Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

Question Number 78620 by mathmax by abdo last updated on 19/Jan/20

$$explicitf\left(x\right)={\int }_0^{+\mathrm{\infty }}ln(1-{xe}^{-t})dtwith\mid x\mid <1$$

Answered by mind is power last updated on 19/Jan/20

$$\ln (1-{\mathrm{xe}}^{-\mathrm{t}})=-\mathrm{\Sigma }\frac{{\left({\mathrm{xe}}^{-\mathrm{t}}\right)}^{\mathrm{k}}}{\mathrm{k}}$$ $${\int }_0^{+\mathrm{\infty }}\ln (1-{\mathrm{xe}}^{-\mathrm{t}})\mathrm{dt}={\int }_0^{+\mathrm{\infty }}-\mathrm{\Sigma }\frac{{\mathrm{x}}^{\mathrm{k}}{\mathrm{e}}^{-\mathrm{kt}}}{\mathrm{k}}\mathrm{dt}$$ $$=\mathrm{\Sigma }[{}_0^{+\mathrm{\infty }}\frac{{\mathrm{e}}^{-\mathrm{kt}}}{{\mathrm{k}}^2}.{\mathrm{x}}^{\mathrm{k}}]=-\sum _{\mathrm{k}⩾1}\frac{{\mathrm{x}}^{\mathrm{k}}}{{\mathrm{k}}^2}={\int }_0^{\mathrm{x}}\frac{\ln (1-\mathrm{t})}{\mathrm{t}}\mathrm{dx}={\mathrm{Li}}_2\left(\mathrm{x}\right)$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com