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Question Number 160979 by amin96 last updated on 10/Dec/21

$$\mathrm{\Omega }={\int }_0^1{\mathbf{x}}^{\mathbf{n}-1}\mathbf{ln}(1-\mathbf{x})\mathbf{dx}=???$$$$\mathbf{n}⩾1$$

Answered by qaz last updated on 10/Dec/21

$$\mathrm{\Omega }=-\underset{\mathrm{k}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{\mathrm{k}}{\int }_0^1{\mathrm{x}}^{\mathrm{n}-1+\mathrm{k}}\mathrm{dx}$$$$=-\underset{\mathrm{k}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{\mathrm{k}(\mathrm{n}+\mathrm{k})}$$$$=-\frac{1}{\mathrm{n}}\underset{\mathrm{k}=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{\mathrm{k}}-\frac{1}{\mathrm{n}+\mathrm{k}})$$$$=-\frac{{\mathrm{H}}_{\mathrm{n}}}{\mathrm{n}}$$

Commented by amin96 last updated on 10/Dec/21

$$bravooooo$$

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