Question-77549

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

Commented by abdomathmax last updated on 08/Jan/20

∫_0 ^1 (dz/(1−xyz)) =∫_0 ^1 (Σ_(n=0) ^∞ x^n y^n z^n )dz =Σ_(n=0) ^∞ x^n y^n ×(1/(n+1)) ⇒∫_0 ^1 (∫_0 ^1 (dz/(1−xyz)))dy =Σ_(n=0) ^∞ x^n ×(1/((n+1)^2 )) ⇒ ∫_0 ^1 (∫_0 ^1 (∫_0 ^1 (dz/(1−xyz)))dy)dx =∫_0 ^1 (Σ_(n=0) ^∞ (x^n /((n+1)^2 )))dx =Σ_(n=0) ^∞ (1/((n+1)^3 )) =Σ_(n=1) ^∞ (1/n^3 ) =ξ(3)

$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dz}}{\mathrm{1}−{xyz}}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\sum_{{n}=\mathrm{0}} ^{\infty} {x}^{{n}} {y}^{{n}} \:{z}^{{n}} \right){dz} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:{x}^{{n}} \:{y}^{{n}} \:×\frac{\mathrm{1}}{{n}+\mathrm{1}}\:\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \left(\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dz}}{\mathrm{1}−{xyz}}\right){dy} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} {x}^{{n}} ×\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \left(\int_{\mathrm{0}} ^{\mathrm{1}} \left(\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dz}}{\mathrm{1}−{xyz}}\right){dy}\right){dx}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{x}^{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\right){dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} }\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{3}} }\:=\xi\left(\mathrm{3}\right) \\ $$

Commented by BK last updated on 08/Jan/20

$$\mathrm{thanks} \\ $$

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Question-77549

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