Let $K(k),E(k)$ be the complete elliptic integral of the first kind and second kind respectively, where $k$ is the elliptic modulus. Consider four integrals,
$$\begin{aligned}
&I_1=\int_{0}^{1} \frac{K(k)^3-\frac{\pi^3}{8} }{k}
\text{d}k,\\
&I_2=\int_{0}^{1} \frac{K(k)^2E(k)-\frac{\pi^3}{8} }{k}
\text{d}k,\\
&I_3=\int_{0}^{1} \frac{K(k)E(k)^2-\frac{\pi^3}{8} }{k}
\text{d}k,\\
&I_4=\int_{0}^{1} \frac{E(k)^3-\frac{\pi^3}{8} }{k}
\text{d}k.
\end{aligned}$$
$I_1$ is computed to be
$$
\int_{0}^{1} \frac{K(k)^3-\frac{\pi^3}{8} }{k}
\text{d}k=\frac{\Gamma\left ( \frac{1}{4} \right )^8}{3200\pi^2}
-\frac{12}{5}\beta(4)+\frac{\pi^3}{4}\ln(2),
$$
where we utilize
$$
\int_{0}^{1}\left ( \frac{K^\prime}{K} \right )^{s-1}
\left[ \frac{K(k)\left ( K(k)^2-\frac{\pi^2}{4} \right ) }{k}
-\frac{k}{5}K(k)^3 \right]\text{d}k
=\frac{\pi^{4-s}}{20}\Gamma(s)\zeta(s)\left [ \beta(s-4)+5\beta(s-2) \right ].
$$
By differentiatng $(1-k^2)K(k)^3$, we have
$$
\int_{0}^{1} \left ( kK(k)^3-\frac{3K(k)^2\left ( K(k)-E(k) \right ) }{k} \right )
\text{d} k=\left [ (1-k^2)K(k)^3 \right ] \Big|^{1}_0=-\frac{\pi^3}{8}.
$$
Hence the evaluation, $I_2$ is
$$
\int_{0}^{1} \frac{K(k)^2E(k)-\frac{\pi^3}{8} }{k}
\text{d}k=-\frac{\Gamma\left ( \frac{1}{4} \right )^8}{4800\pi^2}
-\frac{12}{5}\beta(4)+\frac{\pi^3}{4}\ln(2)-\frac{\pi^3}{24}.
$$
While I have met the tedious part, differentiating $k^2(1-k^2)K(k)^3,E(k)^3,(1-k^2)K(k)^2E(k)$ and $(1-k^2)K(k)E(k)$. They generate the following relations:
$$\begin{aligned}
&(1)3kK^2E-k^3K^3\equiv0,\\
&(2)E^3/k-KE^2/k\equiv0,\\
&(3)E^3/k+KE^2/k+2kK^2E-3kKE^2\equiv0,\\
&(4)2KE^2/k-kK^2E\equiv0,
\end{aligned}$$
in which we say $f(k)$ is "equal" to $0$ if the regularized value of $\int_{0}^{1}f(k)\text{d}k$ has known explicit closed-forms. These four equations have five unknowns and it seems that the fifth can be constructed normally, though, I was stuck here.
Question. Whether we can find the fifth relation or the closed-forms of integrals the title comprised?
2026-03-26 06:31:53.1774506713