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Question Number 104773 by mathmax by abdo last updated on 23/Jul/20

let A_n =∫∫_([0,n[^2 ) (e^(−x^2 −y^2 ) /(√(x^2 +y^2 )))dxdy 1) calculste A_n interm of n 2) find lim_(n→+∞) A_n

$$\mathrm{let}\:\mathrm{A}_{\mathrm{n}} =\int\int_{\left[\mathrm{0},\mathrm{n}\left[^{\mathrm{2}} \right.\right.} \:\:\:\:\frac{\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} } }{\sqrt{\mathrm{x}^{\mathrm{2}} \:+\mathrm{y}^{\mathrm{2}} }}\mathrm{dxdy} \\ $$$$\left.\mathrm{1}\right)\:\mathrm{calculste}\:\mathrm{A}_{\mathrm{n}} \:\mathrm{interm}\:\mathrm{of}\:\mathrm{n} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{find}\:\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{A}_{\mathrm{n}} \\ $$

Answered by mathmax by abdo last updated on 24/Jul/20

1) we do the changement { ((x =rcosθ)),((y =rsinθ)) :} 0≤x<n and 0≤y<n ⇒0≤x^2 +y^2 <2n^2 ⇒0≤r^2 <2n^2 ⇒0≤r<n(√2) ⇒ A_n =∫_0 ^(n(√2)) ∫_0 ^(π/2) (e^(−r^2 ) /r) r dr dθ =(π/2) ∫_0 ^(n(√2)) e^(−r^2 ) dr 2) lim_(n→+∞) A_n =(π/2) ∫_0 ^∞ e^(−r^2 ) dr =(π/2)×((√π)/2) =((π(√π))/4)

$$\left.\mathrm{1}\right)\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\begin{cases}{\mathrm{x}\:=\mathrm{rcos}\theta}\\{\mathrm{y}\:=\mathrm{rsin}\theta}\end{cases} \\ $$$$\mathrm{0}\leqslant\mathrm{x}<\mathrm{n}\:\mathrm{and}\:\mathrm{0}\leqslant\mathrm{y}<\mathrm{n}\:\Rightarrow\mathrm{0}\leqslant\mathrm{x}^{\mathrm{2}} \:+\mathrm{y}^{\mathrm{2}} <\mathrm{2n}^{\mathrm{2}} \:\Rightarrow\mathrm{0}\leqslant\mathrm{r}^{\mathrm{2}} <\mathrm{2n}^{\mathrm{2}} \:\Rightarrow\mathrm{0}\leqslant\mathrm{r}<\mathrm{n}\sqrt{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{A}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\mathrm{n}\sqrt{\mathrm{2}}} \:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{e}^{−\mathrm{r}^{\mathrm{2}} } }{\mathrm{r}}\:\mathrm{r}\:\mathrm{dr}\:\mathrm{d}\theta\:=\frac{\pi}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{n}\sqrt{\mathrm{2}}} \:\:\mathrm{e}^{−\mathrm{r}^{\mathrm{2}} } \mathrm{dr} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\mathrm{A}_{\mathrm{n}} =\frac{\pi}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{r}^{\mathrm{2}} } \mathrm{dr}\:=\frac{\pi}{\mathrm{2}}×\frac{\sqrt{\pi}}{\mathrm{2}}\:=\frac{\pi\sqrt{\pi}}{\mathrm{4}} \\ $$

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