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Question Number 72024 by mathmax by abdo last updated on 23/Oct/19

$$calculate{lim}_{n\to +\mathrm{\infty }}{\int }_0^n{(1-\frac{x}{n})}^{n}ln(1+x)dx$$

Commented by mathmax by abdo last updated on 31/Oct/19

$$A_n={\int }_0^n{(1-\frac{x}{n})}^{n}ln(1+x)dx\Rightarrow A_n={\int }_R{(1-\frac{x}{n})}^{n}ln(1+x){\chi }_{[0,n[}\left(x\right)dx$$$$let{f}_n\left(x\right)={(1-\frac{x}{n})}^{n}ln(1+x){\chi }_{[0,n[}\left(x\right)dxwehave$$$$f_n{\to }^{cs}f\left(x\right)=e^{-x}ln(1+x)on[0,+\mathrm{\infty }[and\mid f_n\left(x\right)\mid <f\left(x\right)$$$$theoremofconvergencedomineegive$$$${lim}_{n\to +\mathrm{\infty }}{A}_n={\int }_R{lim}_{n\to +\mathrm{\infty }}{f}_n\left(x\right)dx={\int }_0^{\mathrm{\infty }}{e}^{-x}ln(1+x)dx$$$${=}_{1+x=t}{\int }_1^{+\mathrm{\infty }}{e}^{-(t-1)}ln\left(t\right)dt=e{\int }_1^{+\mathrm{\infty }}{e}^{-t}ln\left(t\right)dt$$$$=e({\int }_1^0{e}^{-t}ln\left(t\right)dt+{\int }_0^{+\mathrm{\infty }}{e}^{-t}ln\left(t\right)dt)$$$$=-e{\int }_0^1{e}^{-t}ln\left(t\right)dt+e{\int }_0^{\mathrm{\infty }}{e}^{-t}ln\left(t\right)dt$$$$=-e\gamma -e{\int }_0^1{e}^{-t}ln\left(t\right)dt.....becontinued...$$$$$$

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