Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

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....

$$\left.\mathrm{1}\right)\:{let}\:{use}\:{the}\:{diffeomorphism}\:{x}\:={rcos}\theta\:{and}\:{y}\:={rsin}\theta\:\:{we}\:{have}\: \\ $$$$\frac{\mathrm{1}}{{n}}\leqslant{x}\leqslant{n}\:\:{and}\:\frac{\mathrm{1}}{{n}}\leqslant{y}\leqslant{n}\:\Rightarrow\frac{\mathrm{2}}{{n}^{\mathrm{2}} }\:\leqslant{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \leqslant\mathrm{2}{n}^{\mathrm{2}} \:\Rightarrow\frac{\mathrm{2}}{{n}^{\mathrm{2}} }\:\leqslant{r}^{\mathrm{2}} \leqslant\mathrm{2}{n}^{\mathrm{2}} \:\Rightarrow \\ $$$$\frac{\sqrt{\mathrm{2}}}{{n}}\:\leqslant{r}\:\leqslant{n}\sqrt{\mathrm{2}}\:\Rightarrow{A}_{{n}} =\int\int_{\frac{\sqrt{\mathrm{2}}}{{n}}\leqslant{r}\leqslant{n}\sqrt{\mathrm{2}}\:{and}\:\:\mathrm{0}\leqslant\theta\leqslant\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{arctan}\left({r}^{\mathrm{2}} \right)}{{r}^{\mathrm{2}} }\:{rdrd}\theta \\ $$$$=\int_{\frac{\sqrt{\mathrm{2}}}{{n}}\:\:\:\:} ^{{n}\sqrt{\mathrm{2}}} \:\:\:\:\:\frac{{arctan}\left({r}^{\mathrm{2}} \right)}{{r}}\:{dr}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{d}\theta\:=\frac{\pi}{\mathrm{2}}\:\int_{\frac{\sqrt{\mathrm{2}}}{{n}}} ^{{n}\sqrt{\mathrm{2}}} \:\:\:\:\frac{{arctan}\left({r}^{\mathrm{2}} \right)}{{r}}{dr}\:\:{let} \\ $$$${w}_{{n}} \left({x}\right)\:=\:\int_{\frac{\sqrt{\mathrm{2}}}{{n}}} ^{{n}\sqrt{\mathrm{2}}} \:\:\:\frac{{arctan}\left({xr}^{\mathrm{2}} \right)}{{r}}\:{dr}\:\left({x}>\mathrm{0}\right)\:\:\:\Rightarrow{w}_{{n}} ^{'} \left({x}\right)\:=\int_{\frac{\sqrt{\mathrm{2}}}{{n}}} ^{{n}\sqrt{\mathrm{2}}} \:\:\:\frac{{r}^{\mathrm{2}} }{{r}\left(\mathrm{1}+{x}^{\mathrm{2}} {r}^{\mathrm{4}} \right)}{dr} \\ $$$$=\int_{\frac{\sqrt{\mathrm{2}}}{{n}}} ^{{n}\sqrt{\mathrm{2}}} \:\:\:\:\frac{{rdr}}{\mathrm{1}+{x}^{\mathrm{2}} {r}^{\mathrm{4}} }\:\:{let}\:{decompose}\:{F}\left({r}\right)\:=\frac{{r}}{{x}^{\mathrm{2}} {r}^{\mathrm{4}} +\mathrm{1}}\:=\frac{{r}}{\left(\sqrt{{x}}{r}\right)^{\mathrm{4}} \:+\mathrm{1}} \\ $$$$=\frac{{r}}{\left(\mathrm{1}+{xr}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}{xr}^{\mathrm{2}} }\:=\frac{{r}}{\left(\mathrm{1}+{xr}^{\mathrm{2}} −\sqrt{\mathrm{2}{x}}{r}\right)\left(\mathrm{1}+{xr}^{\mathrm{2}} +\sqrt{\mathrm{2}{x}}{r}\right)} \\ $$$$=\frac{{ar}+{b}}{{xr}^{\mathrm{2}} −\sqrt{\mathrm{2}{x}}{r}\:+\mathrm{1}}\:+\frac{{cr}\:+{d}}{{xr}^{\mathrm{2}} \:+\sqrt{\mathrm{2}{x}}{r}\:+\mathrm{1}} \\ $$$${F}\left(−{r}\right)={F}\left({r}\right)\:\Rightarrow{c}=−{a}\:\:{and}\:\:{b}={d}\:\Rightarrow \\ $$$${F}\left({r}\right)\:=\:\frac{{ar}+{b}}{{xr}^{\mathrm{2}} −\sqrt{\mathrm{2}{x}}{r}\:+\mathrm{1}}\:+\frac{−{ar}\:+{b}}{{xr}^{\mathrm{2}} \:+\sqrt{\mathrm{2}{x}}{r}\:+\mathrm{1}}\:....{be}\:{continued}.... \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com