Question Number 205134 by universe last updated on 09/Mar/24
Answered by pi314 last updated on 09/Mar/24
![nx=y ⇔A=lim_(n→∞) ∫_0 ^n ((f((y/n)))/((1+y^2 )))dy=lim_(n→∞) ∫_0 ^n (M/(1+x^2 ));M=sup f_([0,1]) M exist since f is continus over Compact A=lim_(n→∞) ∫_0 ^n ((f((y/n)))/(1+y^2 ))dy≤∫_0 ^∞ (M/(1+x^2 ))=(π/2)M we have uniforme Cv ⇒A=∫_0 ^∞ lim_(n→∞) ((f((y/n)))/(1+y^2 ))dy =∫_0 ^∞ ((f(lim_(n→∞) (y/n)))/(1+y^2 ))dy=∫_0 ^∞ ((f(0))/(1+x^2 ))dx=(π/2)f(0)](https://www.tinkutara.com/question/Q205137.png)
$${nx}={y} \\ $$$$\Leftrightarrow{A}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{{n}} \frac{{f}\left(\frac{{y}}{{n}}\right)}{\left(\mathrm{1}+{y}^{\mathrm{2}} \right)}{dy}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{{n}} \frac{{M}}{\mathrm{1}+{x}^{\mathrm{2}} };{M}={sup}\:{f}_{\left[\mathrm{0},\mathrm{1}\right]} \\ $$$${M}\:{exist}\:{since}\:{f}\:{is}\:{continus}\:{over}\:{Compact} \\ $$$${A}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{{n}} \frac{{f}\left(\frac{{y}}{{n}}\right)}{\mathrm{1}+{y}^{\mathrm{2}} }{dy}\leqslant\int_{\mathrm{0}} ^{\infty} \frac{{M}}{\mathrm{1}+{x}^{\mathrm{2}} }=\frac{\pi}{\mathrm{2}}{M} \\ $$$${we}\:{have}\:{uniforme}\:{Cv}\:\Rightarrow{A}=\int_{\mathrm{0}} ^{\infty} \underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{f}\left(\frac{{y}}{{n}}\right)}{\mathrm{1}+{y}^{\mathrm{2}} }{dy} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{{f}\left(\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{y}}{{n}}\right)}{\mathrm{1}+{y}^{\mathrm{2}} }{dy}=\int_{\mathrm{0}} ^{\infty} \frac{{f}\left(\mathrm{0}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\frac{\pi}{\mathrm{2}}{f}\left(\mathrm{0}\right) \\ $$$$ \\ $$