Multiple integrals problem

Show that! $$\int\limits_0^1 \int\limits_0^1 \int\limits_0^1 \int\limits_0^1 \int\limits_0^1 \int\limits_0^1 \frac{\,\mathrm du \,\mathrm dv \,\mathrm dw \,\mathrm dx \,\mathrm dy \,\mathrm dz}{1-uvwxyz}=\frac{\pi^6}{945}$$

I have no idea about this problem. How to solve this multiple integrals? I have strucked on this problem, please anyone solve it for me.

• Where exactly did you get stuck in the problem? – David H Dec 7 '14 at 11:21

Notice that you can write $$\frac{1}{1-uvwxyz}=\sum_{k=0}^{\infty} (uvwxyz)^k$$ since $0<u,v,w,x,y,z<1$.
So the integral is: $$\sum_{k=0}^{\infty} \int_0^1 \int_0^1 \int_0^1 \int_0^1 \int_0^1 \int_0^1(uvwxyz)^k\,du\,dv\,dw\,dx\,dy\,dz=\sum_{k=0}^{\infty} \frac{1}{(k+1)^6}=\zeta(6)$$
$$=\boxed{\dfrac{\pi^6}{945}}$$
\begin{align} \mathcal{I} &=\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\frac{\mathrm{d}u\mathrm{d}v\mathrm{d}w\mathrm{d}x\mathrm{d}y\mathrm{d}z}{1-uvwxyz}\\ &=\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\frac{\operatorname{Li}_{0}{\left(uvwxyz\right)}}{uvwxyz}\mathrm{d}u\mathrm{d}v\mathrm{d}w\mathrm{d}x\mathrm{d}y\mathrm{d}z\\ &=\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\frac{\operatorname{Li}_{1}{\left(vwxyz\right)}}{vwxyz}\mathrm{d}v\mathrm{d}w\mathrm{d}x\mathrm{d}y\mathrm{d}z\\ &=\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\frac{\operatorname{Li}_{2}{\left(wxyz\right)}}{wxyz}\mathrm{d}w\mathrm{d}x\mathrm{d}y\mathrm{d}z\\ &=\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\frac{\operatorname{Li}_{3}{\left(xyz\right)}}{xyz}\mathrm{d}x\mathrm{d}y\mathrm{d}z\\ &=\int_{0}^{1}\int_{0}^{1}\frac{\operatorname{Li}_{4}{\left(yz\right)}}{yz}\mathrm{d}y\mathrm{d}z\\ &=\int_{0}^{1}\frac{\operatorname{Li}_{5}{\left(z\right)}}{z}\mathrm{d}z\\ &=\operatorname{Li}_{6}{\left(1\right)}\\ &=\frac{\pi^6}{945}\\ \end{align}