For what $a$ and $b$ are there explicit expressions for $I(a, b) =\int_0^1 \int_0^1 \dfrac{dx\,dy}{1-x^ay^b} $? For what $a$ and $b$
are there explicit expressions for
$I(a, b)
=\int_0^1 \int_0^1 \dfrac{dx\,dy}{1-x^ay^b}
$?
This is inspired by
the answer to
https://www.quora.com/What-is-displaystyle-int_-0-1-int_-0-1-frac-1-1-xy-3-mathrm-d-x-mathrm-d-y
where it is shown that
$I(1, 3)
=\dfrac34\left(\ln(3)+\dfrac{\pi\sqrt{3}}{9}\right)
$.
Here's what I've done
so far.
$\begin{array}\\
I(a, b)
&=\int_0^1 \int_0^1 \dfrac{dx\,dy}{1-x^ay^b}\\
&=\int_0^1 \int_0^1 dx\,dy\sum_{n=0}^{\infty} (x^ay^b)^n\\
&=\sum_{n=0}^{\infty} \int_0^1 \int_0^1 dx\,dy(x^ay^b)^n\\
&=\sum_{n=0}^{\infty} \int_0^1 x^{an}dx \int_0^1 y^{bn}dy\\
&=\sum_{n=0}^{\infty} \dfrac{x^{an+1}}{an+1}\big|_0^1 \dfrac{y^{bn+1}}{bn+1}\big|_0^1\\
&=\sum_{n=0}^{\infty} \dfrac{1}{an+1} \dfrac{1}{bn+1}\\
&=ab\sum_{n=0}^{\infty} \dfrac{1}{abn+b} \dfrac{1}{abn+a}\\
\\
I(a.a)
&=\sum_{n=0}^{\infty} \dfrac{1}{(an+1)^2}\\
&=\dfrac1{a^2}\sum_{n=0}^{\infty} \dfrac{1}{(n+1/a)^2}\\
&=\dfrac1{a^2}\psi^{(1)}(1/a)\\
\text{If } a \ne b\\
I(a, b)
&=\dfrac{ab}{a-b}\sum_{n=0}^{\infty} \left(\dfrac{1}{abn+b} -\dfrac{1}{abn+a}\right)\\
&=\dfrac{ab}{a-b}\sum_{n=0}^{\infty} \int_0^1 \left(x^{abn+b-1} -x^{abn+a-1}\right)dx\\
&=\dfrac{ab}{a-b} \int_0^1 \left(\sum_{n=0}^{\infty}x^{abn+b-1} -\sum_{n=0}^{\infty}x^{abn+a-1}\right)dx\\
&=\dfrac{ab}{a-b} \int_0^1 \left(x^{b-1}\sum_{n=0}^{\infty}x^{abn} -x^{a-1}\sum_{n=0}^{\infty}x^{abn}\right)dx\\
&=\dfrac{ab}{a-b} \int_0^1 \left(x^{b-1}\dfrac1{1-x^{ab}} -x^{a-1}\dfrac1{1-x^{ab}}\right)dx\\
&=\dfrac{ab}{a-b} \int_0^1 \left(\dfrac{x^{b-1} -x^{a-1}}{1-x^{ab}}\right)dx\\
&=\dfrac{ab}{a-b} \int_0^1 x^{b-1}\left(\dfrac{1 -x^{a-b}}{1-x^{ab}}\right)dx\\
\end{array}
$
But,
in general,
I can't go further.
Special cases can be handled.
For example,
if $b=1$
and $a$ is a positive integer,
we can show that
$I(a, 1)
=\dfrac{a}{(a-1)^2} \left(\ln(a)-\int_0^1 \left(\dfrac{\sum_{k=0}^{a-3}(a-2-k)x^k}{\sum_{k=0}^{a-1}x^k}\right)dx\right)
$.
This gives,
as above,
$I(3, 1)
=\dfrac34\left(\ln(3)-\int_0^1 \left(\dfrac{1}{1+x+x^2}\right)dx\right)
$
and completing the square
gives the answer above.
Also
$I(2, 1)
=2\ln(2)
$.
Similarly,
$\begin{array}\\
I(4, 1)
&=\dfrac49\left(\ln(4)-\int_0^1 \left(\dfrac{2+x}{1+x+x^2+x^3}\right)dx\right)\\
&=\dfrac49\left(\ln(4)- \dfrac{3 π + \ln(4)}{8}\right)
\quad\text{(According to Wolfy)}\\
&=\dfrac49\left(\dfrac{7\ln(4)-3 π}{8}\right)\\
\end{array}
$
We could get
$I(5, 1)$
since Wolfy gives
$$\dfrac1{1+x+x^2+x^3+x^4}
=\dfrac{-2 x + \sqrt{5} - 1}{\sqrt{5} (2 x^2 - (\sqrt{5}-1) x + 2)} + \dfrac{2 x + \sqrt{5} + 1}{\sqrt{5} (2 x^2 + (\sqrt{5}+1) x + 2)}
$$
but I'm not going to bother
to work it out.

May be, we could start with
$$I(a, b)=\int_0^1 \int_0^1 \dfrac{dx\,dy}{1-x^ay^b}=\int_0^1 \, _2F_1\left(1,\frac{1}{b};\frac{1}{b}+1;x^a\right)\,dx$$ and consider
$$J_a=\int_0^1 \, _2F_1\left(1,\frac{1}{b};\frac{1}{b}+1;x^a\right)\,dx$$ and discarding the ones which contain logarithms of complex arguments, we can obtain a few of them
$$J_1=\frac{\psi \left(\frac{1}{b}\right)+\gamma }{1-b}$$
$$J_2=\frac{\psi \left(\frac{1}{b}\right)+\gamma +\log (4)}{2-b}$$
$$J_3=\frac{6 \psi \left(\frac{1}{b}\right)+\sqrt{3} \pi +6 \gamma +9 \log   (3)}{18-6 b}$$
$$J_4=\frac{2 \left(\psi \left(\frac{1}{b}\right)+\gamma \right)+\pi +\log (64)}{8-2 b}$$
$$J_6=\frac{2 \psi \left(\frac{1}{b}\right)+\sqrt{3} \pi +2 \gamma +\log
   (432)}{12-2 b}$$
$$J_8=\frac{2 \psi \left(\frac{1}{b}\right)+\sqrt{2} \pi +\pi +2 \gamma +\log
   (256)+2 \sqrt{2} \coth ^{-1}\left(\sqrt{2}\right)}{16-2 b}$$
$$J_{12}=\frac{2 \psi \left(\frac{1}{b}\right)+\sqrt{3} \pi +2 \pi +2 \gamma +2
   \sqrt{3} \log \left(2+\sqrt{3}\right)+\log (1728)}{24-2 b}$$
For the particular case where $b=a$, the result seems to be quite simpler
$$K_a=\int_0^1 \, _2F_1\left(1,\frac{1}{a};\frac{1}{a}+1;x^a\right)\,dx=\frac 1 {a^2}\psi ^{(1)}\left(\frac{1}{a}\right)$$
