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Question Number 104644 by Rohit@Thakur last updated on 22/Jul/20

$${\int }_0^1\frac{log(1-x+x^2-x^3+x^4)dx}{x}=-\frac{{\pi }^2}{15}$$

Answered by mathmax by abdo last updated on 23/Jul/20

$$\mathrm{I}={\int }_0^1\frac{\ln (1-\mathrm{x}+{\mathrm{x}}^2-{\mathrm{x}}^3+{\mathrm{x}}^4)}{\mathrm{x}}\mathrm{dx}\Rightarrow \mathrm{I}={\int }_0^1\frac{\ln \left(\frac{1-{(-\mathrm{x})}^5}{1-(-\mathrm{x})}\right)}{\mathrm{x}}\mathrm{dx}$$$$={\int }_0^1\frac{\ln (1+{\mathrm{x}}^5)-\ln (1+\mathrm{x})}{\mathrm{x}}\mathrm{dx}={\int }_0^1\frac{\ln (1+{\mathrm{x}}^5)}{\mathrm{x}}\mathrm{dx}-{\int }_0^1\frac{\ln (1+\mathrm{x})}{\mathrm{x}}\mathrm{dx}$$$$\mathrm{we}\mathrm{have}{\ln }^{{}^{\prime }}(1+\mathrm{u})=\frac{1}{1+\mathrm{u}}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{u}}^{\mathrm{n}}\mathrm{for}\mid \mathrm{u}\mid <1\Rightarrow$$$$\ln (1+\mathrm{u})=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}{\mathrm{u}}^{\mathrm{n}+1}}{\mathrm{n}+1}+\mathrm{c}(\mathrm{c}=0)=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}-1}{\mathrm{u}}^{\mathrm{n}}}{\mathrm{n}}\Rightarrow$$$$\frac{\ln (1+\mathrm{x})}{\mathrm{x}}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}-1}}{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}-1}\Rightarrow {\int }_0^1\frac{\ln (1+\mathrm{x})}{\mathrm{x}}\mathrm{dx}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}-1}}{{\mathrm{n}}^2}$$$$=-\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{{\mathrm{n}}^2}=-\delta \left(2\right)=-(2^{1-2}-1)\xi \left(2\right)=-(-\frac{1}{2})\times \frac{{\pi }^2}{6}=\frac{{\pi }^2}{12}$$$$\ln (1+{\mathrm{x}}^5)=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}-1}}{\mathrm{n}}{\mathrm{x}}^{5\mathrm{n}}\Rightarrow \frac{\ln (1+{\mathrm{x}}^5)}{\mathrm{x}}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}-1}}{\mathrm{n}}{\mathrm{x}}^{5\mathrm{n}-1}\Rightarrow$$$${\int }_0^1\frac{\ln (1+{\mathrm{x}}^5)}{\mathrm{x}}\mathrm{dx}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}-1}}{\mathrm{n}\left(5\mathrm{n}\right)}=-\frac{1}{5}\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{{\mathrm{n}}^2}=\frac{1}{5}\times \frac{{\pi }^2}{12}=\frac{{\pi }^2}{60}$$$$\Rightarrow \mathrm{I}=\frac{{\pi }^2}{60}-\frac{{\pi }^2}{12}=(\frac{1}{60}-\frac{1}{12}){\pi }^2=\frac{1-5}{60}\times {\pi }^2=-\frac{4{\pi }^2}{60}=-\frac{{\pi }^2}{15}$$$$\mathrm{the}\mathrm{result}\mathrm{is}\mathrm{proved}.$$$$$$

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