Question Number 77549 by BK last updated on 07/Jan/20
Commented by abdomathmax last updated on 08/Jan/20
$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dz}}{\mathrm{1}−{xyz}}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\sum_{{n}=\mathrm{0}} ^{\infty} {x}^{{n}} {y}^{{n}} \:{z}^{{n}} \right){dz} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:{x}^{{n}} \:{y}^{{n}} \:×\frac{\mathrm{1}}{{n}+\mathrm{1}}\:\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \left(\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dz}}{\mathrm{1}−{xyz}}\right){dy} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} {x}^{{n}} ×\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \left(\int_{\mathrm{0}} ^{\mathrm{1}} \left(\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dz}}{\mathrm{1}−{xyz}}\right){dy}\right){dx}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{x}^{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\right){dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} }\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{3}} }\:=\xi\left(\mathrm{3}\right) \\ $$
Commented by BK last updated on 08/Jan/20
$$\mathrm{thanks} \\ $$