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Question Number 89936 by ~blr237~ last updated on 20/Apr/20

Prove that Σ_(p≥1,q≥1) (1/(pq(p+q−1))) =(π^2 /3)

$${Prove}\:{that}\:\underset{{p}\geqslant\mathrm{1},{q}\geqslant\mathrm{1}} {\sum}\:\:\frac{\mathrm{1}}{{pq}\left({p}+{q}−\mathrm{1}\right)}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{3}}\: \\ $$

Answered by maths mind last updated on 20/Apr/20

Σ_(p,q≥2) ∫_0 ^1 (x^(p+q−2) /(pq))dx =∫_(0 ) ^1 (Σ_(q≥1) (x^(q−1) /q))(Σ_(p≥1) (x^(p−1) /p))dx =∫_0 ^1 (((ln^2 (1−x))/x^2 ))dx ln(1−x)=−u x=1−e^(−u) =∫_0 ^(+∞) ((u^2 e^(−u) )/((1−e^(−u) )^2 )) (1/(1−e^(−u) ))=Σ_(k≥0) e^(−ku) ⇒(e^(−u) /((1−e^(−u) )^2 ))=Σ_(k≥1) ke^(−ku) =Σ_(k≥1) ∫_0 ^(+∞) ku^2 e^(−ku) du =Σ_(k≥1) ∫_0 ^(+∞) ((w^2 e^(−w) )/k^2 )dw =Σ_(k≥1) (1/k^2 ).∫_0 ^∞ w^2 e^(−w) dw=Γ(3).Σ_(k≥1) (1/k^2 )=Γ(3)ζ(2)=2.(π^2 /6)=(π^2 /3)

$$\underset{{p},{q}\geqslant\mathrm{2}} {\sum}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{p}+{q}−\mathrm{2}} }{{pq}}{dx} \\ $$$$=\int_{\mathrm{0}\:} ^{\mathrm{1}} \left(\underset{{q}\geqslant\mathrm{1}} {\sum}\frac{{x}^{{q}−\mathrm{1}} }{{q}}\right)\left(\underset{{p}\geqslant\mathrm{1}} {\sum}\frac{{x}^{{p}−\mathrm{1}} }{{p}}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{{ln}^{\mathrm{2}} \:\left(\mathrm{1}−{x}\right)}{{x}^{\mathrm{2}} }\right){dx} \\ $$$${ln}\left(\mathrm{1}−{x}\right)=−{u} \\ $$$${x}=\mathrm{1}−{e}^{−{u}} \\ $$$$=\int_{\mathrm{0}} ^{+\infty} \frac{{u}^{\mathrm{2}} {e}^{−{u}} }{\left(\mathrm{1}−{e}^{−{u}} \right)^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{\mathrm{1}−{e}^{−{u}} }=\underset{{k}\geqslant\mathrm{0}} {\sum}{e}^{−{ku}} \\ $$$$\Rightarrow\frac{{e}^{−{u}} }{\left(\mathrm{1}−{e}^{−{u}} \right)^{\mathrm{2}} }=\underset{{k}\geqslant\mathrm{1}} {\sum}{ke}^{−{ku}} \\ $$$$=\underset{{k}\geqslant\mathrm{1}} {\sum}\int_{\mathrm{0}} ^{+\infty} {ku}^{\mathrm{2}} {e}^{−{ku}} {du} \\ $$$$=\underset{{k}\geqslant\mathrm{1}} {\sum}\int_{\mathrm{0}} ^{+\infty} \frac{{w}^{\mathrm{2}} {e}^{−{w}} }{{k}^{\mathrm{2}} }{dw} \\ $$$$=\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{k}^{\mathrm{2}} }.\int_{\mathrm{0}} ^{\infty} {w}^{\mathrm{2}} {e}^{−{w}} {dw}=\Gamma\left(\mathrm{3}\right).\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{k}^{\mathrm{2}} }=\Gamma\left(\mathrm{3}\right)\zeta\left(\mathrm{2}\right)=\mathrm{2}.\frac{\pi^{\mathrm{2}} }{\mathrm{6}}=\frac{\pi^{\mathrm{2}} }{\mathrm{3}} \\ $$

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