While trying to find several references to answer Pranav's problem, I encounter the following multiple integrals
$$I=\int_0^1\int_0^1\int_0^1\frac{\left(1-x^y\right)\left(1-x^z\right)\ln x}{(1-x)^3}\,\mathrm dx\;\mathrm dy\;\mathrm dz$$
Question :
Does the above integral have a closed form? If it has a closed form, how to obtain it?
According to Chriss'sis in the chat room, the integral has a nice closed form $$\frac{2\ln 2 \pi -2 \gamma -5}{4}$$ but no proof given so far. Since the term $\ln 2 \pi$ appears in the close form, I have a hunch that it would involve Stirling's approximation. After trying to evaluate it for hours, I think it is about time to ask it on the main page. Here is my attempt so far: \begin{align} I&=\int_0^1\int_0^1\left(1-x^y\right)\;\mathrm dy\int_0^1\left(1-x^z\right)\;\mathrm dz\frac{\ln x}{(1-x)^3}\,\mathrm dx\\[7pt] &=\int_0^1\frac{\ln x+1-x}{\ln x}\cdot\frac{\ln x+1-x}{\ln x}\cdot\frac{\ln x}{(1-x)^3}\,\mathrm dx\\[7pt] &=\int_0^1\frac{\left(\ln x+1-x\right)^2}{(1-x)^3\ln x}\,\mathrm dx\\ \end{align} I'm stuck in the latter expression. I'm thinking of the following parametric integral $$I(s)=\int_0^1\frac{x^s\left(\ln x+1-x\right)^2}{(1-x)^3\ln x}\,\mathrm dx\quad\implies\quad I'(s)=\int_0^1\frac{x^s\left(\ln x+1-x\right)^2}{(1-x)^3}\,\mathrm dx$$ then expanding the quadratic term and using series representation of $$\frac{1}{(1-x)^3}=\frac{1}{2}\sum_{n=1}^\infty n(n-1)x^{n-2}$$ but the calculation would be cumbersome. Is there a better way? Thanks in advance for your help.