Question Number 88413 by 675480065 last updated on 10/Apr/20
$$\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } \mathrm{dx} \\ $$$$ \\ $$
Commented by abdomathmax last updated on 10/Apr/20
$${x}=\sqrt{{t}}\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:{e}^{−{x}^{\mathrm{2}} } {dx}\:=\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}} \:\frac{{dt}}{\mathrm{2}\sqrt{{t}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:{t}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:{e}^{−{t}} \:{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:{t}^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} \:{e}^{−{t}} \:{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\sqrt{\pi}}{\mathrm{2}} \\ $$
Commented by Tony Lin last updated on 10/Apr/20
$${let}\:{I}=\int_{−\infty} ^{\infty} {e}^{−{x}^{\mathrm{2}} } {dx} \\ $$$${I}^{\mathrm{2}} =\int_{−\infty} ^{\infty} \int_{−\infty} ^{\infty} {e}^{−\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)} {dxdy} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \int_{\mathrm{0}} ^{\infty} {re}^{−{r}^{\mathrm{2}} } {drd}\theta \\ $$$$=\mathrm{2}\pi\left(−\frac{{e}^{−{x}^{\mathrm{2}} } }{\mathrm{2}}\right)\mid_{\mathrm{0}} ^{\infty} \\ $$$$=\mathrm{2}\pi×\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$=\pi \\ $$$$\Rightarrow{I}=\sqrt{\pi} \\ $$$$\int_{\mathrm{0}} ^{\infty} {e}^{−{x}^{\mathrm{2}} } {dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{\infty} {e}^{−{x}^{\mathrm{2}} } {dx}=\frac{\sqrt{\pi}}{\mathrm{2}} \\ $$
Commented by mathmax by abdo last updated on 10/Apr/20
![let A_n =∫∫_([(1/n),n[^2 ) e^(−x^2 −y^2 ) dxdy we use the diffeomorphism x =rcosθ and y =rsinθ we have (1/n)≤x<n and (1/n)≤y<n ⇒(2/n^2 )≤r^2 <2n^2 ⇒((√2)/n)≤r<n(√2)⇒ A_n =∫_((√2)/n) ^(n(√2)) ∫_0 ^(π/2) e^(−r^2 ) rdr dθ =(π/2)[−(1/2)e^(−r^2 ) ]_((√2)/n) ^(n(√2)) =−(π/4) { e^(−2n^2 ) −e^(−(2/n^2 )) } we have (∫_0 ^∞ e^(−x^2 ) dx)^2 =lim_(n→+∞) A_n =(π/4) ⇒ ∫_0 ^∞ e^(−x^2 ) dx =((√π)/2)](https://www.tinkutara.com/question/Q88481.png)
$${let}\:{A}_{{n}} =\int\int_{\left[\frac{\mathrm{1}}{{n}},{n}\left[^{\mathrm{2}} \right.\right.} \:\:{e}^{−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} } {dxdy}\:\:\:{we}\:{use}\:{the}\:{diffeomorphism} \\ $$$${x}\:={rcos}\theta\:\:{and}\:{y}\:={rsin}\theta\:\:\:\:{we}\:{have}\:\:\:\frac{\mathrm{1}}{{n}}\leqslant{x}<{n}\:\:{and}\:\frac{\mathrm{1}}{{n}}\leqslant{y}<{n} \\ $$$$\Rightarrow\frac{\mathrm{2}}{{n}^{\mathrm{2}} }\leqslant{r}^{\mathrm{2}} <\mathrm{2}{n}^{\mathrm{2}} \:\Rightarrow\frac{\sqrt{\mathrm{2}}}{{n}}\leqslant{r}<{n}\sqrt{\mathrm{2}}\Rightarrow\:{A}_{{n}} =\int_{\frac{\sqrt{\mathrm{2}}}{{n}}} ^{{n}\sqrt{\mathrm{2}}} \:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:{e}^{−{r}^{\mathrm{2}} } \:{rdr}\:{d}\theta \\ $$$$=\frac{\pi}{\mathrm{2}}\left[−\frac{\mathrm{1}}{\mathrm{2}}{e}^{−{r}^{\mathrm{2}} } \right]_{\frac{\sqrt{\mathrm{2}}}{{n}}} ^{{n}\sqrt{\mathrm{2}}} \:\:=−\frac{\pi}{\mathrm{4}}\:\left\{\:\:{e}^{−\mathrm{2}{n}^{\mathrm{2}} } −{e}^{−\frac{\mathrm{2}}{{n}^{\mathrm{2}} }} \:\:\:\right\} \\ $$$${we}\:{have}\:\left(\int_{\mathrm{0}} ^{\infty} \:{e}^{−{x}^{\mathrm{2}} } {dx}\right)^{\mathrm{2}} \:={lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} =\frac{\pi}{\mathrm{4}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:{e}^{−{x}^{\mathrm{2}} } {dx}\:\:=\frac{\sqrt{\pi}}{\mathrm{2}} \\ $$