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Question Number 205134 by universe last updated on 09/Mar/24

Answered by pi314 last updated on 09/Mar/24

$$nx=y$$$$\iff A=\underset{n\to \mathrm{\infty }}{\lim }{\int }_0^n\frac{f\left(\frac{y}{n}\right)}{(1+y^2)}dy=\underset{n\to \mathrm{\infty }}{\lim }{\int }_0^n\frac{M}{1+x^2};M=sup{f}_{[0,1]}$$$$MexistsincefiscontinusoverCompact$$$$A=\underset{n\to \mathrm{\infty }}{\lim }{\int }_0^n\frac{f\left(\frac{y}{n}\right)}{1+y^2}dy⩽{\int }_0^{\mathrm{\infty }}\frac{M}{1+x^2}=\frac{\pi }{2}M$$$$wehaveuniformeCv\Rightarrow A={\int }_0^{\mathrm{\infty }}\underset{n\to \mathrm{\infty }}{\lim }\frac{f\left(\frac{y}{n}\right)}{1+y^2}dy$$$$={\int }_0^{\mathrm{\infty }}\frac{f\left(\underset{n\to \mathrm{\infty }}{\lim }\frac{y}{n}\right)}{1+y^2}dy={\int }_0^{\mathrm{\infty }}\frac{f\left(0\right)}{1+x^2}dx=\frac{\pi }{2}f\left(0\right)$$$$$$

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