Calculate $I = \int_{0}^{\infty} \int_{0}^{\infty} \frac{\ln(xy)}{(x^2 + x + 1)(y^2 + y + 1)} \,dx \,dy$

Question

Question Evaluate $$I = \int_{0}^{\infty} \int_{0}^{\infty} \frac{\ln(xy)}{(x^2 + x + 1)(y^2 + y + 1)} \,dx \,dy$$

My try $$I = \int_{0}^{\infty} \int_{0}^{\infty} \frac{\ln(x)}{(x^2 + x + 1)(y^2 + y + 1)} \,dx \,dy + \int_{0}^{\infty} \int_{0}^{\infty} \frac{\ln(y)}{(x^2 + x + 1)(y^2 + y + 1)} \,dx \,dy$$

$$+ 2 \int_{0}^{\infty} \int_{0}^{\infty} \frac{\ln(x) \cdot \ln(y)}{(x^2 + x + 1)(y^2 + y + 1)} \,dx \,dy$$

$$= 2 \int_{0}^{\infty} \frac{\ln(x)}{x^2 + x + 1} \,dx \cdot \int_{0}^{\infty} \frac{1}{y^2 + y + 1} \,dy \quad (\text{symmetry})$$

Answer 1

Let's consider $$I^{(1)}_k=\int_0^\infty ...\int_0^\infty\int_0^\infty \frac{\ln(x_1x_2...x_k)}{(x_1^2 + x_1 + 1)(x_2^2 + x_2 + 1)...(x_k^2 + x_k + 1)} \,dx_1 \,dx_2...dx_k$$ $$=\frac{\partial}{\partial s}\,\bigg|_{s=0}\int_0^\infty ...\int_0^\infty\int_0^\infty \frac{(x_1x_2...x_k)^s}{(x_1^2 + x_1 + 1)(x_2^2 + x_2 + 1)...(x_k^2 + x_k + 1)} \,dx_1 \,dx_2...dx_k$$ $$=\frac{d}{d s}\,\bigg|_{s=0}J^k(s)$$ where $$J(s)=\int_0^\infty \frac{x^s}{x^2 + x + 1} \,dx=\int_0^\infty \frac{x^s}{(x-e^{\frac{2\pi i}3})(x-e^{\frac{4\pi i}3})}\,dx$$ Using the keyhole contour with the cut along the positive part of the axis $X$ $$J(s)(1-e^{2\pi is})=2\pi i\left(\frac{e^{\frac{2\pi is}3}}{e^{\frac{4\pi i}3}-e^{\frac{4\pi i}3}}+\frac{e^{\frac{4\pi is}3}}{e^{\frac{4\pi i}3}-e^{\frac{2\pi i}3}}\right)$$ $$J(s)=\frac{2\pi i}{e^{-\pi is}-e^{\pi is}}\left(\frac{e^{-\frac{\pi is}3}}{e^{\frac{\pi i}3}-e^{-\frac{\pi i}3}}-\frac{e^{\frac{\pi is}3}}{e^{\frac{\pi i}3}-e^{\frac{\pi i}3}}\right)$$ $$=\frac{\pi\,\sin\frac{\pi s}3}{\sin\frac\pi3\,\sin\pi s}=\frac{2\,\pi}{\sqrt3}\,\frac{\sin\frac{\pi s}3}{\sin\pi s}$$ $$I^{(1)}_k=\left(\frac{2\,\pi}{\sqrt3}\right)^k\,\frac{d}{d s}\left(\frac{\sin\frac{\pi s}3}{\sin\pi s}\right)^k\,\bigg|_{s=0}=0\,,\,\,k=1,2,...$$ what follows from the decomposition near $s=0$ :$\quad\left(\frac{\sin\frac{\pi s}3}{\sin\pi s}\right)^k=\left(\frac13+as^2+bs^4+...\right)^k$

In the same way we can evaluate, for example, $$I^{(2)}_k=\int_0^\infty ...\int_0^\infty\int_0^\infty \frac{\ln^2(x_1x_2...x_k)}{(x_1^2 + x_1 + 1)(x_2^2 + x_2 + 1)...(x_k^2 + x_k + 1)} \,dx_1 \,dx_2...dx_k$$ $$=\left(\frac{2\,\pi}{\sqrt3}\right)^k\,\frac{d^2}{d s^2}\left(\frac{\sin\frac{\pi s}3}{\sin\pi s}\right)^k\,\bigg|_{s=0}=\left(\frac{2\,\pi}{\sqrt3}\right)^k\frac{8\, k}{27\cdot3^k}\,,\,\,k=1,2,...$$

Answer 2

Here's a guide. I won't show all the steps, but I'll give you a place to start.

1. Convince yourself that $I=2\int_0^{\infty}\int_0^{\infty} \frac{\log(x)}{(x^2+x+1)(y^2+y+1)}\,dxdy$. I don't quite buy your attempt, but you're going in the right direction.
2. Convince yourself that the $y$ part of the integral is some number $c$ (you can find it by completing the square). Now to focus on $I=2c\int_0^{\infty} \frac{\log(x)}{(x^2+x+1)}\,dx$.
3. Convince yourself that this is improperly integrable, say on $(0,1)$ and then on $(1,\infty)$. At this point you should believe that $I$ is a real number.
4. Carefully make the substitution $x=z^{-1}$. What does this accomplish?