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Question Number 65398 by mathmax by abdo last updated on 29/Jul/19
1) calculate  A_n =∫∫_([1,n[^2 )      sin(x^2  +3y^2 ) e^(−x^2 −3y^2 ) dxdy  2) determine lim_(n→+∞)  A_n
$$\left.\mathrm{1}\right)\:{calculate}\:\:{A}_{{n}} =\int\int_{\left[\mathrm{1},{n}\left[^{\mathrm{2}} \right.\right.} \:\:\:\:\:{sin}\left({x}^{\mathrm{2}} \:+\mathrm{3}{y}^{\mathrm{2}} \right)\:{e}^{−{x}^{\mathrm{2}} −\mathrm{3}{y}^{\mathrm{2}} } {dxdy} \\ $$$$\left.\mathrm{2}\right)\:{determine}\:{lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} \\ $$
Commented by mathmax by abdo last updated on 31/Jul/19
1) let consider thediffeomorphism (r,θ)→(x,y)=(rcosθ,(r/( (√3)))sinθ)  ϕ(r,θ) =(ϕ_1 ,ϕ_2 ) / ϕ_1 (r,θ)=rcosθ and ϕ_2 (r,θ)=(r/( (√3)))sinθ  M_j (ϕ) = ((((∂ϕ_1 /∂r)         (∂ϕ_1 /∂θ))),(((∂ϕ_2 /∂r)            (∂ϕ_2 /∂θ))) )  = (((cosθ            −rsinθ)),((((sinθ)/( (√3)))                 (r/( (√3)))cosθ)) )  and detM_j (ϕ) =(r/( (√3)))  we have   1≤x<n  and  1≤y<n ⇒  1≤x^2 <n^2  and3≤3y^2 <3n^2  ⇒4≤x^2  +3y^2 <4n^(2 )  ⇒2≤r<2n  A_n =∫∫_(2≤r<2n and 0≤θ≤(π/2))     sin(r^2 )e^(−r^2 ) (r/( (√3))) drdθ  =(π/2) ∫_2 ^(2n)  r sin(r^2 )e^(−r^2 ) dr  by parts u^′  =r sin(r^2 ) and v =e^(−r^2 )  ⇒  ∫_2 ^(2n)   rsin(r^2 )e^(−r^2 ) dr =[−(1/2)cos(r^2 )e^(−r^2 ) ]_2 ^(2n)  −∫_2 ^(2n) (−(1/2)cos(r^2 ))(−2r)e^(−r^2 ) dr  =−(1/2){cos(4n^2 )e^(−4n^2 )  −cos(4)e^(−4) }−∫_2 ^(2n)  r cos(r^2 ) e^(−r^2 ) dr  again by parts  u^′  =rcos(r^2 ) and v =e^(−r^2 ) ⇒  ∫_2 ^(2n)  rcos(r^2 )e^(−r^2 ) dr =[(1/2)sin(r^2 )e^(−r^2 ) ]_2 ^(2n)  −∫_2 ^(2n) (1/2)sin(r^2 )(−2r)e^(−r^2 ) dr  =(1/2){sin(4n^2 )e^(−4n^2 ) −sin4 e^(−4) } +∫_2 ^(2n)  rsin(r^2 )e^(−r^2 ) dr ⇒  2∫_2 ^(2n)  rsin(r^2 )e^(−r^2 ) dr =(1/2){cos(4)e^(−4) −cos(4n^2 )e^(−4n^2 ) }  +(1/2){sin(4)e^(−4) −sin(4n^2 )e^(−4n^2 ) } ⇒  A_n =(π/8){  e^(−4) (cos(4)+sin4)−e^(−4n^2 ) (cos(4n^2 )+sin(4n^2 )}  2) lim_(n→+∞)   A_n =((πe^(−4) )/8){ cos(4)+sin(4)}
