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Question Number 129975 by Algoritm last updated on 21/Jan/21

Answered by Olaf last updated on 21/Jan/21

$$\mathrm{\Omega }={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-{\left(\frac{x}{a}\right)}^2}dx$$$$\mathrm{\Omega }=2\mid a\mid {\int }_0^{+\mathrm{\infty }}{e}^{-u^2}du$$$$\mathrm{\Omega }=2\mid a\mid \frac{\sqrt{\pi }}{2}$$$$\mathrm{\Omega }=\mid a\mid \sqrt{\pi }$$

Answered by MJS_new last updated on 21/Jan/21

$$\underset{\mathbb{R}}{\int }{\mathrm{e}}^{-\frac{x^2}{4}}dx=2\underset{\mathbb{R}}{\int }{\mathrm{e}}^{-x^2}dx=2\sqrt{\pi }$$

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