I see an $n$-dimensional Gaussian integrals like this:$$\int_{-\infty}^\infty d^m le^{-\alpha l^2}=\int_0^\infty dl l^{m-1} S_m e^{-\alpha l^2},\,S_m:=\frac{2\pi^{m/2}}{\Gamma\left(\frac12 m\right)}$$Please show me how they do that?
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I've replaced your images with MathJax versions thereof, but there ia sn error in the first equation: its left-hand side should integrate over $\Bbb R^m$, not $\Bbb R^+$. – J.G. Oct 27 '19 at 21:15
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hi how can I use MathJax like You? – Hossein M Oct 29 '19 at 08:41
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Possibly the most common link here in comments. – J.G. Oct 29 '19 at 09:02
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The first equation transforms an $m$-dimensional integral from Cartesian coordinates to polar ones, so $d^m\vec{l}=l^{m-1}dld\Omega$, where $d\Omega$ abbreviates all dependence on angles. For example, $m=2\implies d\Omega=d\theta$, while $m=3\implies d\Omega=\sin\theta d\theta d\phi$.
In this case, one further piece of information is used: the original integrand is a function of $l:=\vec{l}$ alone, so the polar angle is separable as $\int_0^\infty l^{m-1}e^{-\alpha l^2}dl$ times an integral over $\Omega$. (By the way, usually we write $r,\,\vec{r}$ instead of $l,\,\vec{l}$.)
The $\Omega$ integral is the value of $S_m$: it's the measure of the $(m-1)$-dimensional surface of a unit-radius $m$-dimensional ball. See here for a computation.
J.G.
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