Question Number 214366 by issac last updated on 06/Dec/24
$${f}\left(\alpha\right)=\int_{−\infty} ^{\:\infty} \:{e}^{−\alpha{x}^{\mathrm{2}} } \mathrm{d}{x} \\ $$$$\int\:\:{e}^{−\alpha{t}^{\mathrm{2}} } \mathrm{d}{t}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\pi}{\alpha}}\centerdot\mathrm{erf}\left(\sqrt{\alpha}{t}\right)+\mathrm{Const} \\ $$$$\therefore\int_{−\infty} ^{\:\infty} \:{e}^{−\alpha{t}^{\mathrm{2}} } \mathrm{d}{t}=\sqrt{\frac{\pi}{\alpha}}\:,\:\alpha\in\left(\mathrm{0},\infty\right) \\ $$