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Question Number 140531 by Willson last updated on 09/May/21

$$\mathrm{Prove}\mathrm{that}{\underset{0}{\int }}^{\mathrm{x}}\frac{\mathrm{t}}{{\mathrm{e}}^{\mathrm{t}}-1}\mathrm{dt}=\underset{\mathrm{n}=1}{\sum ^{+\mathrm{\infty }}}\frac{{(1-{\mathrm{e}}^{-\mathrm{x}})}^{\mathrm{n}}}{{\mathrm{n}}^2}$$

Answered by mathmax by abdo last updated on 09/May/21

$${\int }_0^{\mathrm{x}}\frac{\mathrm{t}}{{\mathrm{e}}^{\mathrm{t}}-1}\mathrm{dt}={\int }_0^{\mathrm{x}}\frac{{\mathrm{te}}^{-\mathrm{t}}}{1-{\mathrm{e}}^{-\mathrm{t}}}\mathrm{dt}={\int }_0^{\mathrm{x}}{\mathrm{te}}^{-\mathrm{t}}\sum _{\mathrm{n}.0}^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{nt}}\mathrm{dt}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\int }_0^{\mathrm{x}}\mathrm{t}{\mathrm{e}}^{-(\mathrm{n}+1)\mathrm{t}}\mathrm{dt}{=}_{(\mathrm{n}+1)\mathrm{t}=\mathrm{y}}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\int }_0^{(\mathrm{n}+1)\mathrm{x}}\frac{\mathrm{y}}{\mathrm{n}+1}{\mathrm{e}}^{-\mathrm{y}}\frac{\mathrm{dy}}{\mathrm{n}+1}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{{(\mathrm{n}+1)}^2}{\int }_0^{(\mathrm{n}+1)\mathrm{x}}{\mathrm{ye}}^{-\mathrm{y}}\mathrm{dy}\mathrm{and}$$$${\int }_0^{(\mathrm{n}+1)\mathrm{x}}\mathrm{y}{\mathrm{e}}^{-\mathrm{y}}\mathrm{dy}={[-\mathrm{y}{\mathrm{e}}^{-\mathrm{y}}]}_0^{(\mathrm{n}+1)\mathrm{x}}+{\int }_0^{(\mathrm{n}+1)\mathrm{x}}{\mathrm{e}}^{-\mathrm{y}}\mathrm{dy}$$$$=-(\mathrm{n}+1)\mathrm{x}{\mathrm{e}}^{-(\mathrm{n}+1)\mathrm{x}}+{[-{\mathrm{e}}^{-\mathrm{y}}]}_0^{(\mathrm{n}+1)\mathrm{x}}$$$$=-(\mathrm{n}+1){\mathrm{xe}}^{-(\mathrm{n}+1)\mathrm{x}}+1-{\mathrm{e}}^{-(\mathrm{n}+1)\mathrm{x}}$$$$=-((\mathrm{n}+1)\mathrm{x}+1){\mathrm{e}}^{-(\mathrm{n}+1)\mathrm{x}}+1\Rightarrow$$$${\int }_0^{\mathrm{x}}\frac{\mathrm{t}}{{\mathrm{e}}^{\mathrm{t}}-1}\mathrm{dt}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{{(\mathrm{n}+1)}^2}(1-\left((\mathrm{n}+1)\mathrm{x}\right){\mathrm{e}}^{-(\mathrm{n}+1)\mathrm{x}}+1)$$$$=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{1}{{\mathrm{n}}^2}(1-\mathrm{nx}){\mathrm{e}}^{-\mathrm{nx}}+\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{1}{{\mathrm{n}}^2}$$$$=\frac{{\pi }^2}{6}+\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{(1-\mathrm{nx}){\mathrm{e}}^{-\mathrm{nx}}}{{\mathrm{n}}^2}....$$

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