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Question Number 213999 by issac last updated on 24/Nov/24

$$\int \int ...{\int }_{\mathcal{D}}{e}^{-(z_1^2+z_2^2...+z_n^2)}\mathrm{da}$$$$\mathcal{D}=\underset{\mathbf{n}\mathbf{times}}{[0,\mathrm{\infty })\times [0,\mathrm{\infty })......[0,\mathrm{\infty })}$$$${\int }_0^{\pi }{e}^{-{\sin }^2\left(z\right)}\mathrm{d}z$$$$\mathrm{help}$$

Commented by mr W last updated on 24/Nov/24

$${it}^{\prime }sbetterifyoupostonlyone$$$$questionineachquestion.$$

Answered by mr W last updated on 24/Nov/24

$$Q1$$$$={\int }_0^{\mathrm{\infty }}{e}^{-r^2}\frac{2{\pi }^{\frac{n}{2}}}{\mathrm{\Gamma }\left(\frac{n}{2}\right)}{r}^{n-1}dr$$$$=\frac{2{\pi }^{\frac{n}{2}}}{\mathrm{\Gamma }\left(\frac{n}{2}\right)}{\int }_0^{\mathrm{\infty }}{e}^{-r^2}{r}^{n-1}dr$$$$=\frac{2{\pi }^{\frac{n}{2}}}{\mathrm{\Gamma }\left(\frac{n}{2}\right)}\times \frac{1}{2}\mathrm{\Gamma }\left(\frac{n}{2}\right)$$$$={\pi }^{\frac{n}{2}}$$

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