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Question Number 146202 by lapache last updated on 11/Jul/21

$$\int \frac{lnx}{x-1}dx=....???$$$$Laprimitive$$

Answered by Ar Brandon last updated on 12/Jul/21

$$\mathrm{I}={\int }_0^1\frac{\mathrm{lnx}}{\mathrm{x}-1}\mathrm{dx}$$$$=-{\int }_0^1\frac{\mathrm{lnx}}{1-\mathrm{x}}\mathrm{dx}$$$$={\psi }^{\prime }\left(1\right)=\zeta \left(2\right)=\frac{{\pi }^2}{6}$$

Answered by puissant last updated on 12/Jul/21

$$\mathrm{posons}\mathrm{u}=\mathrm{x}-1\Rightarrow \mathrm{du}=\mathrm{dx}$$$$\mathrm{I}=\int \frac{\ln (1+\mathrm{u})}{\mathrm{u}}\mathrm{du}$$$$\mathrm{or}\frac{1}{1+\mathrm{u}}=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}}{\mathrm{u}}^{\mathrm{n}}\Rightarrow \ln (1+\mathrm{u})=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}}\int {\mathrm{u}}^{\mathrm{n}}\mathrm{du}$$$$=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}}\frac{{\mathrm{u}}^{\mathrm{n}+1}}{\mathrm{n}+1}+\mathrm{k}$$$$\mathrm{u}=0\Rightarrow \ln (1+\mathrm{u})=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}}\frac{{\mathrm{u}}^{\mathrm{n}+1}}{\mathrm{n}+1}$$$$\int \frac{\ln (1+\mathrm{u})}{\mathrm{u}}=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}}\frac{1}{\mathrm{n}+1}\int {\mathrm{u}}^{\mathrm{n}}\mathrm{du}$$$$=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}}\frac{{\mathrm{u}}^{\mathrm{n}+1}}{{(\mathrm{n}+1)}^2}+\mathrm{k}=\mathrm{I}$$$$\mathrm{or}\mathrm{u}=\mathrm{x}-1$$$${\mathrm{d}}^{\prime }\mathrm{ou}\mathrm{I}=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}}\frac{{(\mathrm{x}-1)}^{\mathrm{n}+1}}{{(\mathrm{n}+1)}^2}+\mathrm{k}....$$

Answered by Olaf_Thorendsen last updated on 12/Jul/21

$$\mathrm{F}\left(x\right)=\int \frac{\ln x}{x-1}dx$$$$\mathrm{Soit}{\mathrm{Li}}_2\mathrm{appellee}\mathrm{fonction}\mathrm{de}\mathrm{Spence}\mathrm{ou}$$$$\mathrm{fonction}\mathrm{dilogarithme}:$$$${\mathrm{Li}}_2\left(z\right)=-{\int }_0^z\frac{\ln (1-u)}{u}du$$$${\mathrm{Li}}_2\left(z\right)=-{\int }_1^{1-z}\frac{\ln t}{t-1}dt=-\mathrm{F}(1-z)+\mathrm{F}\left(1\right)$$$$\Rightarrow \mathrm{F}\left(x\right)=-{\mathrm{Li}}_2(1-x)+\mathrm{C}$$

Commented by puissant last updated on 12/Jul/21

$$\mathrm{z}\in \mathbb{C}$$

Commented by puissant last updated on 12/Jul/21

$${\mathrm{Li}}_2\left(\mathrm{z}\right)=\underset{\mathrm{k}=1}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{z}}^{\mathrm{k}}}{{\mathrm{k}}^2}..$$

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