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Question Number 109884 by 4635 last updated on 26/Aug/20

Commented by mohammad17 last updated on 26/Aug/20

$$$$$$set:y={ln}^{n}x\to x=e^{\frac{y}{n}}\to dx=e^{\frac{y}{n}}\frac{dy}{n}$$$$x=1\to y=0,x=e\to y=1$$$$$$$${\int }_1^e{ln}^{n}xdx={\int }_0^{1}y{e}^{\frac{y}{n}}\frac{dy}{n}$$$$$$$$u=y\to u^{{}^{\prime }}=dy,v^{{}^{\prime }}=e^{\frac{y}{n}}\frac{dy}{n}\to v=e^{\frac{y}{n}}$$$$$$$${ye}^{\frac{y}{n}}{\mid }_0^1-{\int }_0^1{e}^{\frac{y}{n}}dy\Rightarrow {(y-\frac{1}{n})}_0^1{\left(e^{\frac{y}{n}}\right)}_0^1=(1-\frac{1}{n}+\frac{1}{n})(e^{\frac{1}{n}}-1)=e^{\frac{1}{n}}-1$$$$$$$$mohammadtaha\text{iddots}\ast$$

Answered by mathmax by abdo last updated on 26/Aug/20

$${\mathrm{A}}_{\mathrm{n}}={\int }_1^{\mathrm{e}}{\ln }^{\mathrm{n}}\mathrm{xdx}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\mathrm{lnx}=\mathrm{t}\Rightarrow \mathrm{x}={\mathrm{e}}^{\mathrm{t}}\Rightarrow$$$${\mathrm{A}}_{\mathrm{n}}={\int }_0^1{\mathrm{t}}^{\mathrm{n}}{\mathrm{e}}^{\mathrm{t}}\mathrm{dt}{=}_{\mathrm{by}\mathrm{parts}}{\left[\frac{{\mathrm{t}}^{\mathrm{n}+1}}{\mathrm{n}+1}{\mathrm{e}}^{\mathrm{t}}\right]}_0^1-{\int }_0^1\frac{{\mathrm{t}}^{\mathrm{n}+1}}{\mathrm{n}+1}{\mathrm{e}}^{\mathrm{t}}\mathrm{dt}$$$$=\frac{\mathrm{e}}{\mathrm{n}+1}-\frac{1}{\mathrm{n}+1}{\mathrm{A}}_{\mathrm{n}+1}\Rightarrow (\mathrm{n}+1){\mathrm{A}}_{\mathrm{n}}=\mathrm{e}-{\mathrm{A}}_{\mathrm{n}+1}\Rightarrow {\mathrm{A}}_{\mathrm{n}+1}=\mathrm{e}-(\mathrm{n}+1){\mathrm{A}}_{\mathrm{n}}$$$$\Rightarrow {\mathrm{A}}_{\mathrm{n}}=\mathrm{e}-{\mathrm{nA}}_{\mathrm{n}-1}(\mathrm{n}>0)\mathrm{let}{\mathrm{u}}_{\mathrm{n}}=\frac{{\mathrm{A}}_{\mathrm{n}}}{\mathrm{n}!}$$$${\mathrm{u}}_{\mathrm{n}+1}+{\mathrm{u}}_{\mathrm{n}}=\frac{{\mathrm{A}}_{\mathrm{n}+1}}{(\mathrm{n}+1)!}+\frac{{\mathrm{A}}_{\mathrm{n}}}{\mathrm{n}!}=\frac{\mathrm{e}-(\mathrm{n}+1){\mathrm{A}}_{\mathrm{n}}}{(\mathrm{n}+1)!}+\frac{{\mathrm{A}}_{\mathrm{n}}}{\mathrm{n}!}$$$$=\frac{\mathrm{e}}{(\mathrm{n}+1)!}\Rightarrow \sum _{\mathrm{k}=0}^{\mathrm{n}}{(-1)}^{\mathrm{k}}({\mathrm{u}}_{\mathrm{k}}+{\mathrm{u}}_{\mathrm{k}+1})=\mathrm{e}\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{(\mathrm{k}+1)!}\Rightarrow$$$${\mathrm{u}}_0+{\mathrm{u}}_1-{\mathrm{u}}_1-{\mathrm{u}}_2+....+{(-1)}^{\mathrm{n}-1}({\mathrm{u}}_{\mathrm{n}-1}+{\mathrm{u}}_{\mathrm{n}})+{(-1)}^{\mathrm{n}}({\mathrm{u}}_{\mathrm{n}}+{\mathrm{u}}_{\mathrm{n}+1})$$$$=\mathrm{e}\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{(\mathrm{k}+1)!}\Rightarrow {\mathrm{u}}_0+{(-1)}^{\mathrm{n}}{\mathrm{u}}_{\mathrm{n}+1}=\mathrm{e}\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{(\mathrm{k}+1)!}\Rightarrow$$$${(-1)}^{\mathrm{n}}{\mathrm{u}}_{\mathrm{n}+1}=\mathrm{e}\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{(\mathrm{k}+1)!}-{\mathrm{u}}_0\Rightarrow$$$${\mathrm{u}}_{\mathrm{n}+1}=\mathrm{e}\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{n}+\mathrm{k}}}{(\mathrm{k}+1)!}-{(-1)}^{\mathrm{n}}{\mathrm{u}}_0=\mathrm{e}\sum _{\mathrm{k}=1}^{\mathrm{n}+1}\frac{{(-1)}^{\mathrm{n}+\mathrm{k}-1}}{\mathrm{k}!}-{(-1)}^{\mathrm{n}}{\mathrm{u}}_0$$$$\Rightarrow {\mathrm{u}}_{\mathrm{n}}=\mathrm{e}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{n}+\mathrm{k}}}{\mathrm{k}!}+{(-1)}^{\mathrm{n}+1}{\mathrm{u}}_0\Rightarrow$$$${\mathrm{A}}_{\mathrm{n}}=\mathrm{n}!{\mathrm{u}}_{\mathrm{n}}=\mathrm{n}!\{\mathrm{e}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{n}+\mathrm{k}}}{\mathrm{k}!}+{(-1)}^{\mathrm{n}+1}{\mathrm{u}}_0\}({\mathrm{u}}_0={\mathrm{A}}_0)$$

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