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Question Number 203964 by mnjuly1970 last updated on 03/Feb/24

Advanced calculus ... Q: If , ∫_0 ^1 ∫_0 ^( 1) (( xln(x)ln^2 (y ))/(1−xy)) dxdy = λ Σ_(n=1) ^∞ (( 1)/(n^( 3) ( n+1 )^2 )) ⇒ Find , λ =?

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{Advanced}\:\:{calculus}\:... \\ $$$$\:\:{Q}:\:\:\:{If}\:,\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{x}\mathrm{ln}\left({x}\right)\mathrm{ln}^{\mathrm{2}} \left({y}\:\right)}{\mathrm{1}−{xy}}\:{dxdy}\:=\:\lambda\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:\mathrm{1}}{{n}^{\:\mathrm{3}} \left(\:{n}+\mathrm{1}\:\right)^{\mathrm{2}} }\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\:{Find}\:,\:\:\:\:\:\lambda\:=? \\ $$$$ \\ $$

Answered by witcher3 last updated on 03/Feb/24

nice Probleme ∀x,y∈[0,1[⇒xy<1 (1/(1−xy))=Σ_(k=0) ^n x^k y^k ∫_0 ^1 ∫_0 ^1 ((xln(x)ln^2 (y))/(1−xy))dxdy=∫_0 ^1 ∫_0 ^1 Σ_(k≥0) x^(k+1) y^k ln(x)ln^2 (y)dy =Σ_(k≥0) ∫_0 ^1 x^(k+1) ln(x)dx∫_0 ^1 y^k ln^2 (y)dy Σ_(k≥0) (−(1/((k+2)^2 ))).((2/((k+1)^3 )));k+1=n;−2Σ_(n≥1) (1/(n^3 (n+1)^2 )) λ=−2

$$\mathrm{nice}\:\mathrm{Probleme} \\ $$$$\forall\mathrm{x},\mathrm{y}\in\left[\mathrm{0},\mathrm{1}\left[\Rightarrow\mathrm{xy}<\mathrm{1}\right.\right. \\ $$$$\frac{\mathrm{1}}{\mathrm{1}−\mathrm{xy}}=\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\mathrm{x}^{\mathrm{k}} \mathrm{y}^{\mathrm{k}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{xln}\left(\mathrm{x}\right)\mathrm{ln}^{\mathrm{2}} \left(\mathrm{y}\right)}{\mathrm{1}−\mathrm{xy}}\mathrm{dxdy}=\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \underset{\mathrm{k}\geqslant\mathrm{0}} {\sum}\mathrm{x}^{\mathrm{k}+\mathrm{1}} \mathrm{y}^{\mathrm{k}} \mathrm{ln}\left(\mathrm{x}\right)\mathrm{ln}^{\mathrm{2}} \left(\mathrm{y}\right)\mathrm{dy} \\ $$$$=\underset{\mathrm{k}\geqslant\mathrm{0}} {\sum}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}^{\mathrm{k}+\mathrm{1}} \mathrm{ln}\left(\mathrm{x}\right)\mathrm{dx}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{y}^{\mathrm{k}} \mathrm{ln}^{\mathrm{2}} \left(\mathrm{y}\right)\mathrm{dy} \\ $$$$\underset{\mathrm{k}\geqslant\mathrm{0}} {\sum}\left(−\frac{\mathrm{1}}{\left(\mathrm{k}+\mathrm{2}\right)^{\mathrm{2}} }\right).\left(\frac{\mathrm{2}}{\left(\mathrm{k}+\mathrm{1}\right)^{\mathrm{3}} }\right);\mathrm{k}+\mathrm{1}=\mathrm{n};−\mathrm{2}\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{3}} \left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\lambda=−\mathrm{2} \\ $$

Commented by mnjuly1970 last updated on 03/Feb/24

thanks alot sir wicher

$${thanks}\:\:{alot}\:{sir}\:{wicher} \\ $$

Commented by witcher3 last updated on 04/Feb/24

withe Pleasur

$$\mathrm{withe}\:\mathrm{Pleasur} \\ $$

Answered by MathematicalUser2357 last updated on 06/Feb/24

λ=−2

$$\lambda=−\mathrm{2} \\ $$

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