With the interests of $$ I(a,b)=\int_{0}^{1}k^a(1-k^2)^bK(k)\text{d}k, $$ where $K(k)$ represents the complete elliptic integral with modulus $k$ and $K^\prime(k)$ its complementary, many $I(a,b)$ for rational $a,b$ are evaluated. The methodologies are as follows:
$\textbf{1.Hypergeometric Representations}$
Expanding $K(k)$, we obtain(conditions omitted)
$$I(a,b)=\frac{\pi}{4} \frac{\Gamma(b+1)\Gamma\left ( \frac{a+1}{2} \right ) }{
\Gamma\left ( \frac{a+3}{2}+b \right ) }
{}_3F_2\left ( \frac{1}{2},\frac12,\frac{a+1}{2};\frac{a+3}{2}+b,1;1 \right ).$$
Note that $I(a,b+1)=I(a,b)-I(a+2,b)$.
Apparently, supposing $\frac{a+3}{2}+b=\frac12$ i.e. $b=-1-\frac{a}{2}$, it simplifies to
$$
I\left ( a,-1-\frac{a}{2} \right )
=\frac{\cos\left ( \frac{\pi a}{2} \right ) }{4\pi}
\Gamma\left ( -\frac{a}{2} \right )^2\Gamma\left ( \frac{a+1}{2} \right )^2.
$$
Also,
$$
I(1,b)=\frac{\pi}{4} \frac{\Gamma\left ( b+1 \right )^2 }{
\Gamma\left ( b+\frac32 \right )^2}.
$$
Some hypergeometric transformations are applicable to the given ${}_3F_2$. For instances, from here we have
$$I(s,0)+I(-s-1,0)
=-\frac{\pi}{4}\tan\left ( \frac{\pi s}{2} \right )
\frac{\Gamma\left ( -\frac{s}{2} \right )^2 }{
\Gamma\left ( \frac{1-s}{2} \right )^2 }.$$
(add on May 20th, 23) Applying Dixon's ${}_3F_2$ Theorem, that allowed us to proceed further. Explicitly, as $a+b=-1/2$,
$$
I(a,b)=\frac{\Gamma\left ( \frac14 \right )^2}{8\sqrt{2\pi} }
\frac{\Gamma\left ( \frac{1+a}{2} \right ) \Gamma\left ( \frac12-a \right )
\Gamma\left ( \frac14-\frac{a}{2} \right ) }{\Gamma\left ( \frac34-\frac{a}{2} \right )
\Gamma\left ( \frac{1-a}{2} \right ) }.
$$
If $a-2b=1,\Re(a)>0$,
$$
I(a,b)=\frac{\Gamma\left ( \frac14 \right )^2}{8\sqrt{\pi} }
\frac{\Gamma\left ( 1+\frac{a}{2} \right )\Gamma\left ( \frac{1+a}{2} \right )^3}{
\Gamma(1+a)\Gamma\left ( \frac{3}{4}+\frac{a}{2} \right )^2 },
$$
which gives the evaluation
$$
I\left ( -\frac{5}{6},-\frac{11}{12} \right )
=\frac{3^{3/4}\left ( 1+\sqrt{3} \right )\Gamma\left ( \frac14 \right )^4 }{4\pi\cdot2^{2/3}}.
$$
$\textbf{2.Contour Integration}$
We have
$$
I(a,b)+\cos(b\pi)I(-a-2b-1,b)
+\sin(b\pi)I\left ( 2b+1,-\frac{a}{2}-b-1 \right )
+\sin\left ( \frac{\pi a}{2} \right )I\left ( a,-\frac{a}{2}-b-1 \right )=0
,$$
and
$$
\cos(b\pi)I\left ( 2b+1,-\frac{a}{2}-b-1 \right )
=\cos\left ( \frac{\pi a}{2} \right )
I\left ( a,-\frac{a}{2}-b-1 \right )+\sin\left ( \pi b \right )
I(-a-2b-1,b).
$$
Setting $a=-5/6,b=-11/12$, producing
\begin{aligned}
I\left ( \frac{5}{3},-\frac{11}{12} \right )
& = \sqrt{3}\left ( \sqrt{3}+1 \right ) I\left ( -\frac56,\frac13 \right )\\
& =\frac{3^{1/4}\left ( \sqrt{3}+1 \right )\Gamma\left ( \frac14 \right )^4 }{4\pi\cdot 2^{1/6}},
\end{aligned}
where the second equality is owing to the former identity for $a+b=-1/2$.
