Question Number 62274 by aliesam last updated on 18/Jun/19
$$\underset{\mathrm{0}} {\overset{\infty} {\int}}{e}^{−{x}^{\mathrm{2}} } \:{dx} \\ $$
Commented by maxmathsup by imad last updated on 19/Jun/19
,n[^2 ) e^(−x^2 −y^2 ) dxdy 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)) e^(−r^2 ) rdrdθ =∫_((√2)/n) ^(n(√2)) r e^(−r^2 ) ∫_0 ^(π/2) dθ =(π/2)[−(1/2)e^(−r^2 ) ]_((√2)/n) ^(n(√2)) =−(π/4)( e^(−2n^2 ) −e^(−(2/n^2 )) ) =(π/4)( e^(−(2/n^2 )) −e^(−2n^2 ) ) ⇒ lim_(n→+∞) A_n =(π/4) = ∫∫_(]0,+∞[) e^(−x^2 −y^2 ) dxdy but A_n =(∫_(1/n) ^n e^(−x^2 ) dx)(∫_(1/n) ^n e^(−y^2 ) dy) →(∫_0 ^(+∞) e^(−x^2 ) dx)^2 ⇒ (∫_0 ^(+∞) e^(−x^2 ) dx)^2 =(π/4) due to ∫_0 ^∞ e^(−x^2 ) dx >0 ⇒∫_0 ^∞ e^(−x^2 ) dx =((√π)/2) .](https://www.tinkutara.com/question/Q62301.png)
$${let}\:\:{A}_{{n}} =\int\int_{\left.\right]\frac{\mathrm{1}}{{n}},{n}\left[^{\mathrm{2}} \right.} \:\:\:\:{e}^{−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} } {dxdy}\:\:\:\:\:{let}\:{use}\:{the}\:{diffeomorphism}\:{x}={rcos}\theta\:{and} \\ $$$${y}\:={rsin}\theta\:\:{we}\:{have}\:\:\:\frac{\mathrm{1}}{{n}}<{x}<{n}\:\:{and}\:\frac{\mathrm{1}}{{n}}<{y}<{n}\:\Rightarrow\frac{\mathrm{2}}{{n}^{\mathrm{2}} }<{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} <\mathrm{2}{n}^{\mathrm{2}} \:\Rightarrow\frac{\mathrm{2}}{{n}^{\mathrm{2}} }<{r}^{\mathrm{2}} <\mathrm{2}{n}^{\mathrm{2}} \:\Rightarrow \\ $$$$\frac{\sqrt{\mathrm{2}}}{{n}}<{r}<{n}\sqrt{\mathrm{2}}\:\Rightarrow\:{A}_{{n}} =\int\int_{\frac{\sqrt{\mathrm{2}}}{{n}}<{r}<{n}\sqrt{\mathrm{2}}{and}\:\:\mathrm{0}<\theta<\frac{\pi}{\mathrm{2}}} \:\:\:{e}^{−{r}^{\mathrm{2}} \:} {rdrd}\theta \\ $$$$=\int_{\frac{\sqrt{\mathrm{2}}}{{n}}} ^{{n}\sqrt{\mathrm{2}}} \:{r}\:{e}^{−{r}^{\mathrm{2}} } \int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {d}\theta\:=\frac{\pi}{\mathrm{2}}\left[−\frac{\mathrm{1}}{\mathrm{2}}{e}^{−{r}^{\mathrm{2}} } \right]_{\frac{\sqrt{\mathrm{2}}}{{n}}} ^{{n}\sqrt{\mathrm{2}}} \:=−\frac{\pi}{\mathrm{4}}\left(\:{e}^{−\mathrm{2}{n}^{\mathrm{2}} } −{e}^{−\frac{\mathrm{2}}{{n}^{\mathrm{2}} }} \right) \\ $$$$=\frac{\pi}{\mathrm{4}}\left(\:{e}^{−\frac{\mathrm{2}}{{n}^{\mathrm{2}} }} \:−{e}^{−\mathrm{2}{n}^{\mathrm{2}} } \right)\:\Rightarrow\:{lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} =\frac{\pi}{\mathrm{4}}\:=\:\int\int_{\left.\right]\mathrm{0},+\infty\left[\right.} \:\:{e}^{−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} } {dxdy}\:\:{but} \\ $$$${A}_{{n}} =\left(\int_{\frac{\mathrm{1}}{{n}}} ^{{n}} \:{e}^{−{x}^{\mathrm{2}} } {dx}\right)\left(\int_{\frac{\mathrm{1}}{{n}}} ^{{n}} \:{e}^{−{y}^{\mathrm{2}} } {dy}\right)\:\rightarrow\left(\int_{\mathrm{0}} ^{+\infty} \:{e}^{−{x}^{\mathrm{2}} } {dx}\right)^{\mathrm{2}} \:\Rightarrow\:\left(\int_{\mathrm{0}} ^{+\infty} \:{e}^{−{x}^{\mathrm{2}} } {dx}\right)^{\mathrm{2}} =\frac{\pi}{\mathrm{4}} \\ $$$${due}\:{to}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{x}^{\mathrm{2}} } {dx}\:>\mathrm{0}\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:{e}^{−{x}^{\mathrm{2}} } {dx}\:=\frac{\sqrt{\pi}}{\mathrm{2}}\:. \\ $$
Commented by aliesam last updated on 19/Jun/19
$${thank}\:{you}\:{sir} \\ $$
Commented by behi83417@gmail.com last updated on 19/Jun/19
$$\mathrm{dear}\:\mathrm{proph}\:\mathrm{abdo}!\:\mathrm{Q}#\mathrm{62227}\:\mathrm{in}\:\mathrm{waiting} \\ $$$$\mathrm{for}\:\mathrm{your}\:\mathrm{attention}.\mathrm{please}. \\ $$
Commented by maxmathsup by imad last updated on 19/Jun/19
$${you}\:{are}\:{welcome}\:. \\ $$
Answered by tanmay last updated on 19/Jun/19
$${t}={x}^{\mathrm{2}} \rightarrow{x}={t}^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:\:{dx}=\frac{\mathrm{1}}{\mathrm{2}}×{t}^{\frac{−\mathrm{1}}{\mathrm{2}}} {dt} \\ $$$$\int_{\mathrm{0}} ^{\infty} {e}^{−{t}} ×\frac{\mathrm{1}}{\mathrm{2}}×{t}^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} {dt} \\ $$$${using}\:{gamma}\:{functoon} \\ $$$$\int_{\mathrm{0}} ^{\infty} {e}^{−{t}} ×{t}^{{n}−\mathrm{1}} {dt}=\lceil\left({n}\right) \\ $$$${so}\:{answer}\:{is}\:\frac{\mathrm{1}}{\mathrm{2}}×\lceil\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}}×\sqrt{\pi}\: \\ $$
Commented by aliesam last updated on 19/Jun/19
$${thank}\:{you}\:{sir} \\ $$