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Question Number 100215 by Rio Michael last updated on 25/Jun/20

$$\mathrm{evaluate}\underset{n\to \mathrm{\infty }}{\lim }{\int }_1^e{x}^n\ln xdx$$

Commented by Dwaipayan Shikari last updated on 25/Jun/20

$$\mathrm{\infty }$$

Answered by Ar Brandon last updated on 25/Jun/20

$$\mathrm{Let}\mathcal{I}={\int }_1^{\mathrm{e}}{\mathrm{x}}^{\mathrm{n}}\mathrm{lnxdx}$$$$\mathrm{let}\{\begin{array}{l}\mathrm{u}=\mathrm{lnx}\\ \mathrm{dv}={\mathrm{x}}^{\mathrm{n}}\mathrm{dx}\end{array}\Rightarrow \{\begin{array}{l}\mathrm{du}=\frac{1}{\mathrm{x}}\mathrm{dx}\\ \mathrm{v}=\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}\end{array}$$$$\Rightarrow \mathcal{I}={[\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}\cdot \mathrm{lnx}]}_1^{\mathrm{e}}-\frac{1}{\mathrm{n}+1}{\int }_1^{\mathrm{e}}{\mathrm{x}}^{\mathrm{n}}\mathrm{dx}=\frac{{\mathrm{e}}^{\mathrm{n}+1}}{\mathrm{n}+1}-{\left[\frac{{\mathrm{x}}^{\mathrm{n}+1}}{{(\mathrm{n}+1)}^2}\right]}_1^{\mathrm{e}}$$$$\Rightarrow \mathcal{I}=\frac{{\mathrm{e}}^{\mathrm{n}+1}}{\mathrm{n}+1}-\frac{{\mathrm{e}}^{\mathrm{n}+1}}{{(\mathrm{n}+1)}^2}+\frac{1}{{(\mathrm{n}+1)}^2}$$$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\left(\mathcal{I}\right)=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\{\frac{{\mathrm{e}}^{\mathrm{n}+1}}{1}-\frac{{\mathrm{e}}^{\mathrm{n}+1}}{2(\mathrm{n}+1)}\}=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\{{\mathrm{e}}^{\mathrm{n}+1}-\frac{{\mathrm{e}}^{\mathrm{n}+1}}{2}\}$$$$=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\left\{\frac{{\mathrm{e}}^{\mathrm{n}+1}}{2}\right\}=+\mathrm{\infty }$$

Commented by DGmichael last updated on 25/Jun/20

�� LIATE

Commented by Ar Brandon last updated on 25/Jun/20

Ouais�� c'est ça.

Answered by mathmax by abdo last updated on 25/Jun/20

$${\mathrm{A}}_{\mathrm{n}}={\int }_1^{\mathrm{e}}{\mathrm{x}}^{\mathrm{n}}\mathrm{lnx}\mathrm{dx}\mathrm{changement}\mathrm{lnx}=\mathrm{t}\mathrm{give}{\mathrm{A}}_{\mathrm{n}}={\int }_0^1{\mathrm{e}}^{\mathrm{nt}}\mathrm{t}{\mathrm{e}}^{\mathrm{t}}\mathrm{dt}$$$$={\int }_0^1\mathrm{t}{\mathrm{e}}^{(\mathrm{n}+1)\mathrm{t}}\mathrm{dt}={\left[\frac{1}{\mathrm{n}+1}\mathrm{t}{\mathrm{e}}^{(\mathrm{n}+1)\mathrm{t}}\right]}_0^1-{\int }_0^1\frac{1}{\mathrm{n}+1}{\mathrm{e}}^{(\mathrm{n}+1)\mathrm{t}}\mathrm{dt}$$$$=\frac{{\mathrm{e}}^{\mathrm{n}+1}}{\mathrm{n}+1}-\frac{1}{{(\mathrm{n}+1)}^2}{\left[{\mathrm{e}}^{)\mathrm{n}+1)\mathrm{t}}\right]}_0^1=\frac{{\mathrm{e}}^{\mathrm{n}+1}}{\mathrm{n}+1}-\frac{1}{{(\mathrm{n}+1)}^2}({\mathrm{e}}^{\mathrm{n}+1}-1)$$$$=\frac{{\mathrm{e}}^{\mathrm{n}+1}}{\mathrm{n}+1}-\frac{{\mathrm{e}}^{\mathrm{n}+1}}{{(\mathrm{n}+1)}^2}+\frac{1}{{(\mathrm{n}+1)}^2}=\frac{(\mathrm{n}+1){\mathrm{e}}^{\mathrm{n}+1}-{\mathrm{e}}^{\mathrm{n}+1}}{{(\mathrm{n}+1)}^2}+\frac{1}{{(\mathrm{n}+1)}^2}=\frac{{\mathrm{ne}}^{\mathrm{n}+1}}{{(\mathrm{n}+1)}^2}+\frac{1}{{(\mathrm{n}+1)}^2}\Rightarrow$$$${\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{A}}_{\mathrm{n}}={\lim }_{\mathrm{n}\to +\mathrm{\infty }}\frac{{\mathrm{ne}}^{\mathrm{n}+1}}{{(\mathrm{n}+1)}^2}=+\mathrm{\infty }$$

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