Question Number 213999 by issac last updated on 24/Nov/24
$$\int\int…\int_{\:\mathcal{D}} \:\:{e}^{−\left({z}_{\mathrm{1}} ^{\mathrm{2}} +{z}_{\mathrm{2}} ^{\mathrm{2}} …+{z}_{{n}} ^{\mathrm{2}} \right)} \mathrm{da} \\ $$$$\mathcal{D}=\underset{\boldsymbol{\mathrm{n}}\:\boldsymbol{\mathrm{times}}} {\left[\mathrm{0},\infty\right)×\left[\mathrm{0},\infty\right)……\left[\mathrm{0},\infty\right)} \\ $$$$\int_{\mathrm{0}} ^{\:\pi} \:{e}^{−\mathrm{sin}^{\mathrm{2}} \left({z}\right)} \mathrm{d}{z} \\ $$$$\mathrm{help} \\ $$
Commented by mr W last updated on 24/Nov/24
$${it}'{s}\:{better}\:{if}\:{you}\:{post}\:{only}\:{one} \\ $$$${question}\:{in}\:{each}\:{question}. \\ $$
Answered by mr W last updated on 24/Nov/24
$${Q}\mathrm{1} \\ $$$$=\int_{\mathrm{0}} ^{\infty} {e}^{−{r}^{\mathrm{2}} } \frac{\mathrm{2}\pi^{\frac{{n}}{\mathrm{2}}} }{\Gamma\left(\frac{{n}}{\mathrm{2}}\right)}{r}^{{n}−\mathrm{1}} {dr} \\ $$$$=\frac{\mathrm{2}\pi^{\frac{{n}}{\mathrm{2}}} }{\Gamma\left(\frac{{n}}{\mathrm{2}}\right)}\int_{\mathrm{0}} ^{\infty} {e}^{−{r}^{\mathrm{2}} } {r}^{{n}−\mathrm{1}} {dr} \\ $$$$=\frac{\mathrm{2}\pi^{\frac{{n}}{\mathrm{2}}} }{\Gamma\left(\frac{{n}}{\mathrm{2}}\right)}×\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{{n}}{\mathrm{2}}\right) \\ $$$$=\pi^{\frac{{n}}{\mathrm{2}}} \\ $$