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Question Number 113756 by Dwaipayan Shikari last updated on 15/Sep/20

$${\int }_0^1\frac{log(x+1)}{x}dx$$

Commented by Dwaipayan Shikari last updated on 15/Sep/20

$${\int }_0^1\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n{x}^{n-1}}{n}dx$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{(-1)}^n{\int }_0^1\frac{x^{n-1}}{n}dx$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{(-1)}^n\frac{1}{n^2}=\frac{{\pi }^2}{12}$$$$$$

Answered by mathdave last updated on 15/Sep/20

$$solution$$$$letx=-x$$$${\int }_0^{-1}\frac{\ln (1-x)}{x}dx=-{Li}_2(-1)=\frac{{\pi }^2}{12}$$

Answered by mindispower last updated on 15/Sep/20

$${\int }_0^1\frac{1}{x}\sum _{k⩾0}{(-1)}^k\frac{x^{k+1}}{k+1}dx$$$$=\sum _{k⩾0}\frac{{(-1)}^k}{k+1}{\int }_0^1{x}^{k}dx=\mathrm{\Sigma }\frac{{(-1)}^k}{{(k+1)}^2}$$$$=\sum _{k⩾0}(\frac{1}{{(2k+1)}^2}-\frac{1}{{(2k+2)}^2})$$$$=\sum _{k⩾0}\frac{1}{{(2k+1)}^2}-\frac{1}{4}\mathrm{\Sigma }\frac{1}{{(k+1)}^2}$$$$(\zeta \left(2\right)-\frac{\zeta \left(2\right)}{4}-\frac{\zeta \left(2\right)}{4})=\frac{\zeta \left(2\right)}{2}=\frac{{\pi }^2}{12}$$$$$$

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