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Question Number 80863 by ~blr237~ last updated on 07/Feb/20

$$LetWthelambertfunctiondefinedasW\left({xe}^x\right)=xx⩾0$$$$Provethat{\int }_0^1\frac{W(-ulnu)}{u}du=\frac{\zeta \left(2\right)}{2}$$

Answered by Kamel Kamel last updated on 08/Feb/20

$$\mathrm{w}\left(\mathrm{x}\right)=-\underset{\mathrm{n}=1}{\sum ^{+\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}}{\mathrm{n}}^{\mathrm{n}-1}}{\mathrm{n}!}{\mathrm{x}}^{\mathrm{n}}$$$$\therefore \mathrm{\Omega }={\int }_0^1\mathrm{w}(-\mathrm{uLn}\left(\mathrm{u}\right))\frac{\mathrm{du}}{\mathrm{u}}=-\underset{\mathrm{n}=1}{\sum ^{+\mathrm{\infty }}}\frac{{\mathrm{n}}^{\mathrm{n}-1}}{\mathrm{n}!}{\int }_0^1{\mathrm{u}}^{\mathrm{n}-1}{\mathrm{Ln}}^{\mathrm{n}}\left(\mathrm{u}\right)\mathrm{du}$$$$\stackrel{\mathrm{u}={\mathrm{e}}^{-\mathrm{t}}}{=}-\underset{\mathrm{n}=1}{\sum ^{+\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}}{\mathrm{n}}^{\mathrm{n}-1}}{\mathrm{n}!}{\int }_0^{+\mathrm{\infty }}{\mathrm{t}}^{\mathrm{n}}{\mathrm{e}}^{-\mathrm{nt}}\mathrm{dt}$$$$=-\underset{\mathrm{n}=1}{\sum ^{+\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}}}{{\mathrm{n}}^2\mathrm{n}!}{\int }_0^{+\mathrm{\infty }}{\mathrm{z}}^{\mathrm{n}}{\mathrm{e}}^{-\mathrm{z}}\mathrm{dz}=-\underset{\mathrm{n}=1}{\sum ^{+\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}}}{{\mathrm{n}}^2}$$$$=-(\frac{1}{4}\zeta \left(2\right)-(\zeta \left(2\right)-\frac{1}{4}\zeta \left(2\right))=\frac{\zeta \left(2\right)}{2}=\frac{{\pi }^2}{12}$$

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