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How would one show the following equality,

$$\int_{\mathbb{R}^d}|x|e^{-\frac\beta 2 |x|^2}dx = c_d\beta^{-\frac{d+1}{2}}.$$

Where $c_d$ is a constant only depending on $d$. This is problem is relatively easy to do when $d=1$ by either substitution or integration by parts, when we know the value of the Gaussian integral. My problem is that I'm not aware of any substitution techniques for higher dimensions and integration by parts (i.e. Greens identities) leads me to having to compute a higher dimensional surface integral, which, in arbitrary dimensions I do not know how to compute.

Finally tried brute forcing this and splitting it into $d$ integrals, each integral having the form, $$ \int_\mathbb R \sqrt{(x^2+a)}e^{-x^2}dx.$$

Which annoyingly wolfram alpha doesn't give me a useful answer for...

THIG
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  • Hints: spherical symmetry and solid angles in $R^d$. – A rural reader Mar 11 '21 at 17:27
  • The volume element for hyperspherical coordinates $(r,\varphi_1,\varphi_2,\dots,\varphi_{N-1})$ is given by $$dV= r^{d-1}\sin^{d-2}(\varphi_1)\sin^{d-3}(\varphi_2)\cdots \sin(\varphi_{d-2}), dr,d\varphi_1 , d\varphi_2\cdots d\varphi_{d-1}.$$ Due to spherical symmetry, integrating over all the angles gives rise to a constant (dependent on $d$, which turns out to be the surface area of a unit $(d-1)$-sphere), so you only need to consider the integral over $r$. – projectilemotion Mar 11 '21 at 17:33
  • I'm about to delete this question because its meant to be an inequality rather than an equality in which case it is easy. But I am curious what do you mean about solid angles? Also its been a very long time since I did any integrals explicitly any chance you could enlighten me on what to do with the spherical symmetry. – THIG Mar 11 '21 at 17:33
  • In that integral, change variables to reduce to the case $\beta=1$. When you do that, see if the multiplier you get is $\beta^{-(d+1)/2}$. – GEdgar Mar 12 '21 at 12:25

1 Answers1

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As @A rural reader and @projectilemotion mentioned, the appropriate way to evaluate this integral is to switch to spherical coordinates

$$I=\int_{\mathbb{R}^d}|x|e^{-\frac\beta 2 |x|^2}dx = \int..\int_{-\infty}^\infty\sqrt{x_1^2+...+x_d^2}\,e^{-\frac\beta 2 (x_1^2+...+x_d^2)}dx_1...dx_d=$$ $$= \int{d}\Omega\int_{0}^\infty{r}\,e^{-\frac\beta 2 r^2}r^{d-1}dr$$

where $\int{d}\Omega$ is integration over all angles in spherical coordinates, and $\int_0^\infty...{d}r$ is integration over radius.

We can find $\int{d}\Omega$ from

$\int..\int_{-\infty}^{\infty}e^{-x_1^2-x_2^2-...-x_d^2}dx_1...dx_d=$$\int{d}\Omega\int_0^{\infty}e^{-r^2}r^{d-1}dr \Rightarrow \pi^{\frac{d}{2}}=$$\frac{1}{2}$$\int{d}\Omega$$\int_0^{\infty}$$e^{-t}t^{\frac{d}{2}-1}dt$

Integral over all angles $\int{d}\Omega=\frac{2\pi^{\frac{d}{2}}}{\Gamma(\frac{d}{2})}$

$$I=\frac{2\pi^{\frac{d}{2}}}{\Gamma(\frac{d}{2})}\int_{0}^\infty{r}\,e^{-\frac\beta 2 r^2}r^{d-1}dr=\frac{2\pi^{\frac{d}{2}}}{\Gamma(\frac{d}{2})}(\frac{\beta}{2})^{-\frac{d+1}{2}}\int_{0}^\infty{e}^{-t^2}t^ddt=$$ $$=\frac{2\pi^{\frac{d}{2}}}{\Gamma(\frac{d}{2})}(\frac{\beta}{2})^{-\frac{d+1}{2}}\frac{1}{2}\int_{0}^\infty{e}^{-x}x^{\frac{d-1}{2}}dx$$

$$I=\pi^{\frac{d}{2}}\frac{\Gamma(\frac{d+1}{2})}{\Gamma(\frac{d}{2})}(\frac{\beta}{2})^{-\frac{d+1}{2}}$$

Svyatoslav
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