In a 1879 work, Glaisher proves the following closed forms $$\int_{0}^{K\left(k\right)}\log\left(\text{sn}\left(z;k\right)\right)dz=-\frac{1}{4}\pi K^{\prime}\left(k\right)-\frac{1}{2}K\left(k\right)\log\left(k\right)$$ $$\int_{0}^{K\left(k\right)}\log\left(\text{cn}\left(z;k\right)\right)dz=-\frac{1}{4}\pi K^{\prime}\left(k\right)+\frac{1}{2}K\left(k\right)\log\left(\frac{k}{k^{\prime}}\right)$$ $$\int_{0}^{K\left(k\right)}\log\left(\text{dn}\left(z;k\right)\right)dz=\frac{1}{2}K\left(k\right)\log\left(k^{\prime}\right)$$ where, $\text{sn}\left(z;k\right),\,\text{cn}\left(z;k\right),\,\text{dn}\left(z;k\right)$ are the Jacobi elliptic functions, $K(k)$ is the complete elliptic integral of the first kind and, as usual, $K^{\prime}(k)=K(k^{\prime})$, where $k^{\prime}=\sqrt{1-k^{2}}.$ For the proof he use a product formula for the elliptic functions; I tried to understand what he did but the steps don't have many explanations and therefore I struggle to understand how to prove these identities.
Question 1. How we can prove the previous identities? This is the link to the paper of Glaisher: https://royalsocietypublishing.org/doi/pdf/10.1098/rspl.1879.0056
I need to understand these identities because I would like to find a closed form for the following definite integrals: $$\int_{0}^{K\left(k\right)/2}\log\left(\text{sn}\left(z;k\right)\right)dz,\,\int_{0}^{K\left(k\right)/2}\log\left(\text{cn}\left(z;k\right)\right)dz,\,\int_{0}^{K\left(k\right)/2}\log\left(\text{dn}\left(z;k\right)\right)dz\tag{1}$$
Question 2. Is it possible to evaluate in a closed form (in the sense of Glaisher) the integrals in $(1)$?
I tried some identites, like half argument formulas, hoping to fall back into one of the cases already considered by Glaisher but it seems that this approach does not work.
Thank you.
Best Answer
Use the infinite product expansion of Jacobi functions ($q=e^{-\pi K'/K}$):$$\tag{*}\text{sn}(\frac{2Kx}{\pi},k) = 2q^{1/4}k^{-1/2}\sin x \prod_{n=1}^\infty \frac{(1-q^{2n}e^{2ix})(1-q^{2n}e^{-2ix})}{(1-q^{2n-1}e^{2ix})(1-q^{2n-1}e^{-2ix})}$$ taking log and integrate on $x$ from $0$ to $\pi/2$. The infinite product integrates to $0$ since its Fourier expansion consists of $\cos 2mx$ only and $\int_{0}^{\pi/2} \cos 2mx dx = 0$. Therefore $$\int_0^{\pi/2} \text{sn}(\frac{2Kx}{\pi},k) dx = \frac{\pi}{2}\log(2q^{1/4}k^{-1/2}) - \frac{\pi}{2}\log 2$$ giving your first formula.
Exactly same reasoning for $\text{cn}, \text{dn}$, using $$\text{cn}(\frac{2Kx}{\pi},k) = 2q^{1/4}k'^{1/2}k^{-1/2}\cos x \prod_{n=1}^\infty \frac{(1+q^{2n}e^{2ix})(1+q^{2n}e^{-2ix})}{(1-q^{2n-1}e^{2ix})(1-q^{2n-1}e^{-2ix})}$$
$$\text{dn}(\frac{2Kx}{\pi},k) = k'^{1/2}\prod_{n=1}^\infty \frac{(1+q^{2n-1}e^{2ix})(1+q^{2n-1}e^{-2ix})}{(1-q^{2n-1}e^{2ix})(1-q^{2n-1}e^{-2ix})}$$
Regarding your second problem, the outlook is somewhat bleaker. It's actually closely related to your previous question $\int_{0}^{1}\frac{\log\left(x\right)}{\sqrt{1+x^{4}}}dx$, which I had a deep impression.
You could integrate $(*)$ from $0$ to $\pi/4$, but now the infinite product term produces a series $S=\sum_{n\geq1}\frac{1}{n^{2}}\frac{q^{n} (-1)^n}{1+q^{n}}$. So integrals $$\int_0^{K/2} \log(\text{an elliptic function})$$ has a lot of incarnations, including $S$ and your $_3F_2$ representation.
I said more regarding them in my three long comments there.