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Question Number 104980 by I want to learn more last updated on 25/Jul/20

Commented by Dwaipayan Shikari last updated on 25/Jul/20

Commented by I want to learn more last updated on 25/Jul/20

$$\mathrm{I}\mathrm{appreciate}\mathrm{sir}$$

Commented by Dwaipayan Shikari last updated on 25/Jul/20

https://youtu.be/q9jbmEGClSk watch this video ��

Answered by profdepo last updated on 25/Jul/20

$$\mathrm{\Lambda }={\int }_0^1\mathrm{In}\left(\mathrm{x}\right)\mathrm{In}(1-\mathrm{x})\mathrm{dx}$$$$\frac{{\partial }^2}{\partial \mathrm{a}\partial \mathrm{b}}{\mid }_{\mathrm{a}=1}\mathrm{\Lambda }={\int }_0^1{\mathrm{x}}^{\mathrm{a}-1}{(1-\mathrm{x})}^{\mathrm{b}-1}\mathrm{dx}$$$$\frac{{\partial }^2}{\partial \mathrm{a}\partial \mathrm{b}}{\mid }_{\mathrm{a}=1}\mathrm{\Lambda }=\beta (\mathrm{a},\mathrm{b})$$$$\mathrm{\Lambda }=2-\zeta \left(2\right).$$$$$$

Commented by I want to learn more last updated on 25/Jul/20

$$\mathrm{I}\mathrm{appreciate}\mathrm{sir}$$

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