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

$$\mathrm{let}{\mathrm{A}}_{\mathrm{n}}=\int {\int }_{[0,\mathrm{n}[{}^2}\frac{{\mathrm{e}}^{-{\mathrm{x}}^2-{\mathrm{y}}^2}}{\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}}\mathrm{dxdy}$$$$1)\mathrm{calculste}{\mathrm{A}}_{\mathrm{n}}\mathrm{interm}\mathrm{of}\mathrm{n}$$$$2)\mathrm{find}{\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{A}}_{\mathrm{n}}$$

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

$$1)\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\{\begin{array}{l}\mathrm{x}=\mathrm{rcos}\theta \\ \mathrm{y}=\mathrm{rsin}\theta \end{array}$$$$0⩽\mathrm{x}<\mathrm{n}\mathrm{and}0⩽\mathrm{y}<\mathrm{n}\Rightarrow 0⩽{\mathrm{x}}^2+{\mathrm{y}}^2<{2\mathrm{n}}^2\Rightarrow 0⩽{\mathrm{r}}^2<{2\mathrm{n}}^2\Rightarrow 0⩽\mathrm{r}<\mathrm{n}\sqrt{2}$$$$\Rightarrow {\mathrm{A}}_{\mathrm{n}}={\int }_0^{\mathrm{n}\sqrt{2}}{\int }_0^{\frac{\pi }{2}}\frac{{\mathrm{e}}^{-{\mathrm{r}}^2}}{\mathrm{r}}\mathrm{r}\mathrm{dr}\mathrm{d}\theta =\frac{\pi }{2}{\int }_0^{\mathrm{n}\sqrt{2}}{\mathrm{e}}^{-{\mathrm{r}}^2}\mathrm{dr}$$$$2){\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{A}}_{\mathrm{n}}=\frac{\pi }{2}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{r}}^2}\mathrm{dr}=\frac{\pi }{2}\times \frac{\sqrt{\pi }}{2}=\frac{\pi \sqrt{\pi }}{4}$$

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