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Question Number 61976 by maxmathsup by imad last updated on 13/Jun/19
let A_n = ∫_(1/n) ^n ((arctan(x^2 +y^2 ))/(x^2 +y^2 )) dxdy 1) calculate A_n 2) find lim_(n→∞) A_n
$${let}\:{A}_{{n}} =\:\int_{\frac{\mathrm{1}}{{n}}} ^{{n}} \:\:\:\frac{{arctan}\left({x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:{dxdy}\:\: \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{A}_{{n}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{lim}_{{n}\rightarrow\infty} \:{A}_{{n}} \\ $$
Commented by maxmathsup by imad last updated on 13/Jun/19
A_n =∫∫_(](1/n),n[^2 ) ((arctan(x^2 +y^2 ))/(x^2 +y^2 ))dxdy
$${A}_{{n}} =\int\int_{\left.\right]\frac{\mathrm{1}}{{n}},{n}\left[^{\mathrm{2}} \right.} \:\:\:\:\:\:\frac{{arctan}\left({x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} }{dxdy}\: \\ $$
Commented by maxmathsup by imad last updated on 14/Jun/19
1) let use the diffeomorphism x =rcosθ and y =rsinθ we have (1/n)≤x≤n and (1/n)≤y≤n ⇒(2/n^2 ) ≤x^2 +y^2 ≤2n^2 ⇒(2/n^2 ) ≤r^2 ≤2n^2 ⇒ ((√2)/n) ≤r ≤n(√2) ⇒A_n =∫∫_(((√2)/n)≤r≤n(√2) and 0≤θ≤(π/2)) ((arctan(r^2 ))/r^2 ) rdrdθ =∫_(((√2)/n) ) ^(n(√2)) ((arctan(r^2 ))/r) dr ∫_0 ^(π/2) dθ =(π/2) ∫_((√2)/n) ^(n(√2)) ((arctan(r^2 ))/r)dr let w_n (x) = ∫_((√2)/n) ^(n(√2)) ((arctan(xr^2 ))/r) dr (x>0) ⇒w_n ^′ (x) =∫_((√2)/n) ^(n(√2)) (r^2 /(r(1+x^2 r^4 )))dr =∫_((√2)/n) ^(n(√2)) ((rdr)/(1+x^2 r^4 )) let decompose F(r) =(r/(x^2 r^4 +1)) =(r/(((√x)r)^4 +1)) =(r/((1+xr^2 )^2 −2xr^2 )) =(r/((1+xr^2 −(√(2x))r)(1+xr^2 +(√(2x))r))) =((ar+b)/(xr^2 −(√(2x))r +1)) +((cr +d)/(xr^2 +(√(2x))r +1)) F(−r)=F(r) ⇒c=−a and b=d ⇒ F(r) = ((ar+b)/(xr^2 −(√(2x))r +1)) +((−ar +b)/(xr^2 +(√(2x))r +1)) ....be continued....
