6
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ This is a difficult topic and you may refer Cayley's An Elementary Treatise on Elliptic Functions. This is the most accessible explanation of Jacobi's Fundamenta Nova. I wrote a few blog posts using this book. Unfortunately the blog post contains formula related to $\operatorname {dn} (u, k) $ only. I will try to see if I can present a self contained answer using modern notation. $\endgroup$
    – Paramanand Singh
    Mar 31, 2022 at 17:38

1 Answer 1

5
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ +1 Nice and simple proof. I was hoping to reproduce the proof by Glaisher which uses the transformation theory for elliptic integrals given by Jacobi. $\endgroup$
    – Paramanand Singh
    Apr 1, 2022 at 2:27
  • $\begingroup$ (+1) Thank you, now the proof is clear! Yes, actually this question is related to my previous question about $\int_{0}^{1}\frac{\log\left(x\right)}{\sqrt{1+x^{4}}}dx$. Glaisher's article provides many such identities but, if I am not mistaken, they are all based on the three main ones I wrote. Therefore, I think it will be unlikely to achieve a closed form in the $K/2$ case by exploiting some classical symmetry or transformation. $\endgroup$ Apr 1, 2022 at 6:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .