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Question Number 200802 by mnjuly1970 last updated on 23/Nov/23

Answered by witcher3 last updated on 23/Nov/23

$${\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{x}}^2}\mathrm{dx}=\frac{1}{2}{\int }_0^{\mathrm{\infty }}{\mathrm{t}}^{\frac{1}{2}-1}{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt},{\mathrm{x}}^2=\mathrm{t}$$$$=\frac{\mathrm{\Gamma }\left(\frac{1}{2}\right)}{2}$$$$\mathrm{\varnothing }={\left(\frac{\mathrm{\Gamma }\left(\frac{1}{2}\right)}{2}\right)}^3=\frac{\pi \sqrt{\pi }}{8}$$

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