Integrals of Gradshteyn and Ryzhik: 5.134 – Integrals of Squares of Jacobi Elliptic Functions – Revised

In this post, we make use of results from Relationships Between Squares of Jacobi Elliptic Functions, Derivatives of Jacobi Elliptic Functions, and Derivatives of Jacobi Elliptic Functions – Part 2. We will refer to equations in the first and third of these posts with i.e 12s and 12d2.

Note that only the first three integrals appear explicitly in Gradshteyn and Ryzhik. The other 9 appear implicitly as part of solutions of elliptic integrals. Thus we establish results here and label them for use in future posts. These 9 integrals can be found in Handbook of Elliptic Integrals for Engineers and Scientists by Byrd and Friedman, sections 310 – 321.

Addendum – Previously, we used the integrals of products of pairs of Jacobi elliptic functions to evaluate some of the integrals in this post. This method was somewhat dubious as it was due to working backwards from known results. We have revised the work so that we only require knowledge of derivatives of single Jacobi elliptic functions and thus no need to “look ahead”. We now do this below and rearrange the order so that it is clear how all integrals were evaluated.

GR 5.134.3

\begin{equation}
\int \mathrm{dn}^{2}(u,k)\,du = \mathrm{E}(\mathrm{am}\,u,k) = \mathrm{E}(u)
\end{equation}
this is a definition of the incomplete elliptic integral of the second kind. \(\mathrm{am}\,u\) is the amplitude function.

GR 5.134.1

\begin{equation}
\int \mathrm{sn}^{2} = \int \frac{1}{k^2}\left[ 1 – \mathrm{dn}^{2}\right] = \frac{1}{k^2}\left[u -\mathrm{E}\right]
\end{equation}
We used GR 5.134.3.

GR 5.134.2

\begin{align}
\int \mathrm{cn}^{2} &= \int [1-\mathrm{sn}^{2}] = u-\frac{u}{k^2}+\frac{\mathrm{E}}{k^2} \\
&= \frac{1}{k^2}\left[\mathrm{E} \,-{k^{\prime}}^{2}u \right]
\end{align}
We used GR 5.134.3.

BF 316.02

We begin with
\begin{equation}
\frac{\partial}{\partial u}\mathrm{sn} = \mathrm{cn} \, \mathrm{dn}
\end{equation}
Rearranging and multiplying both sides by \(\mathrm{dn}\) yields
\begin{equation}
\int \mathrm{dn}^{2} = \int \mathrm{dc} \,\frac{\partial}{\partial u}\mathrm{sn}
\end{equation}
Integrating by parts, we have
\begin{equation}
\int \mathrm{dn}^{2} = \mathrm{dc} \,\mathrm{sn} – {k^{\prime}}^{2} \int \mathrm{sc}^{2}
\end{equation}
Solving for the integral on the right yields
\begin{align}
\int \mathrm{sc}^{2} &= \frac{1}{{k^{\prime}}^{2}}\mathrm{dc} \,\mathrm{sn} \,- \frac{1}{{k^{\prime}}^{2}} \int \mathrm{dn}^{2} \\
&= \frac{1}{{k^{\prime}}^{2}} [\mathrm{dn} \,\mathrm{sc}-E]
\end{align}
We used GR 5.134.3.

BF 313.02

\begin{align}
\int \mathrm{nc}^{2} &= \int [1 + \mathrm{sc}^{2}] = u + \frac{1}{{k^{\prime}}^{2}} [\mathrm{dn} \,\mathrm{sc}-E] \\
&= \frac{1}{{k^{\prime}}^{2}} [{k^{\prime}}^{2} u + \mathrm{dn} \,\mathrm{sc}-E]
\end{align}
We used BF 316.02.

BF 321.02

\begin{equation}
\int \mathrm{dc}^{2} = \int [1+{k^{\prime}}^{2}\mathrm{sc}^{2}] = u + \mathrm{dn}\,\mathrm{sc}\,-\mathrm{E}
\end{equation}
We used 12s and BF 316.02.

