $$ Note: I used the elliptic modulus $k$ (and not the parameter $m = k^2$ commonly seen in Mathematica). That is: \begin{align} K(k) &:= \int_0^1\frac{dt}{\sqrt{(1-t^2)(1-k^2 t^2)}} \\ K'(k) &:= K(\sqrt{1-k^2}\;) \end{align}

Answer 1

Let $x= K'(k)/K(k)$, then $\frac{dx}{dk} = -\frac{\pi}{2kk'^{2}K^{2}}$. Let $\tau = ix$, then $$k = \frac{\vartheta_2^2(\tau)}{\vartheta_3^2(\tau)}\qquad k' = \frac{\vartheta_4^2(\tau)}{\vartheta_3^2(\tau)}\qquad K=\frac{\pi}{2}\vartheta_3^2(\tau)\qquad iK'=\frac{\pi}{2}\tau\vartheta_3^2(\tau)$$ where $\vartheta_i$ are Jacobi theta functions. So $$I = \int_0^1 \frac{K'(k)^4}{K(k)^2} k dk = \frac{\pi^3}{8}\int_0^\infty x^4 \vartheta_2^4(\tau) \vartheta_4^4(\tau) dx = 4\pi^3 \int_0^\infty x^4 f(ix) dx =\frac{3}{\pi^2}L(5,f)$$ where $f(z) = \vartheta_2^4(2z) \vartheta_4^4(2z)$ is a weight-$4$ modular form of $\Gamma_1(4)$. There is no cusp form in $M_4(\Gamma_1(4))$, so we can immediately conclude $I$ can be expressed in terms of Dirichlet $L$-functions (because Fourier coefficients of Eisenstein series are given by divisor-sum functions, and their $L$-series are products of degree $1$ $L$-functions).

This answer explicitly computes $L(s,f)$: $$L(s,f) = 4^{2-s} (2^s-16)(2^s-1) \zeta (s-3) \zeta (s)$$ so $I = 31\zeta(5)/8$ as desired.