$$\left.\mathrm{1}\right)\:{let}\:{consider}\:{thediffeomorphism}\:\left({r},\theta\right)\rightarrow\left({x},{y}\right)=\left({rcos}\theta,\frac{{r}}{\:\sqrt{\mathrm{3}}}{sin}\theta\right) \\ $$$$\varphi\left({r},\theta\right)\:=\left(\varphi_{\mathrm{1}} ,\varphi_{\mathrm{2}} \right)\:/\:\varphi_{\mathrm{1}} \left({r},\theta\right)={rcos}\theta\:{and}\:\varphi_{\mathrm{2}} \left({r},\theta\right)=\frac{{r}}{\:\sqrt{\mathrm{3}}}{sin}\theta \\ $$$${M}_{{j}} \left(\varphi\right)\:=\begin{pmatrix}{\frac{\partial\varphi_{\mathrm{1}} }{\partial{r}}\:\:\:\:\:\:\:\:\:\frac{\partial\varphi_{\mathrm{1}} }{\partial\theta}}\\{\frac{\partial\varphi_{\mathrm{2}} }{\partial{r}}\:\:\:\:\:\:\:\:\:\:\:\:\frac{\partial\varphi_{\mathrm{2}} }{\partial\theta}}\end{pmatrix} \\ $$$$=\begin{pmatrix}{{cos}\theta\:\:\:\:\:\:\:\:\:\:\:\:−{rsin}\theta}\\{\frac{{sin}\theta}{\:\sqrt{\mathrm{3}}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{r}}{\:\sqrt{\mathrm{3}}}{cos}\theta}\end{pmatrix} \\ $$$${and}\:{detM}_{{j}} \left(\varphi\right)\:=\frac{{r}}{\:\sqrt{\mathrm{3}}}\:\:{we}\:{have}\:\:\:\mathrm{1}\leqslant{x}<{n}\:\:{and}\:\:\mathrm{1}\leqslant{y}<{n}\:\Rightarrow \\ $$$$\mathrm{1}\leqslant{x}^{\mathrm{2}} <{n}^{\mathrm{2}} \:{and}\mathrm{3}\leqslant\mathrm{3}{y}^{\mathrm{2}} <\mathrm{3}{n}^{\mathrm{2}} \:\Rightarrow\mathrm{4}\leqslant{x}^{\mathrm{2}} \:+\mathrm{3}{y}^{\mathrm{2}} <\mathrm{4}{n}^{\mathrm{2}\:} \:\Rightarrow\mathrm{2}\leqslant{r}<\mathrm{2}{n} \\ $$$${A}_{{n}} =\int\int_{\mathrm{2}\leqslant{r}<\mathrm{2}{n}\:{and}\:\mathrm{0}\leqslant\theta\leqslant\frac{\pi}{\mathrm{2}}} \:\:\:\:{sin}\left({r}^{\mathrm{2}} \right){e}^{−{r}^{\mathrm{2}} } \frac{{r}}{\:\sqrt{\mathrm{3}}}\:{drd}\theta \\ $$$$=\frac{\pi}{\mathrm{2}}\:\int_{\mathrm{2}} ^{\mathrm{2}{n}} \:{r}\:{sin}\left({r}^{\mathrm{2}} \right){e}^{−{r}^{\mathrm{2}} } {dr} \\ $$$${by}\:{parts}\:{u}^{'} \:={r}\:{sin}\left({r}^{\mathrm{2}} \right)\:{and}\:{v}\:={e}^{−{r}^{\mathrm{2}} } \:\Rightarrow \\ $$$$\int_{\mathrm{2}} ^{\mathrm{2}{n}} \:\:{rsin}\left({r}^{\mathrm{2}} \right){e}^{−{r}^{\mathrm{2}} } {dr}\:=\left[−\frac{\mathrm{1}}{\mathrm{2}}{cos}\left({r}^{\mathrm{2}} \right){e}^{−{r}^{\mathrm{2}} } \right]_{\mathrm{2}} ^{\mathrm{2}{n}} \:−\int_{\mathrm{2}} ^{\mathrm{2}{n}} \left(−\frac{\mathrm{1}}{\mathrm{2}}{cos}\left({r}^{\mathrm{2}} \right)\right)\left(−\mathrm{2}{r}\right){e}^{−{r}^{\mathrm{2}} } {dr} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left\{{cos}\left(\mathrm{4}{n}^{\mathrm{2}} \right){e}^{−\mathrm{4}{n}^{\mathrm{2}} } \:−{cos}\left(\mathrm{4}\right){e}^{−\mathrm{4}} \right\}−\int_{\mathrm{2}} ^{\mathrm{2}{n}} \:{r}\:{cos}\left({r}^{\mathrm{2}} \right)\:{e}^{−{r}^{\mathrm{2}} } {dr} \\ $$$${again}\:{by}\:{parts}\:\:{u}^{'} \:={rcos}\left({r}^{\mathrm{2}} \right)\:{and}\:{v}\:={e}^{−{r}^{\mathrm{2}} } \Rightarrow \\ $$$$\int_{\mathrm{2}} ^{\mathrm{2}{n}} \:{rcos}\left({r}^{\mathrm{2}} \right){e}^{−{r}^{\mathrm{2}} } {dr}\:=\left[\frac{\mathrm{1}}{\mathrm{2}}{sin}\left({r}^{\mathrm{2}} \right){e}^{−{r}^{\mathrm{2}} } \right]_{\mathrm{2}} ^{\mathrm{2}{n}} \:−\int_{\mathrm{2}} ^{\mathrm{2}{n}} \frac{\mathrm{1}}{\mathrm{2}}{sin}\left({r}^{\mathrm{2}} \right)\left(−\mathrm{2}{r}\right){e}^{−{r}^{\mathrm{2}} } {dr} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{{sin}\left(\mathrm{4}{n}^{\mathrm{2}} \right){e}^{−\mathrm{4}{n}^{\mathrm{2}} } −{sin}\mathrm{4}\:{e}^{−\mathrm{4}} \right\}\:+\int_{\mathrm{2}} ^{\mathrm{2}{n}} \:{rsin}\left({r}^{\mathrm{2}} \right){e}^{−{r}^{\mathrm{2}} } {dr}\:\Rightarrow \\ $$$$\mathrm{2}\int_{\mathrm{2}} ^{\mathrm{2}{n}} \:{rsin}\left({r}^{\mathrm{2}} \right){e}^{−{r}^{\mathrm{2}} } {dr}\:=\frac{\mathrm{1}}{\mathrm{2}}\left\{{cos}\left(\mathrm{4}\right){e}^{−\mathrm{4}} −{cos}\left(\mathrm{4}{n}^{\mathrm{2}} \right){e}^{−\mathrm{4}{n}^{\mathrm{2}} } \right\} \\ $$$$+\frac{\mathrm{1}}{\mathrm{2}}\left\{{sin}\left(\mathrm{4}\right){e}^{−\mathrm{4}} −{sin}\left(\mathrm{4}{n}^{\mathrm{2}} \right){e}^{−\mathrm{4}{n}^{\mathrm{2}} } \right\}\:\Rightarrow \\ $$$${A}_{{n}} =\frac{\pi}{\mathrm{8}}\left\{\:\:{e}^{−\mathrm{4}} \left({cos}\left(\mathrm{4}\right)+{sin}\mathrm{4}\right)−{e}^{−\mathrm{4}{n}^{\mathrm{2}} } \left({cos}\left(\mathrm{4}{n}^{\mathrm{2}} \right)+{sin}\left(\mathrm{4}{n}^{\mathrm{2}} \right)\right\}\right. \\ $$$$\left.\mathrm{2}\right)\:{lim}_{{n}\rightarrow+\infty} \:\:{A}_{{n}} =\frac{\pi{e}^{−\mathrm{4}} }{\mathrm{8}}\left\{\:{cos}\left(\mathrm{4}\right)+{sin}\left(\mathrm{4}\right)\right\} \\ $$$$ \\ $$

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