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Question Number 33349 by caravan msup abdo. last updated on 14/Apr/18

$$provethat$$$${\int }_0^{\mathrm{\infty }}x(x-ln(e^x-1))dx=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^3}$$

Commented by math khazana by abdo last updated on 18/Apr/18

$$letputI={\int }_0^{\mathrm{\infty }}x(x-ln(e^x-1))dx$$$$I={\int }_0^{\mathrm{\infty }}x(x-ln\left(e^x(1-e^{-x})\right)dx$$$$=-{\int }_0^{\mathrm{\infty }}xln(1-e^{-x})dxbutwehave$$$$\frac{1}{1-u}=\sum _{n=0}^{\mathrm{\infty }}{u}^n\Rightarrow -ln(1-u)=\sum _{n=0}^{\mathrm{\infty }}\frac{u^{n+1}}{n+1}$$$$=\sum _{n=1}^{\mathrm{\infty }}\frac{u^n}{n}\Rightarrow -ln(1-e^{-x})=\sum _{n=1}^{\mathrm{\infty }}\frac{e^{-nx}}{n}$$$$I={\int }_0^{\mathrm{\infty }}\left(\sum _{n=1}^{\mathrm{\infty }}\frac{e^{-nx}}{n}\right)xdx$$$$=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}{\int }_0^{\mathrm{\infty }}x{e}^{-nx}dx$$$${\int }_0^{\mathrm{\infty }}x{e}^{-nx}dx{=}_{nx=t}{\int }_0^{\mathrm{\infty }}\frac{t}{n}{e}^{-t}\frac{dt}{n}$$$$=\frac{1}{n^2}{\int }_0^{\mathrm{\infty }}t{e}^{-t}dtandbyparts$$$${\int }_0^{\mathrm{\infty }}t{e}^{-t}dt={[-t{e}^{-t}]}_0^{+\mathrm{\infty }}+{\int }_0^{\mathrm{\infty }}{e}^{-t}dt$$$$={[-e^{-t}]}_0^{+\mathrm{\infty }}=1\Rightarrow I=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^3}.$$

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