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Question Number 77549 by BK last updated on 07/Jan/20

Commented by abdomathmax last updated on 08/Jan/20

$${\int }_0^1\frac{dz}{1-xyz}={\int }_0^1\left(\sum _{n=0}^{\mathrm{\infty }}{x}^n{y}^n{z}^n\right)dz$$$$=\sum _{n=0}^{\mathrm{\infty }}{x}^n{y}^n\times \frac{1}{n+1}\Rightarrow {\int }_0^1\left({\int }_0^1\frac{dz}{1-xyz}\right)dy$$$$=\sum _{n=0}^{\mathrm{\infty }}{x}^n\times \frac{1}{{(n+1)}^2}\Rightarrow$$$${\int }_0^1\left({\int }_0^1\left({\int }_0^1\frac{dz}{1-xyz}\right)dy\right)dx={\int }_0^1\left(\sum _{n=0}^{\mathrm{\infty }}\frac{x^n}{{(n+1)}^2}\right)dx$$$$=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(n+1)}^3}=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^3}=\xi \left(3\right)$$

Commented by BK last updated on 08/Jan/20

$$\mathrm{thanks}$$

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