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Question Number 207648 by SANOGO last updated on 22/May/24

$$f_n\left(x\right)=\frac{n}{1+x^2}sin\left(\frac{x}{n}\right)/x\in [0,1],n⩾1$$$$calculerlimn\to \mathrm{\infty }{\int }_0^1{f}_n\left(x\right)dx$$

Commented by mr W last updated on 22/May/24

$$why{don}^{\prime }tyouuse+\mathrm{\infty }insteadof$$$$+oo?$$$$why{don}^{\prime }tyouuse\underset{n\to +\mathrm{\infty }}{\lim }insteadof$$$$limn\to +oo?$$

Answered by Berbere last updated on 22/May/24

$$lemmasin\left(x\right)⩽x;\mathrm{\forall }x⩾0$$$$sin\left(x\right)={\int }_0^{x}cos\left(t\right)dt⩽{\int }_0^{x}1dt=x$$$$f_n\left(x\right)⩽\frac{n}{1+x^2}.\frac{x}{n}=\frac{x}{1+x^2}=G\left(x\right)$$$$G\left(x\right)integrableover[0,1]$$$$\underset{n\to \mathrm{\infty }}{\lim }{\int }_0^1\frac{n}{1+x^2}.sin\left(\frac{x}{n}\right)dx={\int }_0^1\underset{n\to \mathrm{\infty }}{\lim }.\frac{nsin\left(\frac{x}{n}\right)}{1+x^2}dx$$$$={\int }_0^1\frac{sin\left(\frac{x}{n}\right).x}{\frac{x}{n}(1+x^2)}dx;\underset{x\to 0}{\lim }\frac{sin\left(a\right)}{a}=1$$$$={\int }_0^1\frac{x}{1+x^2}dx={\left[\frac{ln(1+x^2)}{2}\right]}_0^1=\frac{ln\left(2\right)}{2}=ln\left(\sqrt{2}\right)$$

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