Integrals of Pairs of Products of Squares of Jacobi Elliptic Functions

We use combinations of the following to evaluate integrals in this post:

The order of presentation here is done to avoid look ahead bias, thus avoiding circularity when referencing previous results.

All results can be found in section 361 of Handbook of Elliptic Integrals for Engineers and Scientists by Byrd and Friedman. We use their equation numbering scheme.

Note that this post will be continuously updated as results are required for evaluating other integrals.

BF 361.02

We begin with
\begin{equation}
\frac{\partial}{\partial u}\mathrm{cn} = -\,\mathrm{sn} \, \mathrm{dn}
\end{equation}
Squaring both sides, integrating by parts, and then making substitutions for \(\mathrm{cn}^{2}\), we have
\begin{align}
\int \mathrm{sn}^{2} \mathrm{dn}^{2} &= \int \left(\frac{\partial}{\partial u}\mathrm{cn}\right) \left(\frac{\partial}{\partial u}\mathrm{cn}\right) \\
&= -\,\mathrm{cn} \,\mathrm{dn}\,\mathrm{sn} + \int \mathrm{cn}^{2} \mathrm{dn}^{2} – k^{2}\int \mathrm{cn}^{2} \mathrm{sn}^{2} \\
&= -\,\mathrm{cn} \,\mathrm{dn}\,\mathrm{sn} + \int (1 – \,\mathrm{sn}^{2})\,\mathrm{dn}^{2} – \int (\mathrm{dn}^{2} – {k^{\prime}}^{2} ) \mathrm{sn}^{2} \\
&= \frac{1}{3 k^{2}}\left[{k^{\prime}}^{2}u + (2k^{2} – 1)E \,-\, k^{2}\,\mathrm{cn} \,\mathrm{dn}\,\mathrm{sn} \right]
\end{align}

BF 361.01

From BF 361.02 we have
\begin{equation}
\int \mathrm{sn}^{2} ({k^{\prime}}^{2} + k^{2}\mathrm{cn}^{2}) = \frac{1}{3 k^{2}}\left[{k^{\prime}}^{2}u + (2k^{2} – 1)E \,-\, k^{2}\,\mathrm{cn} \,\mathrm{dn}\,\mathrm{sn} \right]
\end{equation}
Rearranging yields
\begin{equation}
\int \mathrm{sn}^{2} \mathrm{cn}^{2} = \frac{1}{3 k^{4}}\left[(2 -\,k^{2})E \,-\,2{k^{\prime}}^{2}u \,-\, k^{2}\,\mathrm{cn} \,\mathrm{dn}\,\mathrm{sn} \right]
\end{equation}

BF 361.03

From BF 361.02 we have
\begin{equation}
\int (1 \,-\,\mathrm{cn}^{2}) \mathrm{dn}^{2} = \frac{1}{3 k^{2}}\left[{k^{\prime}}^{2}u + (2k^{2} – 1)E \,-\, k^{2}\,\mathrm{cn} \,\mathrm{dn}\,\mathrm{sn} \right]
\end{equation}
Rearranging yields
\begin{equation}
\int \mathrm{dn}^{2} \mathrm{cn}^{2} = \frac{1}{3 k^{2}}\left[(1+k^{2})E \,-\,{k^{\prime}}^{2}u + k^{2}\,\mathrm{cn} \,\mathrm{dn}\,\mathrm{sn} \right]
\end{equation}

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