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Question Number 105155 by mohammad17 last updated on 26/Jul/20

Answered by mathmax by abdo last updated on 26/Jul/20

$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\pi {\mathrm{e}}^{-\frac{{\alpha }^2}{2}}\mathrm{d}\alpha =\pi {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{\mathrm{e}}^{-\frac{{\alpha }^2}{2}}\mathrm{d}\alpha {=}_{\frac{\alpha }{\sqrt{2}}=\mathrm{x}}\pi {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{x}}^2}\sqrt{2}\mathrm{dx}$$$$=\pi \sqrt{2}\times \sqrt{\pi }=\pi \sqrt{2\pi }$$

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