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Question Number 100514 by mathmax by abdo last updated on 27/Jun/20

$${\mathrm{calculatelim}}_{\mathrm{n}\to +\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{(1-\frac{\mathrm{x}}{\mathrm{n}})}^{\mathrm{n}}\ln (1+2\mathrm{x})\mathrm{dx}$$

Answered by mathmax by abdo last updated on 27/Jun/20

$${\mathrm{A}}_{\mathrm{n}}={\int }_0^{\mathrm{\infty }}{(1-\frac{\mathrm{x}}{\mathrm{n}})}^{\mathrm{n}}\ln (2\mathrm{x}+1)\mathrm{dx}={\int }_{\mathrm{R}}{(1-\frac{\mathrm{x}}{\mathrm{n}})}^{\mathrm{n}}\ln (2\mathrm{x}+1){\chi }_{[0,+\mathrm{\infty }[}\left(\mathrm{x}\right)\mathrm{dx}$$$$={\int }_{\mathrm{R}}{\mathrm{f}}_{\mathrm{n}}\left(\mathrm{x}\right)\mathrm{dx}\mathrm{with}{\mathrm{f}}_{\mathrm{n}}\left(\mathrm{x}\right)={(1-\frac{\mathrm{x}}{\mathrm{n}})}^{\mathrm{n}}\ln (2\mathrm{x}+1)$$$${\mathrm{f}}_{\mathrm{n}}{\to }^{\mathrm{cs}}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^{-\mathrm{x}}\ln (2\mathrm{x}+1)\mathrm{and}\mid {\mathrm{f}}_{\mathrm{n}}\mid ⩽\mathrm{f}\left(\mathrm{x}\right)\mathrm{integrable}\mathrm{on}[0,+\mathrm{\infty }[$$$$\mathrm{tbeorem}\mathrm{of}\mathrm{convegence}\mathrm{dominee}\mathrm{give}{\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{A}}_{\mathrm{n}}={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{x}}\ln (2\mathrm{x}+1)\mathrm{dx}=\mathrm{L}$$$$\mathrm{by}\mathrm{parts}\mathrm{L}={[-{\mathrm{e}}^{-\mathrm{x}}\ln (2\mathrm{x}+1)]}_0^{+\mathrm{\infty }}+{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{x}}\frac{2}{2\mathrm{x}+1}\mathrm{dx}$$$$=2{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\mathrm{x}}}{2\mathrm{x}+1}\mathrm{dx}{=}_{2\mathrm{x}+1=\mathrm{t}}2{\int }_1^{+\mathrm{\infty }}\frac{{\mathrm{e}}^{-\left(\frac{\mathrm{t}-1}{2}\right)}}{\mathrm{t}}\times \frac{\mathrm{dt}}{2}=\sqrt{\mathrm{e}}{\int }_1^{+\mathrm{\infty }}\frac{{\mathrm{e}}^{-\frac{\mathrm{t}}{2}}}{\mathrm{t}}\mathrm{dt}...\mathrm{be}\mathrm{continued}...$$$$$$

Answered by maths mind last updated on 28/Jun/20

$$e^{-x}⩾1-x\iff ,x>0$$$$\frac{e^{-x}-1}{x}⩾-1...truebymeanvalutth$$$$f\left(t\right)=e^{-t}over[0,x]x>0\mathrm{\exists }c\in [0,x]$$$$\Rightarrow \frac{e^{-x}-1}{x}=-e^{-c}⩾-1$$$$\Rightarrow e^{-\frac{x}{n}}⩾1-\frac{x}{n}\Rightarrow {(1-\frac{x}{n})}^{n}ln(1+2x)⩽e^{-x}ln(1+2x)$$$${\int }_0^{+\mathrm{\infty }}{e}^{-x}ln(1+2x)dx<\mathrm{\infty }true$$$$\Rightarrow convergencetheorem$$$$\Rightarrow \underset{n\to \mathrm{\infty }}{\lim }{\int }_0^{\mathrm{\infty }}{(1-\frac{x}{n})}^{n}ln(1+2x)dx={\int }_0^{+\mathrm{\infty }}\underset{n\to \mathrm{\infty }}{\lim }{(1-\frac{x}{n})}^{n}ln(1+2x)dx$$$$={\int }_0^{+\mathrm{\infty }}{e}^{-x}ln(1+2x)dx$$$$[-e^{-x}ln(1+2x)]+{\int }_0^{+\mathrm{\infty }}\frac{e^{-x}}{1+2x}dx$$$$={\int }_0^{+\mathrm{\infty }}\frac{e^{-x}}{1+2x}dx={\int }_1^{+\mathrm{\infty }}\frac{e^{-\left(\frac{u-1}{2}\right)}}{2u}du$$$$=\frac{\sqrt{e}}{2}{\int }_1^{+\mathrm{\infty }}\frac{e^{-\frac{u}{2}}}{2u}du=\frac{\sqrt{e}}{2}{\int }_{\frac{1}{2}}^{+\mathrm{\infty }}\frac{e^{-t}}{2t}$$$$=\frac{\sqrt{e}}{4}{\int }_{\frac{1}{2}}^{+\mathrm{\infty }}\frac{e^{-t}}{t}dt=\frac{\sqrt{e}}{4}E\left(\frac{1}{2}\right)$$$$$$

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