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Let-W-the-lambert-function-defined-as-W-xe-x-x-x-0-Prove-that-0-1-W-ulnu-u-du-2-2-

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Question Number 80863 by ~blr237~ last updated on 07/Feb/20
Let W the lambert function defined as W(xe^x )=x x≥0 Prove that ∫_0 ^1 (( W(−ulnu))/u)du=((ζ(2))/2)
$$\:{Let}\:{W}\:{the}\:{lambert}\:{function}\:{defined}\:{as}\:{W}\left({xe}^{{x}} \right)={x}\:\:\:{x}\geqslant\mathrm{0} \\ $$$${Prove}\:{that}\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\:{W}\left(−{ulnu}\right)}{{u}}{du}=\frac{\zeta\left(\mathrm{2}\right)}{\mathrm{2}}\:\: \\ $$
Answered by Kamel Kamel last updated on 08/Feb/20
w(x)=−Σ_(n=1) ^(+∞) (((−1)^n n^(n−1) )/(n!))x^n ∴ Ω=∫_0 ^1 w(−uLn(u))(du/u)=−Σ_(n=1) ^(+∞) (n^(n−1) /(n!))∫_0 ^1 u^(n−1) Ln^n (u)du =^(u=e^(−t) ) −Σ_(n=1) ^(+∞) (((−1)^n n^(n−1) )/(n!))∫_0 ^(+∞) t^n e^(−nt) dt = −Σ_(n=1) ^(+∞) (((−1)^n )/(n^2 n!))∫_0 ^(+∞) z^n e^(−z) dz=−Σ_(n=1) ^(+∞) (((−1)^n )/n^2 ) =−((1/4)ζ(2)−(ζ(2)−(1/4)ζ(2))=((ζ(2))/2)=(π^2 /(12))
$$\mathrm{w}\left(\mathrm{x}\right)=−\underset{\mathrm{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{n}^{\mathrm{n}−\mathrm{1}} }{\mathrm{n}!}\mathrm{x}^{\mathrm{n}} \\ $$$$\therefore\:\Omega=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{w}\left(−\mathrm{uLn}\left(\mathrm{u}\right)\right)\frac{\mathrm{du}}{\mathrm{u}}=−\underset{\mathrm{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{n}^{\mathrm{n}−\mathrm{1}} }{\mathrm{n}!}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{u}^{\mathrm{n}−\mathrm{1}} \mathrm{Ln}^{\mathrm{n}} \left(\mathrm{u}\right)\mathrm{du} \\ $$$$\:\:\:\:\:\:\:\:\overset{\mathrm{u}=\mathrm{e}^{−\mathrm{t}} } {=}\:\:\:−\underset{\mathrm{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{n}^{\mathrm{n}−\mathrm{1}} }{\mathrm{n}!}\int_{\mathrm{0}} ^{+\infty} \mathrm{t}^{\mathrm{n}} \mathrm{e}^{−\mathrm{nt}} \mathrm{dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:\:\:−\underset{\mathrm{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}^{\mathrm{2}} \mathrm{n}!}\int_{\mathrm{0}} ^{+\infty} \mathrm{z}^{\mathrm{n}} \mathrm{e}^{−\mathrm{z}} \mathrm{dz}=−\underset{\mathrm{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:=−\left(\frac{\mathrm{1}}{\mathrm{4}}\zeta\left(\mathrm{2}\right)−\left(\zeta\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{4}}\zeta\left(\mathrm{2}\right)\right)=\frac{\zeta\left(\mathrm{2}\right)}{\mathrm{2}}=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\right. \\ $$

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