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Question Number 116806 by Backer last updated on 07/Oct/20

Hi  Prove that:   ∫_(-∞) ^(+∞) -e^(-x^2 ) dx=(√π)  Thanks beforehand

$$\mathrm{Hi} \\ $$$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\:\int_{-\infty} ^{+\infty} -\mathrm{e}^{-\mathrm{x}^{\mathrm{2}} } \mathrm{dx}=\sqrt{\pi} \\ $$$$\mathrm{Thanks}\:\mathrm{beforehand} \\ $$$$ \\ $$

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)

$${let}\:{A}_{\xi} =\int\int_{\left.\right]−\xi,\xi\left[^{\mathrm{2}} \right.} \:\:{e}^{−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} } {dxdy} \\ $$$${we}\:{hsve}\:{lim}_{\xi\rightarrow+\infty} \:{A}_{\xi} =\left(\int_{−\infty} ^{\infty} \:{e}^{−{x}^{\mathrm{2}} } {dx}\right)^{\mathrm{2}} \\ $$$${let}\:{use}\:{the}\:{diffeomorphism} \\ $$$$\begin{cases}{{x}\:={rcos}\theta}\\{{y}\:={rsin}\theta\:\:\:\:\:{we}\:{have}\:−\xi\leqslant{x}\leqslant\xi}\end{cases} \\ $$$${and}\:−\xi\leqslant{y}\leqslant\xi\:\Rightarrow\mathrm{0}\leqslant{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \leqslant\mathrm{2}\xi^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{0}\leqslant{r}\leqslant\xi\sqrt{\mathrm{2}}\:\Rightarrow \\ $$$${A}_{\xi} =\int_{\mathrm{0}} ^{\xi\sqrt{\mathrm{2}}} {r}\:{e}^{−{r}^{\mathrm{2}} } {dr}\:\int_{−\pi} ^{\pi} \:{d}\theta \\ $$$$=\mathrm{2}\pi\:\left[−\frac{\mathrm{1}}{\mathrm{2}}{e}^{−{r}^{\mathrm{2}} } \right]_{\mathrm{0}} ^{\xi\sqrt{\mathrm{2}}} \\ $$$$=\pi\left(\mathrm{1}−{e}^{−\mathrm{2}\xi^{\mathrm{2}} } \right)\:\Rightarrow{lim}_{\xi\rightarrow+\infty} {A}_{\xi} =\pi \\ $$$$=\left(\int_{−\infty} ^{+\infty} \:{e}^{−{x}^{\mathrm{2}} } {dx}\right)^{\mathrm{2}} \:\:{but}\int_{−\infty} ^{\infty} \:{e}^{−{x}^{\mathrm{2}} } {dx}>\mathrm{0} \\ $$$$\Rightarrow\int_{−\infty} ^{+\infty} \:{e}^{−{x}^{\mathrm{2}} } {dx}\:=\sqrt{\pi} \\ $$

Answered by bobhans last updated on 07/Oct/20

I=∫_(−∞) ^∞ e^(−x^2 )  dx ⇒I^2 =∫_(−∞) ^∞ ∫_(−∞) ^∞ e^(−x^2 −y^2 )  dx dy  I^2  = ∫_0 ^∞   ∫_0 ^(2π)  r e^(−r^2 )  dθ dr =∫_0 ^∞ ∣(re^(−r^2 )  (θ))∣_0 ^(2π)  )dr  I^2  =2π ∫_0 ^∞ re^(−r^2 )  dr = π ∫_0 ^∞  e^(−r^2 )  d(r^2 )  I^2  = −π ∣(e^(−r^2 ) )∣_0 ^∞ = −π(0−1)=π  Hence I = (√π)

$$\mathrm{I}=\underset{−\infty} {\overset{\infty} {\int}}\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } \:\mathrm{dx}\:\Rightarrow\mathrm{I}^{\mathrm{2}} =\underset{−\infty} {\overset{\infty} {\int}}\underset{−\infty} {\overset{\infty} {\int}}\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} } \:\mathrm{dx}\:\mathrm{dy} \\ $$$$\left.\mathrm{I}^{\mathrm{2}} \:=\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\:\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\int}}\:\mathrm{r}\:\mathrm{e}^{−\mathrm{r}^{\mathrm{2}} } \:\mathrm{d}\theta\:\mathrm{dr}\:=\underset{\mathrm{0}} {\overset{\infty} {\int}}\mid\left(\mathrm{re}^{−\mathrm{r}^{\mathrm{2}} } \:\left(\theta\right)\right)\mid_{\mathrm{0}} ^{\mathrm{2}\pi} \:\right)\mathrm{dr} \\ $$$$\mathrm{I}^{\mathrm{2}} \:=\mathrm{2}\pi\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\mathrm{re}^{−\mathrm{r}^{\mathrm{2}} } \:\mathrm{dr}\:=\:\pi\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\mathrm{e}^{−\mathrm{r}^{\mathrm{2}} } \:\mathrm{d}\left(\mathrm{r}^{\mathrm{2}} \right) \\ $$$$\mathrm{I}^{\mathrm{2}} \:=\:−\pi\:\mid\left(\mathrm{e}^{−\mathrm{r}^{\mathrm{2}} } \right)\mid_{\mathrm{0}} ^{\infty} =\:−\pi\left(\mathrm{0}−\mathrm{1}\right)=\pi \\ $$$$\mathrm{Hence}\:\mathrm{I}\:=\:\sqrt{\pi} \\ $$

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