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Prove-that-0-1-ln-2-1-x-ln-x-x-dx-1-2-4-Without-using-the-Beta-function-m-n-




Question Number 143609 by mnjuly1970 last updated on 16/Jun/21
      Prove that::    Ω:=∫_0 ^( 1) ((ln^2 (1−x).ln(x))/x)dx=((−1)/2) ζ (4 )  Without using the “Beta function”    m.n
$$\:\:\: \\ $$$$\:{Prove}\:{that}:: \\ $$$$\:\:\Omega:=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right).{ln}\left({x}\right)}{{x}}{dx}=\frac{−\mathrm{1}}{\mathrm{2}}\:\zeta\:\left(\mathrm{4}\:\right) \\ $$$${Without}\:{using}\:{the}\:“{Beta}\:{function}'' \\ $$$$\:\:{m}.{n} \\ $$
Answered by mindispower last updated on 16/Jun/21
∫((ln(1−x))/x)dx=−li_2 (x)  =−[li_2 (x)ln(x)ln(1−x)]_0 ^1 −∫_0 ^1 ((li_2 (x))/(1−x))ln(x)dx  +∫_0 ^1 ((li_2 (x)ln(1−x))/x)dx  =−∫_0 ^1 ((li_2 (x)ln(x))/(1−x))+∫_0 ^1 li_2 (x).(−d(li_2 (x))  =−∫_0 ^1 ((li_2 (x)ln(x))/(1−x))dx−(1/2)ζ^2 (2)    −∫_0 ^1 ((li_2 (x)ln(x))/(1−x))dx=−Σ_(n≥1) Σ_(k≥0) ∫_0 ^1 (x^n /n^2 ).x^k ln(x)dx  =Σ_(n≥1) Σ_(k≥0) (1/n^2 )∫_0 ^1 x^(n+k) (−ln(x))dx  Σ_(n≥1) Σ_(k≥0) (1/(n^2 (n+k+1)))=Σ_(n≥1) Σ_(k≥n+1) (1/(k^2 n^2 ))=S  start ζ(2).ζ(2)=Σ_(n≥1) Σ_(k≥1) (1/(n^2 k^2 ))=Σ_(n≥1) Σ_(k≥n+1) (1/(n^2 k^2 ))+Σ_(n≥1) (1/(n^2 .n^2 ))  +Σ_(n≥2 ) Σ_(k≤n−1) (1/(n^2 k^2 ))  ⇒2Σ_(n≥1) Σ_(k≥n+1) (1/(n^2 k^2 ))+ζ(4)=ζ^2 (2)  ⇒S=−((ζ(4))/2)+((ζ^2 (2))/2)  Ω=−((ζ(4))/2)+((ζ^2 (2))/2)−((ζ^2 (2))/2)=((ζ(4))/2)
$$\int\frac{{ln}\left(\mathrm{1}−{x}\right)}{{x}}{dx}=−{li}_{\mathrm{2}} \left({x}\right) \\ $$$$=−\left[{li}_{\mathrm{2}} \left({x}\right){ln}\left({x}\right){ln}\left(\mathrm{1}−{x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{li}_{\mathrm{2}} \left({x}\right)}{\mathrm{1}−{x}}{ln}\left({x}\right){dx} \\ $$$$+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{li}_{\mathrm{2}} \left({x}\right){ln}\left(\mathrm{1}−{x}\right)}{{x}}{dx} \\ $$$$=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{li}_{\mathrm{2}} \left({x}\right){ln}\left({x}\right)}{\mathrm{1}−{x}}+\int_{\mathrm{0}} ^{\mathrm{1}} {li}_{\mathrm{2}} \left({x}\right).\left(−{d}\left({li}_{\mathrm{2}} \left({x}\right)\right)\right. \\ $$$$=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{li}_{\mathrm{2}} \left({x}\right){ln}\left({x}\right)}{\mathrm{1}−{x}}{dx}−\frac{\mathrm{1}}{\mathrm{2}}\zeta^{\mathrm{2}} \left(\mathrm{2}\right) \\ $$$$ \\ $$$$−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{li}_{\mathrm{2}} \left({x}\right){ln}\left({x}\right)}{\mathrm{1}−{x}}{dx}=−\underset{{n}\geqslant\mathrm{1}} {\sum}\underset{{k}\geqslant\mathrm{0}} {\sum}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{n}} }{{n}^{\mathrm{2}} }.{x}^{{k}} {ln}\left({x}\right){dx} \\ $$$$=\underset{{n}\geqslant\mathrm{1}} {\sum}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}+{k}} \left(−{ln}\left({x}\right)\right){dx} \\ $$$$\underset{{n}\geqslant\mathrm{1}} {\sum}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{{n}^{\mathrm{2}} \left({n}+{k}+\mathrm{1}\right)}=\underset{{n}\geqslant\mathrm{1}} {\sum}\underset{{k}\geqslant{n}+\mathrm{1}} {\sum}\frac{\mathrm{1}}{{k}^{\mathrm{2}} {n}^{\mathrm{2}} }={S} \\ $$$${start}\:\zeta\left(\mathrm{2}\right).\zeta\left(\mathrm{2}\right)=\underset{{n}\geqslant\mathrm{1}} {\sum}\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}^{\mathrm{2}} {k}^{\mathrm{2}} }=\underset{{n}\geqslant\mathrm{1}} {\sum}\underset{{k}\geqslant{n}+\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}^{\mathrm{2}} {k}^{\mathrm{2}} }+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}^{\mathrm{2}} .{n}^{\mathrm{2}} } \\ $$$$+\underset{{n}\geqslant\mathrm{2}\:} {\sum}\underset{{k}\leqslant{n}−\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}^{\mathrm{2}} {k}^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{2}\underset{{n}\geqslant\mathrm{1}} {\sum}\underset{{k}\geqslant{n}+\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}^{\mathrm{2}} {k}^{\mathrm{2}} }+\zeta\left(\mathrm{4}\right)=\zeta^{\mathrm{2}} \left(\mathrm{2}\right) \\ $$$$\Rightarrow{S}=−\frac{\zeta\left(\mathrm{4}\right)}{\mathrm{2}}+\frac{\zeta^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{2}} \\ $$$$\Omega=−\frac{\zeta\left(\mathrm{4}\right)}{\mathrm{2}}+\frac{\zeta^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{2}}−\frac{\zeta^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{2}}=\frac{\zeta\left(\mathrm{4}\right)}{\mathrm{2}} \\ $$$$ \\ $$

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