How to evaluate $\int_{0}^{1} \int_{0}^{1} \frac{{(1 + x) \cdot \log(x) - (1 + y) \cdot \log(y)}}{{x - y}} \cdot (1 + \log(xy)) \,dy \,dx$

Question

Question: How to evaluate this integral $$\int_{0}^{1} \int_{0}^{1} \frac{{(1 + x) \cdot \log(x) - (1 + y) \cdot \log(y)}}{{x - y}} \cdot (1 + \log(xy)) \,dy \,dx$$

My messy try

$$\int_{0}^{1} \int_{0}^{1} \frac{{(1 + x) \cdot \log(x) - (1 + y) \cdot \log(y)}}{{x - y}} \cdot (1 + \log(xy)) \,dy \,dx$$

$$ \begin{array}{r} I=\int_0^1 \int_0^1 \frac{(1+x) \ln (x)-(1+y) \ln (y)}{x-y}(1+\ln (x y)) d y d x \\ I=\int_0^1 \int_0^{\frac{1}{x}} \frac{(1+x) \ln (x)-(1+x t)(\ln (x)+\ln (t))}{1-t}(1+2 \ln (x)+\ln (t)) d t d x \\ =\int_0^1 \int_0^{\frac{1}{x}} x \ln (x)(1+2 \ln (x)+\ln (t)) d t d x- \\ -\int_0^1 \int_0^{\frac{1}{x}} \frac{(1+x t) \ln (t)}{(1-t)}(1+2 \ln (x)+\ln (t)) d t d x \\ =\int_0^1 x \ln (x)\left(\frac{1+2 \ln (x)-\ln (x)-1}{x}\right) d x-\int_0^1 \int_0^1 \frac{(1+x t) \ln (t)}{(1-t)}(1+2 \ln (x) \ln (t)) d x d t + \end{array} $$ $$- \int_{0}^{\infty} \int_{0}^{\frac{1}{t}} \frac{{(1 + xt) \cdot \ln(t)}}{{1 - t}} \cdot (1 + 2 \cdot \ln(x) + \ln(t)) \,dx \,dt$$

$$I_1 = \int_{0}^{1} x \ln(x) \left( \frac{{1 + 2 \ln(x) - \ln(x) - 1}}{{x}} \right) \,dx - \int_{0}^{1} \int_{0}^{1} \frac{{(1 + xt) \ln(t)}}{{1 - t}} \cdot (1 + 2 \ln(x) + \ln(t)) \,dx \,dt $$

$$I_2 = -\int_{0}^{\infty} \int_{0}^{1/t} \frac{{(1 + xt) \ln(t)}}{{1 - t}} \cdot (1 + 2 \ln(x) + \ln(t)) \,dx \,dt$$

I need hints for figuring out $I_1$ and $I_2$.

Answer 1

$$I = \int_{0}^{1}\int_{0}^{1}\frac{\left(1+x\right)\ln x-\left(1+y\right)\ln y}{x-y}\left(1+\ln\left(xy\right)\right)dydx$$ $$I = I_{1} +I_{2} + I_{3} + I_{4}$$ where $$I_{1} = \int_{0}^{1}\int_{0}^{1}\frac{\ln^{2}x-\ln^{2}y}{x-y}dydx, I_{2} = \int_{0}^{1}\int_{0}^{1}\frac{\ln x-\ln y}{x-y}dydx,$$$$ I_{3} = \int_{0}^{1}\int_{0}^{1}\frac{x\ln x-y\ln y}{x-y}dydx, I_{4} = \int_{0}^{1}\int_{0}^{1}\frac{x\ln x-y\ln y}{x-y}\left(\ln x+\ln y\right)dydx$$.

Every one of these integrands don't behave well when y = x, so split the bounds for x < y and x > y. For example:

$$I_{1} = \int_{0}^{1}\int_{0}^{x}\frac{\ln^{2}x-\ln^{2}y}{x\left(1-\frac{y}{x}\right)}dydx-\int_{0}^{1}\int_{x}^{1}\frac{\ln^{2}x-\ln^{2}y}{y\left(1-\frac{x}{y}\right)}dydx$$

Now you can preform a geometric series expansion on the denominators. This is a very long and tedious calculation but is not very difficult so I will omit it.