$$\left.\mathrm{1}\right)\:{let}\:{use}\:{the}\:{diffeomorphism}\:{x}\:={rcos}\theta\:{and}\:{y}\:={rsin}\theta\:\:{we}\:{have}\: \\ $$$$\frac{\mathrm{1}}{{n}}\leqslant{x}\leqslant{n}\:\:{and}\:\frac{\mathrm{1}}{{n}}\leqslant{y}\leqslant{n}\:\Rightarrow\frac{\mathrm{2}}{{n}^{\mathrm{2}} }\:\leqslant{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \leqslant\mathrm{2}{n}^{\mathrm{2}} \:\Rightarrow\frac{\mathrm{2}}{{n}^{\mathrm{2}} }\:\leqslant{r}^{\mathrm{2}} \leqslant\mathrm{2}{n}^{\mathrm{2}} \:\Rightarrow \\ $$$$\frac{\sqrt{\mathrm{2}}}{{n}}\:\leqslant{r}\:\leqslant{n}\sqrt{\mathrm{2}}\:\Rightarrow{A}_{{n}} =\int\int_{\frac{\sqrt{\mathrm{2}}}{{n}}\leqslant{r}\leqslant{n}\sqrt{\mathrm{2}}\:{and}\:\:\mathrm{0}\leqslant\theta\leqslant\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{arctan}\left({r}^{\mathrm{2}} \right)}{{r}^{\mathrm{2}} }\:{rdrd}\theta \\ $$$$=\int_{\frac{\sqrt{\mathrm{2}}}{{n}}\:\:\:\:} ^{{n}\sqrt{\mathrm{2}}} \:\:\:\:\:\frac{{arctan}\left({r}^{\mathrm{2}} \right)}{{r}}\:{dr}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{d}\theta\:=\frac{\pi}{\mathrm{2}}\:\int_{\frac{\sqrt{\mathrm{2}}}{{n}}} ^{{n}\sqrt{\mathrm{2}}} \:\:\:\:\frac{{arctan}\left({r}^{\mathrm{2}} \right)}{{r}}{dr}\:\:{let} \\ $$$${w}_{{n}} \left({x}\right)\:=\:\int_{\frac{\sqrt{\mathrm{2}}}{{n}}} ^{{n}\sqrt{\mathrm{2}}} \:\:\:\frac{{arctan}\left({xr}^{\mathrm{2}} \right)}{{r}}\:{dr}\:\left({x}>\mathrm{0}\right)\:\:\:\Rightarrow{w}_{{n}} ^{'} \left({x}\right)\:=\int_{\frac{\sqrt{\mathrm{2}}}{{n}}} ^{{n}\sqrt{\mathrm{2}}} \:\:\:\frac{{r}^{\mathrm{2}} }{{r}\left(\mathrm{1}+{x}^{\mathrm{2}} {r}^{\mathrm{4}} \right)}{dr} \\ $$$$=\int_{\frac{\sqrt{\mathrm{2}}}{{n}}} ^{{n}\sqrt{\mathrm{2}}} \:\:\:\:\frac{{rdr}}{\mathrm{1}+{x}^{\mathrm{2}} {r}^{\mathrm{4}} }\:\:{let}\:{decompose}\:{F}\left({r}\right)\:=\frac{{r}}{{x}^{\mathrm{2}} {r}^{\mathrm{4}} +\mathrm{1}}\:=\frac{{r}}{\left(\sqrt{{x}}{r}\right)^{\mathrm{4}} \:+\mathrm{1}} \\ $$$$=\frac{{r}}{\left(\mathrm{1}+{xr}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}{xr}^{\mathrm{2}} }\:=\frac{{r}}{\left(\mathrm{1}+{xr}^{\mathrm{2}} −\sqrt{\mathrm{2}{x}}{r}\right)\left(\mathrm{1}+{xr}^{\mathrm{2}} +\sqrt{\mathrm{2}{x}}{r}\right)} \\ $$$$=\frac{{ar}+{b}}{{xr}^{\mathrm{2}} −\sqrt{\mathrm{2}{x}}{r}\:+\mathrm{1}}\:+\frac{{cr}\:+{d}}{{xr}^{\mathrm{2}} \:+\sqrt{\mathrm{2}{x}}{r}\:+\mathrm{1}} \\ $$$${F}\left(−{r}\right)={F}\left({r}\right)\:\Rightarrow{c}=−{a}\:\:{and}\:\:{b}={d}\:\Rightarrow \\ $$$${F}\left({r}\right)\:=\:\frac{{ar}+{b}}{{xr}^{\mathrm{2}} −\sqrt{\mathrm{2}{x}}{r}\:+\mathrm{1}}\:+\frac{−{ar}\:+{b}}{{xr}^{\mathrm{2}} \:+\sqrt{\mathrm{2}{x}}{r}\:+\mathrm{1}}\:….{be}\:{continued}…. \\ $$

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