BF 317.02

We begin with
\begin{equation}
\frac{\partial}{\partial u}\mathrm{cn} = -\mathrm{sn} \, \mathrm{dn}
\end{equation}
Rearranging and multiplying both sides by \(\mathrm{dn}\), then integrating by parts, we have
\begin{align}
-\int \mathrm{dn}^{2} &= \int \mathrm{ds} \,\frac{\partial}{\partial u}\mathrm{cn} \\
&= \mathrm{cn} \, \mathrm{ds} + \int \mathrm{cn}^{2} \mathrm{ns}^{2} \\
&= \mathrm{cs} \, \mathrm{dn} + \int \mathrm{cs}^{2}
\end{align}
Solving for the integral on the right, and using GR 5.134.3 for the integral on the left, yields
\begin{equation}
\int \mathrm{cs}^{2} = -\,E \,- \mathrm{cs} \, \mathrm{dn}
\end{equation}

BF 311.02

\begin{equation}
\int \mathrm{ns}^{2} = \int [1 + \mathrm{cs}^{2}] = u -\,E \,- \mathrm{cs} \, \mathrm{dn}
\end{equation}
We used BF 317.02.

BF 319.02

\begin{equation}
\int \mathrm{ds}^{2} = \int [\mathrm{ns}^{2}-k^{2}] = {k^{\prime}}^{2}u\,-\mathrm{E}\,-\mathrm{dn}\,\mathrm{cs}
\end{equation}
We used BF 311.02.

BF 315.02

We begin with
\begin{equation}
\frac{\partial}{\partial u}\mathrm{sd} = \mathrm{cn} \, \mathrm{nd}^{2}
\end{equation}
Rearranging yields
\begin{equation}
\int \mathrm{nd}^{2} = \int \mathrm{nc} \, \frac{\partial}{\partial u}\mathrm{sd}
\end{equation}
Integrating by parts, we have
\begin{align}
\int \mathrm{nd}^{2} &= \mathrm{nc} \, \mathrm{sd} \,- \int \mathrm{sd}\,\mathrm{dc}\,\mathrm{sc} \\
&= \mathrm{nc} \, \mathrm{sd} \,- \int \mathrm{sc}^{2} \\
&= \mathrm{nc} \, \mathrm{sd} \,- \frac{1}{{k^{\prime}}^{2}} \mathrm{dn}\,\mathrm{sc} + \frac{1}{{k^{\prime}}^{2}} E \\
&= \frac{1}{{k^{\prime}}^{2}} [E – \mathrm{dn}\,\mathrm{sc} + {k^{\prime}}^{2}\mathrm{nc} \, \mathrm{sd}] \\
&= \frac{1}{{k^{\prime}}^{2}} [E – k^{2} \mathrm{sn}\,\mathrm{cd}]
\end{align}
We used BF 316.02.

BF 318.02

\begin{equation}
\int \mathrm{sd}^{2} = \frac{1}{k^2} \int [\mathrm{nd}^{2}-1] = \frac{1}{k^{2}{k^{\prime}}^{2}} [\mathrm{E}\,-k^{2}\mathrm{sn}\,\mathrm{cd}\,-{k^{\prime}}^{2}u]
\end{equation}
We used BF 315.02.

BF 320.02

\begin{align}
\int \mathrm{cd}^{2} &= \int [\mathrm{nd}^{2}-\mathrm{sd}^{2}] \\
&= \frac{1}{k^{2}{k^{\prime}}^{2}} [-{k^{\prime}}^{2}\mathrm{E} + {k^{\prime}}^{2}u + {k^{\prime}}^{2}k^{2}\mathrm{sn}\,\mathrm{cd} ] \\
&= \frac{1}{k^{2}} [u\,-\mathrm{E}+k^{2}\mathrm{sn}\,\mathrm{cd} ]
\end{align}
We used BF 315.02 and BF 318.02.

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