Closed form of $\int_0^1\int_0^1\int_0^1\frac{\left(1-x^y\right)\left(1-x^z\right)\ln x}{(1-x)^3}\,\mathrm dx\;\mathrm dy\;\mathrm dz$

Solution 1:

Continuing from Chris's sis answer, consider the partial sum:

$$\frac{1}{2}\sum_{n=1}^m n(n+1)\left(\ln(n)-2\ln(n+1)+\ln(n+2)\right)+1-\frac{1}{n}$$ $$=\frac{m-H_m}{2}+\frac{1}{2}\sum_{n=1}^m\left( \ln\left(\frac{n}{n+1}\right)^{n(n+1)}+\ln\left(\frac{n+2}{n+1}\right)^{n(n+1)}\right)$$ $$=\frac{m-H_m}{2}+\frac{1}{2}\left(\ln\left(\dfrac{\displaystyle \prod_{k=1}^m k^{2k}}{(m+1)^{m(m+1)}}\right)+\ln\left(\dfrac{(m+2)^{m(m+1)}}{\displaystyle \prod_{k=1}^m (k+1)^{2k}}\right)\right)$$ $$=\frac{m-H_m}{2}+\frac{1}{2}\ln\left(\frac{(m+2)^{m(m+1)}}{(m+1)^{m(m+1)}}\cdot \frac{(m!)^2}{(m+1)^{2m}}\right)$$

Use Stirling's approximation and rewrite the expression as:

$$\frac{1}{2}\left(\ln(2\pi)-H_m+\ln m+\ln\left(\left(\frac{m+2}{m+1}\right)^{m(m+1)}\left(\frac{m}{m+1}\right)^{2m}\frac{1}{e^m}\right)\right)$$

I still need to evaluate the logarithmic limit by hand but wolfram alpha gives $-5/2$ i.e the final result is: $$\frac{1}{2}\left(\ln(2\pi)-\gamma-\frac{5}{2}\right)$$

Solution 2:

Using series representation you specified above, I got that your integral gets reduced to $$\sum _{n=1}^{\infty } \frac{(n+1) n^2 (\log (n)-2 \log (n+1)+\log (n+2))+n-1}{2 n}$$

Can you take it from here?