Multiple Integrals

Changing variables to $u = x$ , $v = x+y$ , $w = x+y+z$ , the integral becomes $\begin{array}{rcl} \displaystyle I & = & \displaystyle \int_0^1dw \int_0^w dv \int_0^v du \frac{u^2(v-u)^2(w-v)^2 \ln^2(1-w)}{w^9} \\ & = & \displaystyle \left(\int_0^1 u^2(1-u)^2\,du\right)\int_0^1dw \int_0^w dv \frac{v^5(w-v)^2 \ln^2(1-w)}{w^9} \\ & = & \displaystyle B(3,3)\left(\int_0^1 v^5(1-v)^2\,dv\right) \int_0^1 \frac{w^8\ln^2(1-w)}{w^9}\,dw \\ & = & \displaystyle B(3,3)B(6,3) \int_0^1 \frac{\ln^2w}{1-w}\,dw \\ & = & \displaystyle \frac{1}{7!}\sum_{n=0}^\infty \int_0^1 w^n \ln^2w\,dw \; = \; \frac{2}{7!}\sum_{n=0}^\infty \frac{1}{(n+1)^3} \; = \; \frac{\zeta(3)}{2520} \end{array}$ making the answer $\frac{2520}{3} = \boxed{840}$ . This method is, of course, essentially the proof of the result Aditya quotes.

The fact lies in the famous Drichlet's theorem and Liouville's extension of the theorem that is,

$\displaystyle \int \int \int_{V}^{} f(x+y+z)x^{a-1}y^{b-1}z^{c-1} dx dy dz = \frac{\Gamma(a)\Gamma(b)\Gamma(c)}{\Gamma(a+b+c)}\int_{R_1}^{R_2} f(u) u^{a+b+c-1} du$ where $R_1<x,y,z<R_2$ extend over all positive values.

Applying the theorem yields if $I$ is our desired integral, $\displaystyle I = \frac{\Gamma^3 (3)}{\Gamma(9)} \int_{0}^{1} \frac{\ln^2 (1-u)}{u}du$

the latter integral is $\displaystyle \int_{0}^{1}\frac{\ln^2 x}{1-x}dx = \sum_{n=0}^{\infty}\int_{0}^{1} x^n\ln^2 x dx =\sum_{n=0}^{\infty} \frac{2}{(n+1)^3} = 2\zeta(3)$

So our integral is $\displaystyle I=\frac{\zeta(3)}{2520}$ which makes $\displaystyle \frac{Q}{P} = 840$