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Question Number 208306 by SANOGO last updated on 10/Jun/24

$$calcul/limn\to +\mathrm{\infty }{\int }_0^{+\mathrm{\infty }}{f}_n\left(x\right)$$$$f_n\left(x\right)=arctan\left(\frac{x}{n}\right)e^{-x}dx$$

Commented by SANOGO last updated on 11/Jun/24

$$thankyou$$

Answered by Berbere last updated on 10/Jun/24

$$f_n\left(x\right)⩽\frac{\pi }{2}{e}^{-x}x\to \frac{\pi }{2}{e}^{-x}integrableover\mathbb{R}+$$$${\underset{n\to \mathrm{\infty }}{\lim tan}}^{-1}\left(\frac{x}{n}\right)e^{-x}=0$$$$\Rightarrow \underset{n\to \mathrm{\infty }}{\lim }{\int }_0^{\mathrm{\infty }}{f}_n\left(x\right)={\int }_0^{\mathrm{\infty }}{\underset{n\to \mathrm{\infty }}{\lim tan}}^{-1}\left(\frac{x}{n}\right)e^{-x}dx={\int }_0^{\mathrm{\infty }}0dx=0$$

Answered by mathzup last updated on 10/Jun/24

$${arctanu)}^{{}^{\prime }}=\frac{1}{1+u^2}=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{u}^{2n}$$$$\Rightarrow arctanu=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n{u}^{2n+1}}{2n+1}+c(c=0)$$$$=u-\frac{u^3}{3}+\frac{u^5}{5}-...\Rightarrow arctanu⩽u\mathrm{\forall }u>0$$$$\Rightarrow \mid f_n\left(x\right)\mid ⩽\frac{x}{n}{e}^{-x}\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\mid f_n\left(x\right)\mid dx⩽\frac{1}{n}{\int }_0^{\mathrm{\infty }}x{e}^{-x}dx$$$$but{\int }_0^{\mathrm{\infty }}{xe}^{-x}dx=\mathrm{\Gamma }\left(2\right)=1!=1\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\mid f_n\left(x\right)\mid dx⩽\frac{1}{n}\to 0(n\to +\mathrm{\infty })\Rightarrow$$$${lim}_{n\to +\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{f}_n\left(x\right)dx=0$$

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