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Question Number 118928 by mnjuly1970 last updated on 20/Oct/20

Commented by prakash jain last updated on 20/Oct/20

$$\mathrm{Suggestion}:\mathrm{attach}\mathrm{image}\mathrm{to}$$$$\mathrm{original}\mathrm{question}\mathrm{as}\mathrm{answer}.$$

Commented by mnjuly1970 last updated on 20/Oct/20

$$thankyouforyour$$$$suggestion..$$

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

$$\mathrm{another}\mathrm{way}\mathrm{\Phi }={\int }_0^1\frac{\ln (1+\mathrm{x})\ln (1-\mathrm{x})}{\mathrm{x}}\mathrm{dx}$$$$\mathrm{we}\mathrm{have}{\ln }^{{}^{\prime }}(1+\mathrm{x})=\frac{1}{1+\mathrm{x}}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}\Rightarrow \ln (1+\mathrm{x})=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}$$$$=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}-1}{\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}}\mathrm{and}\ln (1-\mathrm{x})=-\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}{(-\mathrm{x})}^{\mathrm{n}}}{\mathrm{n}}$$$$=-\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}}\Rightarrow \frac{\ln (1+\mathrm{x})\ln (1-\mathrm{x})}{\mathrm{x}}=-\left(\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}-1}{\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}}\right)\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}}$$$$=\left(\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}\right)\left(\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}}\right)=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{c}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}$$$${\mathrm{c}}_{\mathrm{n}}=\sum _{\mathrm{i}+\mathrm{j}=\mathrm{n}}{\mathrm{a}}_{\mathrm{i}}{\mathrm{b}}_{\mathrm{j}}=\sum _{\mathrm{i}+\mathrm{j}=\mathrm{n}}\frac{{(-1)}^{\mathrm{i}}}{\mathrm{i}}.\frac{1}{\mathrm{j}}=\sum _{\mathrm{i}=1}^{\mathrm{n}-1}\frac{{(-1)}^{\mathrm{i}}}{\mathrm{i}(\mathrm{n}-\mathrm{i})}\Rightarrow$$$$\mathrm{\Phi }={\int }_0^1\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\left(\sum _{\mathrm{i}=1}^{\mathrm{n}-1}\frac{{(-1)}^{\mathrm{i}}}{\mathrm{i}(\mathrm{n}-\mathrm{i})}\right){\mathrm{x}}^{\mathrm{n}-1}$$$$=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\left(\sum _{\mathrm{i}=1}^{\mathrm{n}-1}\frac{{(-1)}^{\mathrm{i}}}{\mathrm{i}(\mathrm{n}-\mathrm{i})}\right).\frac{1}{\mathrm{n}}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\left(\frac{1}{\mathrm{n}}\sum _{\mathrm{i}=1}^{\mathrm{n}-1}\frac{{(-1)}^{\mathrm{i}}}{\mathrm{i}(\mathrm{n}-\mathrm{i})}\right)$$$$....\mathrm{rest}\mathrm{to}\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{this}\mathrm{sum}....\mathrm{be}\mathrm{continued}....$$

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