Question Number 51994 by maxmathsup by imad last updated on 01/Jan/19 | ||
$${let}\:{D}_{{n}} =\:\left\{\left({x},{y}\right)\in{R}^{\mathrm{2}} \:\:/\left({x},{y}\right)\in\left[\frac{\mathrm{1}}{{n}}\:,{n}\left[\:\right\}\right.\right. \\ $$$$\left.\mathrm{1}\right)\:{find}\:{the}\:{value}\:{of}\:\int\int_{{D}_{{n}} } \:\:\:\:\:{e}^{−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} } {dxdy} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{+\infty} \:{e}^{−{x}^{\mathrm{2}} } {dx}\:. \\ $$ | ||
Commented by Abdo msup. last updated on 03/Jan/19 | ||
$${changement}\:{x}={r}\:{cos}\theta\:{and}\:{y}={r}\:{sin}\theta\:{give} \\ $$$${x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} ={r}^{\mathrm{2}} \:{but}\:\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\leqslant{x}^{\mathrm{2}} \leqslant{n}^{\mathrm{2}} \:{and}\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\leqslant{y}^{\mathrm{2}} \leqslant{n}^{\mathrm{2}} \:\Rightarrow \\ $$$$\frac{\mathrm{2}}{{n}^{\mathrm{2}} }\leqslant{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \leqslant\mathrm{2}{n}^{\mathrm{2}} \:\Rightarrow\frac{\mathrm{2}}{{n}^{\mathrm{2}} }\:\leqslant{r}^{\mathrm{2}} \:\leqslant\mathrm{2}{n}^{\mathrm{2}} \:\: \\ $$$$\frac{\sqrt{\mathrm{2}}}{{n}}\:\leqslant{r}\leqslant\sqrt{\mathrm{2}}{n}\:\:{also}\:{x}\geqslant\mathrm{0}\:{and}\:{y}\geqslant\mathrm{0}\:\Rightarrow\mathrm{0}\leqslant\theta\leqslant\frac{\pi}{\mathrm{2}} \\ $$$$\int\int_{{D}_{{n}} \:} \:\:\:{e}^{−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} } {dxdy}\:=\int\int_{\frac{\sqrt{\mathrm{2}}}{{n}}\leqslant{r}<\sqrt{\mathrm{2}}{n}\:\:{and}\:\mathrm{0}\leqslant\theta\leqslant\frac{\pi}{\mathrm{2}}} \:\:\:{e}^{−{r}^{\mathrm{2}} } {rdr}\:{d}\theta \\ $$$$=\frac{\pi}{\mathrm{2}}\:\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\left[\:{e}^{−{r}^{\mathrm{2}} } \right]_{\frac{\sqrt{\mathrm{2}}}{{n}}} ^{\sqrt{\mathrm{2}}{n}} =\frac{\pi}{\mathrm{4}}\left\{\:−{e}^{−\mathrm{2}{n}^{\mathrm{2}} } \:\:+{e}^{−\frac{\mathrm{2}}{{n}^{\mathrm{2}} }} \right\} \\ $$$$\left.\mathrm{2}\right)\:{we}\:{have}\:\int\int_{{D}_{{n}} } \:\:{e}^{−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} } {dxdy} \\ $$$$=\int_{\frac{\mathrm{1}}{{n}}} ^{{n}} \:{e}^{−{x}^{\mathrm{2}} } {dx}.\int_{\frac{\mathrm{1}}{{n}}} ^{{n}} \:{e}^{−{y}^{\mathrm{2}} } {dy}\:=\left(\int_{\frac{\mathrm{1}}{{n}}} ^{{n}} \:\:{e}^{−{x}^{\mathrm{2}} } {dx}\right)^{\mathrm{2}} \:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} \:\left(\int_{\frac{\mathrm{1}}{{n}}} ^{{n}} \:{e}^{−{x}^{\mathrm{2}} } {dx}\right)^{\mathrm{2}} \:={lim}_{{n}\rightarrow+\infty} \:{D}_{{n}} \:=\frac{\pi}{\mathrm{4}}\:\Rightarrow \\ $$$$\left(\int_{\mathrm{0}} ^{+\infty} \:{e}^{−{x}^{\mathrm{2}} } {dx}\right)^{\mathrm{2}} =\frac{\pi}{\mathrm{4}}\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{x}^{\mathrm{2}} } {dx}\:=\frac{\sqrt{\pi}}{\mathrm{2}}\:. \\ $$$$ \\ $$ | ||