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Question Number 118924 by mnjuly1970 last updated on 21/Oct/20

$$...advancedcalculus...$$$$provethat::$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n{H}_n}{n^2}={\int }_0^1\frac{ln(1-x)ln(1+x)}{x}dx$$$$note::H_n=\underset{k=1}{\sum ^n}\frac{1}{k}$$$$therefore:\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n{H}_n}{n^2}=\frac{-5}{8}\zeta \left(3\right)✓$$$$..m.n.july.1970...$$$$$$

Answered by mindispower last updated on 21/Oct/20

$${\int }_0^1{x}^{n-1}ln(1-x)dx=-{\int }_0^1{x}^{n-1}\sum _{k⩾1}\frac{x^k}{k}dx$$$$=-\sum _k\frac{1}{k}{\int }_0^1{x}^{n+k-1}dx=-\sum _{k⩾1}\frac{1}{(n+k)k}$$$$=-\sum _{k⩾1}\frac{1}{n}(\frac{1}{k}-\frac{1}{n+k})=-\frac{1}{n}\sum _{k⩽n}\frac{1}{k}=-\frac{H_n}{n}$$$$\Rightarrow {\int }_0^1{x}^{n-1}ln(1-x)dx=-\frac{H_n}{n}$$$$\frac{1}{n}\int x^{n-1}ln(1-x)dx=-\frac{H_n}{n^2}$$$$\Rightarrow \mathrm{\Sigma }\frac{{(-1)}^{n-1}}{n}{\int }_0^1{x}^{n-1}ln(1-x)dx=\sum _{n⩾1}\frac{{(-1)}^n{H}_n}{n^2}$$$$\Rightarrow {\int }_0^1\left(\frac{1}{x}\mathrm{\Sigma }\frac{{(-1)}^{n-1}{x}^n}{n}\right)ln(1-x)dx=\sum _{n⩾1}\frac{{(-1)}^n{H}_n}{n^2}$$$$\sum _{n⩾1}\frac{{(-1)}^{n-1}{x}^n}{n}=ln(1+x)\iff$$$${\int }_0^1\frac{ln(1+x)ln(1-x)}{x}dx=\mathrm{\Sigma }\frac{{(-1)}^n{H}_n}{n^2}$$$${\int }_0^1\frac{ln(1+x)ln(1-x)}{x}dxdoneinyouranswer$$$$$$

Commented by mnjuly1970 last updated on 21/Oct/20

$$bravobravo$$$$mrpower$$$$gratefulsir...$$

Commented by mindispower last updated on 21/Oct/20

$$withepleasur$$$$$$

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