Double integral related to the Riemann zeta function

Question

Is there a way to compute the double integral $$ I(p,q)=\int_{0}^{\frac{1}{2}}\int_{0}^{x}x^{p}y^{q}\cot\pi x\cot\pi y\,dy\,dx $$ for $p,q\geq 1$ being integers? It is known that the single integrals $\int_{0}^{1/2}x^{p}\cot\pi x\,dx$ can be expressed via zeta values. What about the double case? The same question holds for the integral $\int_{0}^{\frac{1}{2}}\int_{0}^{x}\frac{x^{p}y^{q}}{\sin\pi x\sin\pi y}\,dy\,dx$.

Answer 1

I've only been able to derive a result for $p=q=1$ which is $\int_0^{\frac{1}{2}} \int_0^x x^1 y^1 \cot (\pi x) \cot (\pi y) \, dy \, dx=\frac{\log ^2(2)}{8 \pi ^2}$.

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Note $\int_0^{\frac{1}{2}}\int_0^x x^p y^q \cot(\pi x) \cot(\pi y)\,dy\,dx=\int_0^{\frac{1}{2}} x^p \cot(\pi x)\left(\int_0^x y^q \cot(\pi y)\,dy\right)\,dx$. The inner integral can be evaluated for at least some integer values of $q>1$, but the following table illustrates the result grows more-and-more complex as $q$ increases.

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$ \begin{array}{cc} q & \int_0^x y^q \cot (\pi y) \, dy \\ 1 & \frac{1}{12} i \left(1-\frac{6 \left(\text{Li}_2\left(e^{2 i \pi x}\right)+\pi x \left(\pi x+2 i \log \left(1-e^{2 i \pi x}\right)\right)\right)}{\pi ^2}\right) \\ 2 & \frac{6 i \pi x \text{Li}_2\left(e^{-2 i \pi x}\right)+3 \text{Li}_3\left(e^{-2 i \pi x}\right)+2 \pi ^2 x^2 \left(i \pi x+3 \log \left(1-e^{-2 i \pi x}\right)\right)-3 \zeta (3)}{6 \pi ^3} \\ 3 & \frac{180 \pi x \left(i \pi x \text{Li}_2\left(e^{-2 i \pi x}\right)+\text{Li}_3\left(e^{-2 i \pi x}\right)\right)-90 i \text{Li}_4\left(e^{-2 i \pi x}\right)+\pi ^3 \left(i \left(30 \pi x^4+\pi \right)+120 x^3 \log \left(1-e^{-2 i \pi x}\right)\right)}{120 \pi ^4} \\ 4 & \frac{10 \pi x \left(\pi x \left(2 i \pi x \text{Li}_2\left(e^{-2 i \pi x}\right)+3 \text{Li}_3\left(e^{-2 i \pi x}\right)\right)-3 i \text{Li}_4\left(e^{-2 i \pi x}\right)\right)-15 \text{Li}_5\left(e^{-2 i \pi x}\right)+2 \pi ^4 x^4 \left(i \pi x+5 \log \left(1-e^{-2 i \pi x}\right)\right)+15 \zeta (5)}{10 \pi ^5} \\ 5 & \frac{630 \pi x \left(\pi x \left(\pi x \left(i \pi x \text{Li}_2\left(e^{-2 i \pi x}\right)+2 \text{Li}_3\left(e^{-2 i \pi x}\right)\right)-3 i \text{Li}_4\left(e^{-2 i \pi x}\right)\right)-3 \text{Li}_5\left(e^{-2 i \pi x}\right)\right)+945 i \text{Li}_6\left(e^{-2 i \pi x}\right)+\pi ^5 \left(i \pi \left(42 x^6-1\right)+252 x^5 \log \left(1-e^{-2 i \pi x}\right)\right)}{252 \pi ^6} \\ \end{array}$

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I haven't been able to fully evaluate $\int_0^{\frac{1}{2}}\int_0^x\frac{x^p y^q}{\sin(\pi x) \sin(\pi y)}\,dy\,dx=\int_0^{\frac{1}{2}}\frac{x^p}{\sin(\pi x)}\left(\int_0^x\frac{y^q}{\sin(\pi y)}\,dy\right)\,dx$ for any combination of $p$ and $q$ positive integer values, but the inner integral can be evaluated for at least some positive integer values of q as illustrated in the following table.

