Question Number 116806 by Backer last updated on 07/Oct/20
$$\mathrm{Hi} \\ $$$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\:\int_{-\infty} ^{+\infty} -\mathrm{e}^{-\mathrm{x}^{\mathrm{2}} } \mathrm{dx}=\sqrt{\pi} \\ $$$$\mathrm{Thanks}\:\mathrm{beforehand} \\ $$$$ \\ $$
Answered by Bird last updated on 07/Oct/20
![let A_ξ =∫∫_(]−ξ,ξ[^2 ) e^(−x^2 −y^2 ) dxdy we hsve lim_(ξ→+∞) A_ξ =(∫_(−∞) ^∞ e^(−x^2 ) dx)^2 let use the diffeomorphism { ((x =rcosθ)),((y =rsinθ we have −ξ≤x≤ξ)) :} and −ξ≤y≤ξ ⇒0≤x^2 +y^2 ≤2ξ^2 ⇒0≤r≤ξ(√2) ⇒ A_ξ =∫_0 ^(ξ(√2)) r e^(−r^2 ) dr ∫_(−π) ^π dθ =2π [−(1/2)e^(−r^2 ) ]_0 ^(ξ(√2)) =π(1−e^(−2ξ^2 ) ) ⇒lim_(ξ→+∞) A_ξ =π =(∫_(−∞) ^(+∞) e^(−x^2 ) dx)^2 but∫_(−∞) ^∞ e^(−x^2 ) dx>0 ⇒∫_(−∞) ^(+∞) e^(−x^2 ) dx =(√π)](https://www.tinkutara.com/question/Q116807.png)
$${let}\:{A}_{\xi} =\int\int_{\left.\right]−\xi,\xi\left[^{\mathrm{2}} \right.} \:\:{e}^{−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} } {dxdy} \\ $$$${we}\:{hsve}\:{lim}_{\xi\rightarrow+\infty} \:{A}_{\xi} =\left(\int_{−\infty} ^{\infty} \:{e}^{−{x}^{\mathrm{2}} } {dx}\right)^{\mathrm{2}} \\ $$$${let}\:{use}\:{the}\:{diffeomorphism} \\ $$$$\begin{cases}{{x}\:={rcos}\theta}\\{{y}\:={rsin}\theta\:\:\:\:\:{we}\:{have}\:−\xi\leqslant{x}\leqslant\xi}\end{cases} \\ $$$${and}\:−\xi\leqslant{y}\leqslant\xi\:\Rightarrow\mathrm{0}\leqslant{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \leqslant\mathrm{2}\xi^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{0}\leqslant{r}\leqslant\xi\sqrt{\mathrm{2}}\:\Rightarrow \\ $$$${A}_{\xi} =\int_{\mathrm{0}} ^{\xi\sqrt{\mathrm{2}}} {r}\:{e}^{−{r}^{\mathrm{2}} } {dr}\:\int_{−\pi} ^{\pi} \:{d}\theta \\ $$$$=\mathrm{2}\pi\:\left[−\frac{\mathrm{1}}{\mathrm{2}}{e}^{−{r}^{\mathrm{2}} } \right]_{\mathrm{0}} ^{\xi\sqrt{\mathrm{2}}} \\ $$$$=\pi\left(\mathrm{1}−{e}^{−\mathrm{2}\xi^{\mathrm{2}} } \right)\:\Rightarrow{lim}_{\xi\rightarrow+\infty} {A}_{\xi} =\pi \\ $$$$=\left(\int_{−\infty} ^{+\infty} \:{e}^{−{x}^{\mathrm{2}} } {dx}\right)^{\mathrm{2}} \:\:{but}\int_{−\infty} ^{\infty} \:{e}^{−{x}^{\mathrm{2}} } {dx}>\mathrm{0} \\ $$$$\Rightarrow\int_{−\infty} ^{+\infty} \:{e}^{−{x}^{\mathrm{2}} } {dx}\:=\sqrt{\pi} \\ $$
Answered by bobhans last updated on 07/Oct/20
$$\mathrm{I}=\underset{−\infty} {\overset{\infty} {\int}}\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } \:\mathrm{dx}\:\Rightarrow\mathrm{I}^{\mathrm{2}} =\underset{−\infty} {\overset{\infty} {\int}}\underset{−\infty} {\overset{\infty} {\int}}\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} } \:\mathrm{dx}\:\mathrm{dy} \\ $$$$\left.\mathrm{I}^{\mathrm{2}} \:=\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\:\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\int}}\:\mathrm{r}\:\mathrm{e}^{−\mathrm{r}^{\mathrm{2}} } \:\mathrm{d}\theta\:\mathrm{dr}\:=\underset{\mathrm{0}} {\overset{\infty} {\int}}\mid\left(\mathrm{re}^{−\mathrm{r}^{\mathrm{2}} } \:\left(\theta\right)\right)\mid_{\mathrm{0}} ^{\mathrm{2}\pi} \:\right)\mathrm{dr} \\ $$$$\mathrm{I}^{\mathrm{2}} \:=\mathrm{2}\pi\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\mathrm{re}^{−\mathrm{r}^{\mathrm{2}} } \:\mathrm{dr}\:=\:\pi\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\mathrm{e}^{−\mathrm{r}^{\mathrm{2}} } \:\mathrm{d}\left(\mathrm{r}^{\mathrm{2}} \right) \\ $$$$\mathrm{I}^{\mathrm{2}} \:=\:−\pi\:\mid\left(\mathrm{e}^{−\mathrm{r}^{\mathrm{2}} } \right)\mid_{\mathrm{0}} ^{\infty} =\:−\pi\left(\mathrm{0}−\mathrm{1}\right)=\pi \\ $$$$\mathrm{Hence}\:\mathrm{I}\:=\:\sqrt{\pi} \\ $$