In the end: $$I_{1} = -\frac{2\pi^{2}}{3}-4\sum_{n=1}^{\infty}\frac{1}{n^{3}}$$ $$I_{2} = \frac{\pi^{2}}{3}$$ $$I_{3} = \frac{\pi^{2}}{6}-\frac{3}{2}$$ $$I_{4} = \frac{9}{2}-\frac{\pi^{2}}{6}-2\sum_{n=1}^{\infty}\frac{1}{n^{3}}$$

And so finally after adding up each contribution:

$$I = 3-\frac{\pi^{2}}{3}-6\sum_{n=1}^{\infty}\frac{1}{n^{3}}$$

Answer 2

Continuation

$$\begin{aligned} & -\int_0^{+\infty} \int_0^{\frac{1}{t}} \frac{(1+x t) \ln (t)}{(1-t)}(1+2 \ln (x)+\ln (t)) d x d t \\ & I_1=\int_0^1 x \ln (x)\left(\frac{1+2 \ln (x)-\ln (x)-1}{x}\right) d x- \\ & -\int_0^1 \int_0^1 \frac{(1+x t) \ln (t)}{(1-t)}(1+2 \ln (x)+\ln (t)) d x d t+ \\ & I_2=-\int_0^{+\infty} \int_0^{\frac{1}{t}} \frac{(1+x t) \ln (t)}{(1-t)}(1+2 \ln (x)+\ln (t)) d x d t \\ & I=I_1+I_2 \text { (1) } \\ & \left(\begin{array}{c}\boldsymbol{D}=\left\{(\boldsymbol{x}, \boldsymbol{t}) / \mathbf{0}<x<\mathbf{1}, \mathbf{0}<t<\frac{\mathbf{1}}{\boldsymbol{x}}\right\}= \\ =\{(\boldsymbol{t}, \boldsymbol{x}) / \mathbf{0}<t<\mathbf{1}, \mathbf{0}<x<1\} \cup\left\{(\boldsymbol{t}, \boldsymbol{x}) / \mathbf{1}<t<+\infty, 0<x<\frac{\mathbf{1}}{\boldsymbol{t}}\right\}\end{array}\right) \\ & I_1=\int_0^1 \ln ^2(x) d x-\int_0^1 \frac{\ln (t)}{1-t}[x+2 x \ln (x)-2 x+x \ln (t)]_0^1 d t- \\ & -\int_0^1 \frac{t \ln (t)}{1-t}\left[\frac{1}{2} x^2+x^2 \ln (x)-\frac{1}{2} x^2+\frac{1}{2} x^2 \ln (t)\right]_0^1 d t \\ & =\int_0^1 \ln ^2(t) d t+\int_0^1 \frac{\ln (t)}{1-t} d t-\int_0^1 \frac{\ln ^2(t)}{1-t} d t-\frac{1}{2} \int_0^1 \frac{t \ln ^2(t)}{1-t} d t \\ & =\frac{3}{2} \int_0^1 \ln ^2(t) d t+\int_0^1 \frac{\ln (t)}{1-t}-\frac{3}{2} \int_0^1 \frac{\ln ^2(t)}{1-t} d t=3-\frac{\pi^2}{6}-3 \zeta(3)(2) \\ & I_2=-\int_0^{+\infty} \int_0^{\frac{1}{t}} \frac{(1+x t) \ln (t)}{(1-t)}(1+2 \ln (x)+\ln (t)) d x d t \\ & I_2=-\int_0^{+\infty} \int_0^{\frac{1}{t}} \frac{\ln (t)}{(1-t)}(1+2 \ln (x)+\ln (t)) d x d t- \\ & \end{aligned}$$ $$\begin{gathered}-\int_1^{+\infty} \frac{t \ln (t)}{1-t} \int_0^{\frac{1}{t}}(2 x \ln (x)+x(1+\ln (t)) d x d t \\ \left.I_2=-\int_0^{+\infty} \frac{\ln (t)}{(1-t)}(x+2(x \ln (x)-x)+x \ln (t))\right]_0^{\frac{1}{t}} d t- \\ -\int_0^{+\infty} \frac{t \ln (t)}{1-t}\left[x^2 \ln (x)-\frac{1}{2} x^2+\frac{1}{2} x^2(1+\ln (t))\right]_0^{\frac{1}{t}} d t \\ =-\int_0^{+\infty}\left(-\frac{\ln (t)}{t(1-t)}-\frac{\ln ^2(t)}{t(1-t)} d t+\frac{1}{2} \int_0^{+\infty} \frac{\ln ^2(t)}{t(1-t)} d t\right) \\ =\int_0^{+\infty} \frac{\ln (t)}{t(1-t)} d t+\frac{3}{2} \int_0^{+\infty} \frac{\ln ^2(t)}{t(1-t)} d t\left(t \rightarrow \frac{1}{t}\right)=\int_0^{1-t} d t-\frac{3}{2} \int_0^1 \frac{\ln ^2(t)}{1-t} d t \\ \therefore \int_0^1 \frac{1}{\int_0} \frac{(1+x) \ln (x)-(1+y) \ln (y)}{x-y}(1+\ln (x y)) d y d x=3-\frac{\pi^2}{3}-6 \zeta(3)\end{gathered}$$