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Question Number 135252 by mnjuly1970 last updated on 11/Mar/21

$$....advancedcalculus....$$$$firstprovethat::$$$${\mathit{\varphi }}_1={\int }_0^1\frac{ln(1-x)ln(1+x)}{x}dx=\frac{-5}{8}\zeta \left(3\right)$$$$thenconcludethat:$$$${\mathit{\varphi }}_2={\int }_0^1\frac{{ln}^2(1+x)}{x}dx=\frac{\zeta \left(3\right)}{4}$$$$....m.n...$$

Answered by mathmax by abdo last updated on 11/Mar/21

$${\mathrm{\Phi }}_1={\int }_0^1\frac{\ln (1-\mathrm{x})}{\mathrm{x}}\ln (1+\mathrm{x})\mathrm{dx}\mathrm{we}\mathrm{have}{\ln }^{{}^{\prime }}(1+\mathrm{x})=\frac{1}{1+\mathrm{x}}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}$$$$\Rightarrow \ln (1+\mathrm{x})=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}+\mathrm{c}(\mathrm{c}=0)=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}-1}\frac{{\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}}\Rightarrow$$$$\frac{\ln (1+\mathrm{x})}{\mathrm{x}}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}-1}\frac{{\mathrm{x}}^{\mathrm{n}-1}}{\mathrm{n}}\Rightarrow {\mathrm{\Phi }}_1=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}-1}}{\mathrm{n}}{\int }_0^1{\mathrm{x}}^{\mathrm{n}-1}\ln (1-\mathrm{x})\mathrm{dx}$$$${\mathrm{A}}_{\mathrm{n}}={\int }_0^1{\mathrm{x}}^{\mathrm{n}-1}\ln (1-\mathrm{x})\mathrm{dx}={\left[\frac{{\mathrm{x}}^{\mathrm{n}}-1}{\mathrm{n}}\ln (1-\mathrm{x})\right]}_0^1+{\int }_0^1\frac{{\mathrm{x}}^{\mathrm{n}}-1}{\mathrm{n}(1-\mathrm{x})}\mathrm{dx}$$$$=-\frac{1}{\mathrm{n}}{\int }_0^1\frac{{\mathrm{x}}^{\mathrm{n}}-1}{\mathrm{x}-1}\mathrm{dx}=-\frac{1}{\mathrm{n}}{\int }_0^1(1+\mathrm{x}+{\mathrm{x}}^2+....+{\mathrm{x}}^{\mathrm{n}-1})\mathrm{dx}$$$$=-\frac{1}{\mathrm{n}}{[\mathrm{x}+\frac{{\mathrm{x}}^2}{2}+....+\frac{{\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}}]}_0^1=-\frac{1}{\mathrm{n}}(1+\frac{1}{2}+....+\frac{1}{\mathrm{n}})=-\frac{{\mathrm{H}}_{\mathrm{n}}}{\mathrm{n}}\Rightarrow$$$${\mathrm{\Phi }}_1=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{{\mathrm{n}}^2}{\mathrm{H}}_{\mathrm{n}}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}\frac{{\mathrm{H}}_{\mathrm{n}}}{{\mathrm{n}}^2}$$$$\mathrm{rest}\mathrm{to}\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{this}\mathrm{serie}....\mathrm{be}\mathrm{continued}...$$

Commented by mnjuly1970 last updated on 12/Mar/21

$$thankyousomuch$$$$\mathrm{\Sigma }\frac{{(-1)}^n{H}_n}{n^2}=\frac{-5}{8}\zeta \left(3\right)$$

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