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Let's have integral $$ I = \int \limits_{0}^{\text{T.P.}}dx_{1}...dx_{D}\sqrt{1 - \sum_{i}x_{i}^{a_{i}}}, \quad \text{T.P.}: \quad 1 - \sum_{i}x_{i}^{a_{i}} = 0 $$ By rewriting it in coordinates $x_{i}^{a_{i}} = y_{i}^{2}$ we have $$ \tag 1 I \sim \int \limits_{0}^{T.P.}\sqrt{1 - \sum_{i}y_{i}^{2}}\prod_{i = 1}^{D}dy_{i}y_{i}^{\frac{2 - a_{i}}{a_{i}}} $$ Is it possible to evaluate it by performing integration in n-spherical coordinate system? The problem is that after introducing them I get set of integrals like $$ \int \limits_{0}^{\pi}\sin^{\alpha_{j}}(\varphi_{j})cos^{\beta_{j}}(\varphi_{j})d\varphi_{j} $$ which are complex, so it seems that I have missed something.

Or, maybe, integral $I$ in new variables given by Eq. $(1)$ isn't written correctly, and I need to write something like $|x_{i}|^{a_{i}} = y_{i}^{2}$? Finally, is integral $I$ ill defined in zone where $x_{i} < 0$?

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    $\begingroup$ Consider $v(t)$ the volume of the domain $\sum\limits_{i=1}^D|x_i|^a\leqslant t$ then $v(t)=t^{D/a}v(1)$ by homogeneity and $$I=\int_{0}^{1}\sqrt{1-t}\,dv(t)=v(1)\int_{0}^{1}\sqrt{1-t}\,t^{D/a-1}\,dt=v(1)\,\mathrm{Bin}(1/2,D/a).$$ $\endgroup$
    – Did
    Oct 4, 2015 at 15:25
  • $\begingroup$ Your integration limits are wrong; this is what makes the result complex. You only want to integrate over the first quadrant $y_i\geq 0 \implies 0\leq\varphi_i \leq \frac{\pi}{2}$. $\endgroup$
    – Winther
    Oct 4, 2015 at 15:44
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    $\begingroup$ *Replace Bin$(1/2,D/a)$ in my previous comment by Beta$(3/2,D/a)\cdot D/a$. $\endgroup$
    – Did
    Oct 4, 2015 at 17:22

1 Answer 1

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Yes you can evaluate it using spherical coordinates, but it is a long and tedious calculation. The final result is fairly simple though

$$\color{red}{\int\limits_{0\leq x_i,~\sum_{i=1}^D x_i^{a_i}\leq 1}\sqrt{1-\sum_{i=1}^Dx_i^{a_i}}{\rm d}V = \frac{\sqrt{\pi}}{2}\frac{\prod_{i=1}^{D}\Gamma\left(a_i^{-1}\right)a_i^{-1}}{\Gamma\left(\frac{3}{2} + \sum_{i=1}^Da_i^{-1}\right)}}$$

so there are probably easier ways to derive this (see Did's comments).


Starting with the substitution you propose $x_i = y_i^{\frac{2}{a_i}}\implies {\rm d}x_i = \frac{2}{a_i}y_i^{\frac{2-a_i}{a_i}}{\rm d}y_i$ we get

$$I = \frac{2^D}{\prod_{i=1}^Da_i}\int\sqrt{1-\sum_{i=1}^D y_i^2}\prod_{i=1}^Dy_i^{\frac{2-a_i}{a_i}}{\rm d}y_i$$

Now changing to (hyper)spherical coordinates we find

$$\prod_{i=1}^{D}{\rm d}y_i = r^{D-1}{\rm d}r\prod_{i=1}^{D-1}\sin^{D-1-i}(\theta_i){\rm d}\theta_i$$ $$\prod_{i=1}^{D}y_i^{z_i} = r^{\sum_{j=1}^D z_j}\prod_{i=1}^{D-1}\cos^{z_i}(\theta_i)\sin^{\sum_{j=i+1}^{D}z_j}(\theta_i)$$

Since you only want to integrate over the first quadrant the integration limits for the angular variables are $\theta_i\in[0,\pi/2]$. This gives us the integral(s)

$$I = \frac{2^D}{\prod_{j=1}^Da_j}\int_0^1\sqrt{1-r^2}r^{\left(\sum_{i=1}^D \frac{2}{a_i}\right)-1}\,{\rm d}r\prod_{i=1}^{D-1}\int_0^{\frac{\pi}{2}}\cos^{\frac{2}{a_i}-2}(\theta_i)\sin^{\left(\sum_{j=i+1}^{D}\frac{2}{a_j}\right)-1}(\theta_i)\,{\rm d}\sin\theta_i$$

Perform a change of variables $\sin\theta_i\to z_i$ in the last $D-1$ integrals and evaluate using the $\beta$-function to get

$$I = \frac{2^D}{\prod_{j=1}^Da_j}\left[\frac{\Gamma\left(\frac{3}{2}\right)\Gamma \left(\sum_{i=1}^D \frac{1}{a_i}\right)}{2\Gamma\left(\sum_{i=1}^D \frac{1}{a_i}+\frac{3}{2}\right)}\right]\prod_{i=1}^{D-1}\frac{\Gamma\left(\frac{1}{a_i}\right) \Gamma \left(\sum_{j=i+1}^{D}\frac{1}{a_j}\right)}{2\Gamma\left(\sum_{j=i}^{D}\frac{1}{a_j}\right)}$$

which can be simplified down to

$$I= \frac{\sqrt{\pi}}{2}\frac{\prod_{i=1}^{D}\Gamma\left(a_i^{-1}\right)a_i^{-1}}{\Gamma\left(\frac{3}{2} + \sum_{i=1}^Da_i^{-1}\right)}$$

As a consistency check lets take $a_1=a_2=\ldots=a_D=2$. Then the integral is just $1/2^{D+1}$ of the volume of a $D+1$ dimensional unit sphere. The formula above gives $2^{D+1}I = \frac{\pi^{\frac{D+1}{2}}}{\Gamma\left(\frac{D+1}{2}+1\right)}$ which agrees with the known expression for the volume.

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  • $\begingroup$ well done (+1)! i would guess that some kind of center of mass coordinates simplify the calculations. $\endgroup$
    – tired
    Oct 4, 2015 at 18:27
  • $\begingroup$ @tired Thanks. I haven't tried that, but we loose the symmetry of the problem in that case so I would think that the integrational limits would be hard to deal with. $\endgroup$
    – Winther
    Oct 4, 2015 at 18:51

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