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$\begin{array}{cc} q & \int_0^x \frac{y^q}{\sin(\pi y)}\,dy \\ 1 & \frac{i \left(-8 \text{Li}_2\left(e^{i \pi x}\right)+2 \text{Li}_2\left(e^{2 i \pi x}\right)+\pi \left(\pi +8 i x \tanh ^{-1}\left(e^{i \pi x}\right)\right)\right)}{4 \pi ^2} \\ 2 & -\frac{2 i \pi x \left(4 \text{Li}_2\left(e^{i \pi x}\right)-\text{Li}_2\left(e^{2 i \pi x}\right)\right)-8 \text{Li}_3\left(e^{i \pi x}\right)+\text{Li}_3\left(e^{2 i \pi x}\right)+4 \pi ^2 x^2 \tanh ^{-1}\left(e^{i \pi x}\right)+7 \zeta (3)}{2 \pi ^3} \\ 3 & \frac{i \left(360 \pi x \left(\pi x \left(\text{Li}_2\left(e^{-i \pi x}\right)+\text{Li}_2\left(-e^{i \pi x}\right)\right)-2 i \left(\text{Li}_3\left(e^{-i \pi x}\right)-\text{Li}_3\left(-e^{i \pi x}\right)\right)\right)-720 \text{Li}_4\left(e^{-i \pi x}\right)-720 \text{Li}_4\left(-e^{i \pi x}\right)+\pi ^3 \left(30 \pi x^4-120 i x^3 \left(\log \left(1-e^{-i \pi x}\right)-\log \left(1+e^{i \pi x}\right)\right)+\pi \right)\right)}{120 \pi ^4} \\ 4 & \frac{40 i \pi x \left(\pi x \left(\pi x \left(\text{Li}_2\left(e^{-i \pi x}\right)+\text{Li}_2\left(-e^{i \pi x}\right)\right)-3 i \left(\text{Li}_3\left(e^{-i \pi x}\right)-\text{Li}_3\left(-e^{i \pi x}\right)\right)\right)-6 \left(\text{Li}_4\left(e^{-i \pi x}\right)+\text{Li}_4\left(-e^{i \pi x}\right)\right)\right)-240 \text{Li}_5\left(e^{-i \pi x}\right)+240 \text{Li}_5\left(-e^{i \pi x}\right)+2 \pi ^4 x^4 \left(i \pi x+5 \log \left(1-e^{-i \pi x}\right)-5 \log \left(1+e^{i \pi x}\right)\right)+465 \zeta (5)}{10 \pi ^5} \\ 5 & \frac{1260 \pi x \left(i \pi x \left(\pi x \left(\pi x \left(\text{Li}_2\left(e^{-i \pi x}\right)+\text{Li}_2\left(-e^{i \pi x}\right)\right)-4 i \left(\text{Li}_3\left(e^{-i \pi x}\right)-\text{Li}_3\left(-e^{i \pi x}\right)\right)\right)-12 \left(\text{Li}_4\left(e^{-i \pi x}\right)+\text{Li}_4\left(-e^{i \pi x}\right)\right)\right)+24 \left(\text{Li}_5\left(-e^{i \pi x}\right)-\text{Li}_5\left(e^{-i \pi x}\right)\right)\right)+30240 i \text{Li}_6\left(e^{-i \pi x}\right)+30240 i \text{Li}_6\left(-e^{i \pi x}\right)+\pi ^5 \left(i \pi \left(42 x^6-1\right)+252 x^5 \left(\log \left(1-e^{-i \pi x}\right)-\log \left(1+e^{i \pi x}\right)\right)\right)}{252 \pi ^6} \\ \end{array}$