Another less obvious one is,
$$
\int_{0}^{1} \frac{k^{2/3}K(k)}{\left ( 1-k^2 \right )^{2/3} }
\text{d} k=\frac{3^{3/4}\Gamma\left ( \frac14 \right )^4 }{24\pi\cdot2^{1/6} }.
$$
$\textbf{3.Modular Forms(1)}$
The basic idea is to construct a modular form expressed by Jacobi $\vartheta$ functions and compute its $L$-value. For example, setting $q=\exp(-\pi K^\prime(k)/K(k))$
$$
f(q)=\sum_{m,n\in\mathbb{Z}}
(-1)^m\left [ 3\left ( m+\frac16 \right )^2-n^2 \right ]
q^{3\left ( m+\frac16 \right )^2+n^2 }.$$
We have
$$f(q)=\frac{2^{2/3}k^{1/6}(1-k^2)^{1/12}(1-2k^2)}{
3\pi^3}K(k)^3.$$
And
$$\int_{0}^{\infty}xf(q)\text{d}x
=\frac{1}{\pi^2} \sum_{m,n\in\mathbb{Z}}
\frac{(-1)^m}{\left ( \sqrt{3}\left ( m+\frac16 \right )+ni \right )^2 }.$$
Therefore it's sufficient to show that
$$\int_{0}^{1} \frac{(2k^2-1)K(k)}{k^{5/6}(1-k^2)^{11/12}}\text{d}k
=\frac{3^{7/4}\left ( 1+\sqrt{3} \right )\Gamma\left ( \frac13 \right )^6 }{
16\pi^2\cdot2^{1/3} },$$
and therefore,
$$
I\left ( \frac{7}{6},-\frac{11}{12} \right )
=\frac{3^{3/4}\left ( 1+\sqrt{3} \right )\Gamma\left ( \frac14 \right )^4 }{8\pi\cdot2^{2/3}}
+\frac{3^{7/4}\left ( 1+\sqrt{3} \right )\Gamma\left ( \frac13 \right )^6 }{
32\pi^2\cdot2^{1/3} }.
$$
Which also gives
$$
I\left ( -\frac{5}{6},\frac{1}{12} \right )
=\frac{3^{3/4}\left ( 1+\sqrt{3} \right )\Gamma\left ( \frac14 \right )^4 }{8\pi\cdot2^{2/3}}
-\frac{3^{7/4}\left ( 1+\sqrt{3} \right )\Gamma\left ( \frac13 \right )^6 }{
32\pi^2\cdot2^{1/3} },
$$
with
$$
\int_{0}^{1}\left ( 1-k^{12} \right )^{\frac{1}{12}}
K\left ( k^6 \right )\text{d}k =\frac{3^{3/4}\left ( 1+\sqrt{3} \right )\Gamma\left ( \frac14 \right )^4 }{48\pi\cdot2^{2/3}}
-\frac{3^{3/4}\left ( 1+\sqrt{3} \right )\Gamma\left ( \frac13 \right )^6 }{64\pi^2\cdot2^{1/3} }.
