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Question Number 205775 by SANOGO last updated on 30/Mar/24

$$calcu/limit/n\to +oo$$$${\int }_0^{+oo}arctan\left(\frac{x}{n}\right)e^{-x}dx$$

Answered by Berbere last updated on 30/Mar/24

$$U_n\left(x\right)={\tan }^{-1}\left(\frac{x}{n}\right)e^{-x}⩽\frac{\pi }{2}{e}^{-x};\mathrm{\forall }x\in {\mathbb{R}}_{+}$$$$x\to \frac{\pi }{2}{e}^{-x};isRiemannintegrablover[0,\mathrm{\infty }[$$$${\int }_0^{\mathrm{\infty }}{\tan }^{-1}\left(\frac{x}{n}\right)e^{-x}dx⩽{\int }_0^{\mathrm{\infty }}\frac{\pi }{2}{e}^{-x}=\frac{\pi }{2}$$$$domiatecv[Theorem\Rightarrow$$$$\underset{n\to \mathrm{\infty }}{\lim }{\int }_0^{\mathrm{\infty }}{\tan }^{-1}\left(\frac{x}{n}\right)e^{-x}dx={\int }_0^{\mathrm{\infty }}{\underset{n\to \mathrm{\infty }}{\lim tan}}^{-1}\left(\frac{x}{n}\right)e^{-x}dx=0$$$$orusing0⩽{\tan }^{-1}\left(x\right)⩽x$$$${\tan }^{-1}\left(x\right)={\int }_0^x\frac{dt}{1+t^2}⩽{\int }_0^{x}dt=x$$$$\Rightarrow 0⩽{\tan }^{-1}\left(\frac{x}{n}\right)e^{-x}⩽\frac{x}{n}{e}^{-x}$$$$0⩽{\int }_0^{\mathrm{\infty }}{\tan }^{-1}\left(\frac{x}{n}\right)e^{-x}⩽\frac{1}{n}{\int }_0^{\mathrm{\infty }}{xe}^{-x}dx=\frac{1}{n}\mathrm{\Gamma }{\left(2\right)}_{\to 0}$$$$\underset{n\to \mathrm{\infty }}{\lim }{\int }_0^{\mathrm{\infty }}{\tan }^{-1}\left(\frac{x}{n}\right)e^{-x}dx=0$$$$$$

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