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

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

Answered by Bird last updated on 07/Oct/20

$$let{A}_{\xi }=\int {\int }_{]-\xi ,\xi [{}^2}{e}^{-x^2-y^2}dxdy$$$$wehsve{lim}_{\xi \to +\mathrm{\infty }}{A}_{\xi }={\left({\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-x^2}dx\right)}^2$$$$letusethediffeomorphism$$$$\{\begin{array}{l}x=rcos\theta \\ y=rsin\theta wehave-\xi ⩽x⩽\xi \end{array}$$$$and-\xi ⩽y⩽\xi \Rightarrow 0⩽x^2+y^2⩽2{\xi }^2$$$$\Rightarrow 0⩽r⩽\xi \sqrt{2}\Rightarrow$$$$A_{\xi }={\int }_0^{\xi \sqrt{2}}r{e}^{-r^2}dr{\int }_{-\pi }^{\pi }d\theta$$$$=2\pi {[-\frac{1}{2}{e}^{-r^2}]}_0^{\xi \sqrt{2}}$$$$=\pi (1-e^{-2{\xi }^2})\Rightarrow {lim}_{\xi \to +\mathrm{\infty }}{A}_{\xi }=\pi$$$$={\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-x^2}dx\right)}^{2}but{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-x^2}dx>0$$$$\Rightarrow {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-x^2}dx=\sqrt{\pi }$$

Answered by bobhans last updated on 07/Oct/20

$$\mathrm{I}=\underset{-\mathrm{\infty }}{\stackrel{\mathrm{\infty }}{\int }}{\mathrm{e}}^{-{\mathrm{x}}^2}\mathrm{dx}\Rightarrow {\mathrm{I}}^2=\underset{-\mathrm{\infty }}{\stackrel{\mathrm{\infty }}{\int }}\underset{-\mathrm{\infty }}{\stackrel{\mathrm{\infty }}{\int }}{\mathrm{e}}^{-{\mathrm{x}}^2-{\mathrm{y}}^2}\mathrm{dx}\mathrm{dy}$$$${\mathrm{I}}^2=\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\underset{0}{\stackrel{2\pi }{\int }}\mathrm{r}{\mathrm{e}}^{-{\mathrm{r}}^2}\mathrm{d}\theta \mathrm{dr}=\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\mid \left({\mathrm{re}}^{-{\mathrm{r}}^2}\left(\theta \right)\right){\mid }_0^{2\pi })\mathrm{dr}$$$${\mathrm{I}}^2=2\pi \underset{0}{\stackrel{\mathrm{\infty }}{\int }}{\mathrm{re}}^{-{\mathrm{r}}^2}\mathrm{dr}=\pi \underset{0}{\stackrel{\mathrm{\infty }}{\int }}{\mathrm{e}}^{-{\mathrm{r}}^2}\mathrm{d}\left({\mathrm{r}}^2\right)$$$${\mathrm{I}}^2=-\pi \mid \left({\mathrm{e}}^{-{\mathrm{r}}^2}\right){\mid }_0^{\mathrm{\infty }}=-\pi (0-1)=\pi$$$$\mathrm{Hence}\mathrm{I}=\sqrt{\pi }$$

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