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

$$$$$$Advancedcalculus...$$$$Q:If,{\int }_0^1{\int }_0^1\frac{x\ln \left(x\right){\ln }^2\left(y\right)}{1-xy}dxdy=\lambda \underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^3{(n+1)}^2}$$$$\Rightarrow Find,\lambda =?$$$$$$

Answered by witcher3 last updated on 03/Feb/24

$$\mathrm{nice}\mathrm{Probleme}$$$$\mathrm{\forall }\mathrm{x},\mathrm{y}\in [0,1[\Rightarrow \mathrm{xy}<1$$$$\frac{1}{1-\mathrm{xy}}=\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}}}{\mathrm{x}}^{\mathrm{k}}{\mathrm{y}}^{\mathrm{k}}$$$${\int }_0^1{\int }_0^1\frac{\mathrm{xln}\left(\mathrm{x}\right){\ln }^2\left(\mathrm{y}\right)}{1-\mathrm{xy}}\mathrm{dxdy}={\int }_0^1{\int }_0^1\sum _{\mathrm{k}⩾0}{\mathrm{x}}^{\mathrm{k}+1}{\mathrm{y}}^{\mathrm{k}}\ln \left(\mathrm{x}\right){\ln }^2\left(\mathrm{y}\right)\mathrm{dy}$$$$=\sum _{\mathrm{k}⩾0}{\int }_0^1{\mathrm{x}}^{\mathrm{k}+1}\ln \left(\mathrm{x}\right)\mathrm{dx}{\int }_0^1{\mathrm{y}}^{\mathrm{k}}{\ln }^2\left(\mathrm{y}\right)\mathrm{dy}$$$$\sum _{\mathrm{k}⩾0}(-\frac{1}{{(\mathrm{k}+2)}^2}).\left(\frac{2}{{(\mathrm{k}+1)}^3}\right);\mathrm{k}+1=\mathrm{n};-2\sum _{\mathrm{n}⩾1}\frac{1}{{\mathrm{n}}^3{(\mathrm{n}+1)}^2}$$$$\lambda =-2$$

Commented by mnjuly1970 last updated on 03/Feb/24

$$thanksalotsirwicher$$

Commented by witcher3 last updated on 04/Feb/24

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

Answered by MathematicalUser2357 last updated on 06/Feb/24

$$\lambda =-2$$

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