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Question Number 35675 by abdo imad last updated on 21/May/18

$$calculate{\int }_1^3\frac{x}{e^x-1}dx..$$

Commented by prof Abdo imad last updated on 25/May/18

$$I={\int }_1^3\frac{x{e}^{-x}}{1-e^{-x}}dx={\int }_1^3\left(\sum _{n=0}^{\mathrm{\infty }}{e}^{-nx}\right)x{e}^{-x}dx$$$$=\sum _{n=0}^{\mathrm{\infty }}{\int }_1^{3}x{e}^{-(n+1)x}dx=\sum _{n=0}^{\mathrm{\infty }}{A}_{n}with$$$$A_n={\int }_1^{3}x.e^{-(n+1)x}dx.changement(n+1)x=t$$$$give{A}_n={\int }_{n+1}^{3(n+1)}\frac{t}{n+1}{e}^{-t}\frac{dt}{n+1}$$$$=\frac{1}{{(n+1)}^2}{\int }_{n+1}^{3(n+1)}{t}^{}{e}^{-t}dtbyparts$$$${\int }_{n+1}^{3(n+1)}t{e}^{-t}dt={[-t{e}^{-t}]}_{n+1}^{3(n+1)}+{\int }_{n+1}^{3(n+1)}{e}^{-t}dt$$$$=(n+1)e^{-(n+1)}-3(n+1)e^{-3(n+1)}$$$$+{[-e^{-t}]}_{n+1}^{3(n+1)}$$$$=(n+1)e^{-(n+1)}-3(n+1)e^{-3(n+1)}+e^{-(n+1)}-e^{-3(n+1)}$$$$A_n=\frac{e^{-(n+1)}}{n+1}-3\frac{e^{-3(n+1)}}{n+1}+\frac{e^{-(n+1)}}{{(n+1)}^2}-\frac{e^{-3(n+1)}}{{(n+1)}^2}$$$$\sum _{n=0}^{\mathrm{\infty }}{A}_n=\sum _{n=0}^{\mathrm{\infty }}\frac{e^{-(n+1)}}{n+1}-3\sum _{n=0}^{\mathrm{\infty }}\frac{e^{-3(n+1)}}{n+1}$$$$+\sum _{n=0}^{\mathrm{\infty }}\frac{e^{-(n+1)}}{{(n+1)}^2}-\sum _{n=0}^{\mathrm{\infty }}\frac{e^{-3(n+1)}}{{(n+1)}^2}$$$$=\sum _{n=1}^{\mathrm{\infty }}\frac{e^{-n}}{n}-3\sum _{n=1}^{\mathrm{\infty }}\frac{e^{-3n}}{n}+\sum _{n=1}^{\mathrm{\infty }}\frac{e^{-n}}{n^2}$$$$-\sum _{n=1}^{\mathrm{\infty }}\frac{e^{-3n}}{n^2}....becontinued....$$

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