$$
However, this usually only gives linear equation among two $I(a,b)$. Apart from
$$I\left(-\frac34,0\right)=\frac{\left ( 3+\sqrt{2} \right )
\Gamma\left ( \frac18 \right )^2\Gamma\left ( \frac38 \right )^2 }{
48\pi\sqrt{2} },$$
$$
I\left(-\frac14,0\right)=\frac{\left ( 3-\sqrt{2} \right )
\Gamma\left ( \frac18 \right )^2\Gamma\left ( \frac38 \right )^2 }{
48\pi\sqrt{2} },$$
they can be:
$$
2I\left ( -\frac13,-\frac{11}{12} \right )
-I\left ( \frac53,-\frac{11}{12} \right )
=\frac{3^{7/4}\left ( 1+\sqrt{3} \right )\Gamma\left ( \frac13 \right )^6 }{
8\cdot2^{5/6}\pi^2 },
$$
$$
4I\left ( -\frac23,-\frac{17}{24} \right )
+I\left ( \frac43,-\frac{17}{24} \right )
=\frac{\left ( \sqrt{2}-1 \right )^{\frac32}
\left ( \sqrt{3} +\sqrt{2} \right )^{\frac32}\cdot3^{\frac14}
\left ( 1+\left ( 2-\sqrt{3} \right )
\left ( \sqrt{3} -\sqrt{2} \right ) \right )^2 }{8\cdot2^{\frac13}\pi}
\Gamma\left ( \frac{1}{24} \right )
\Gamma\left ( \frac{5}{24} \right )
\Gamma\left ( \frac{7}{24} \right )
\Gamma\left ( \frac{11}{24} \right ).
$$
Gathering all information we know about $I(a,b)$, we obtain two additional
values:
\begin{aligned}
&I\left ( -\frac13,-\frac{11}{12} \right )
=\frac{3^{1/4}\left ( 1+\sqrt{3} \right )\Gamma\left ( \frac14 \right )^4 }{8\pi\cdot 2^{1/6}}
+\frac{3^{7/4}\left ( 1+\sqrt{3} \right )\Gamma\left ( \frac13 \right )^6 }{
16\cdot2^{5/6}\pi^2 },\\
&I\left ( -\frac13,\frac{1}{12} \right )
=-\frac{3^{1/4}\left ( 1+\sqrt{3} \right )\Gamma\left ( \frac14 \right )^4 }{8\pi\cdot 2^{1/6}}
+\frac{3^{7/4}\left ( 1+\sqrt{3} \right )\Gamma\left ( \frac13 \right )^6 }{
16\cdot2^{5/6}\pi^2 }.
\end{aligned}
$\textbf{4.Modular Forms(2)}$
It's also possible to prove that
\begin{aligned}
&\color{purple}{\left(\frac2\pi\right)^2\int_{0}^{1} \left ( \frac{K^\prime}{K} \right )^{s-1}
\frac{(1-2k^2)K(k)}{\sqrt{k}(1-k^2)^{3/4}}\text{d}k
=2^{2s}\pi^{-s}\Gamma(s)[L_8(s)L_{-8}(s-2)+L_{-8}(s)L_{8}(s-2)]},\\
&\color{purple}{\left(\frac2\pi\right)^2\int_{0}^{1} \left ( \frac{K^\prime}{K} \right )^{s-1}
\frac{\sqrt{k}K(k)}{(1-k^2)^{1/4}}\text{d}k
=2^{2s-1}\pi^{-s}\Gamma(s)[L_8(s)L_{-8}(s-2)-L_{-8}(s)L_{8}(s-2)]}.
\end{aligned}
Also note that $a=-\frac12,b=-\frac34=-1-\frac{a}{2}$, we derive
\begin{aligned}
&I\left ( \frac32,-\frac34 \right )
=\frac{\pi^2}{4\sqrt{2} }+\frac{\Gamma\left ( \frac14 \right )^4 }{8\pi\sqrt{2} },\\
&I\left ( \frac12,-\frac14 \right )
=\frac{\pi^2}{4\sqrt{2} }.
\end{aligned}
$\textbf{5.Fourier-Legendre Expansions}$
Similarly to here. But most results aren't newly-created.
The questions come here:
- Whether we can find more closed-forms for single $I(a,b)$?
- Can we come up with more ways to cope with $I(a,b)$?
Remark: The remained question is $I\left(\frac34,-\frac38\right)$, which seems to be unable to prove in ways listed above.
