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Question Number 214366 by issac last updated on 06/Dec/24

$$f\left(\alpha \right)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\alpha x^2}\mathrm{d}x$$$$\int e^{-\alpha t^2}\mathrm{d}t=\frac{1}{2}\sqrt{\frac{\pi }{\alpha }}\cdot \mathrm{erf}\left(\sqrt{\alpha }t\right)+\mathrm{Const}$$$$\therefore {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\alpha t^2}\mathrm{d}t=\sqrt{\frac{\pi }{\alpha }},\alpha \in (0,\mathrm{\infty })$$

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