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} $?

Question

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.

Answer 1

The following is trivial due to expansion of digamma (formula 14 of this page) $$\sum _{n=0}^{\infty } \frac{1}{(a+n) (b+n)}=\frac{\psi ^{(0)}(a)-\psi ^{(0)}(b)}{a-b}$$ Thus OP's integral equals to $$I(a,b)=\sum _{n=0}^{\infty } \frac{a b}{(a b n+b) (a b n+a)}=\frac{\psi ^{(0)}\left(\frac{1}{b}\right)-\psi ^{(0)}\left(\frac{1}{a}\right)}{a-b}$$ Due to Gauss's digamma theorem (formula 11 of link above) for all $a,b\in \mathbb Q$ the integral can be expressed in terms of log and trig functions (even square roots, when $a,b$ are small). For instance $$I(3,5)=\frac{1}{2} \pi \sqrt{\frac{1}{6} \left(-\sqrt{3+\frac{6}{\sqrt{5}}}+\frac{3}{\sqrt{5}}+2\right)}+\frac{1}{8} \log \left(\frac{3125}{729}\right)+\frac{1}{4} \sqrt{5} \coth ^{-1}\left(\sqrt{5}\right)$$

Answer 2

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 $$K_a=\int_0^1 \, _2F_1\left(1,\frac{1}{a};\frac{1}{a}+1;x^a\right)\,dx=\frac 1 {a^2}\psi ^{(1)}\left(\frac{1}{a}\right)$$