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I have the following $d-$dimensional integral $$\int_{r_{x_1}>0, \lvert r\rvert<r_0} \frac{r_{x_{1}}}{1+e^{\lvert r \rvert-\eta}} d^d\mathbf{r}$$

where $r \in \mathbb{R}^d$, $r_{x_i}$ are the cartesian coordinates, $\eta$ is a constant, $r_0$ is the fixed radius of the ball to integrate upto and $d^d\mathbf{r}$ is the volume element in $d-$dimensional space. Is there a way to convert this integral into just a function of the radius $r$? I want to eliminate out all the integrals over angular coordinates $\phi_1, \phi_2,\cdot\cdot\cdot,\phi_{d-1}$ (which should probably turn out to be a constant) and be just left with an integral over $r$.

Note that $r_{x_1} = r \cos(\phi_1)$. I am not sure how the limits and the integral would work out.

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  • $\begingroup$ What is the meaning of $e^x$ with a vector $x$? $\endgroup$
    – metamorphy
    Commented Mar 15, 2020 at 12:28
  • $\begingroup$ I meant $e^{\lvert x\rvert}$, since $r$ is the radial function so I assumed it was understood. I will edit the question anyway. $\endgroup$ Commented Mar 15, 2020 at 13:19
  • $\begingroup$ @metamorphy My apologies. I missed a $1+$ in the denominator. It is of the form of a PolyLog function but that I can take care of once the angular parts have been eliminated. I am not able to eliminate the angular portion of the integral. Thanks for attempting to answer. $\endgroup$ Commented Mar 15, 2020 at 13:50

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With an "arbitrary" function $f(r)$ in place of $r\mapsto 1/(1+e^{r-\eta})$, representing $\mathbf{r}\in\mathbb{R}^d$ by $(x,\mathbf{y})$, where $x\in\mathbb{R}$ is that very $r_{x_1}$, and $\mathbf{y}\in\mathbb{R}^{d-1}$, the integral is equal to $$I=\int_0^{r_0}\int_{|\mathbf{y}|<\sqrt{r_0^2-x^2}}xf\left(\sqrt{x^2+|\mathbf{y}|^2}\right)\,\mathrm{d}^{d-1}\mathbf{y}\,\mathrm{d}x,$$ and the "angular elimination" is applicable to the integration over $\mathbf{y}$: $$I=S_{d-1}\int_0^{r_0}\int_0^\sqrt{r_0^2-x^2}xy^{d-2}f\left(\sqrt{x^2+y^2}\right)\mathrm{d}y\,\mathrm{d}x,$$ where $S_n$ is the area of the unit sphere in $\mathbb{R}^n$ (for this step to be correct in the case $d=2$, we must agree that $S_1=2$). Substituting "back" $\sqrt{x^2+y^2}=z$, we get \begin{align}I&=S_{d-1}\int_0^{r_0}\int_x^{r_0}xz(z^2-x^2)^{(d-3)/2}f(z)\,\mathrm{d}z\,\mathrm{d}x\\\color{gray}{[\text{exchange integrations}]}\quad&=S_{d-1}\int_0^{r_0}\int_0^z xz(z^2-x^2)^{(d-3)/2}f(z)\,\mathrm{d}x\,\mathrm{d}z\\\color{gray}{[\text{integrate over $x$}]}\quad&=\frac{S_{d-1}}{d-1}\int_0^{r_0}z^d f(z)\,\mathrm{d}z.\end{align} Up to diving into special functions, this is the most closed form I believe.

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    $\begingroup$ Thanks for the lucid answer. I was wondering if instead of $r_{x_1}$ in the numerator of integrand, we had $\frac{r_{x_1}}{\lvert r \rvert}$, would the final result just have a $z^{d}$ instead of $z^{d+1}$? $\endgroup$ Commented Mar 15, 2020 at 16:32
  • $\begingroup$ Also just as a sanity check, I tried solving the integral in $2-$dimensions and the result doesn't look as expected. In $2-$D, wouldn't the integral just be $\int_{|r|<r_0}\int_{-\pi/2}^{\pi/2} \frac{r^2 \cos(\theta)}{1+e^{r-\eta}}d\theta dr = 2\int_{0}^{r_0}\frac{r^2 dr}{1+e^{r-\eta}} $. Doesn't it look different from your result? $\endgroup$ Commented Mar 15, 2020 at 16:40
  • $\begingroup$ Regarding $|r|$ in the denominator - yes. (I should have written the answer considering an arbitrary radial function - I think I have to edit it.) As for the 2D case, under the agreement that $S_1=2$ (forget the geometry...), the result is half of yours, and this is correct (why "$|r|<r_0$" and not $0<r<r_0$?). $\endgroup$
    – metamorphy
    Commented Mar 15, 2020 at 17:03
  • $\begingroup$ But I have also mentioned the condition that $r_{x_1}>0$. Anyway I think I found a tiny error in your solution, which has caused this problem. When you write simplification of $d^{d-1}\mathbf{y}$, it should be $d^{d-1}\mathbf{y} = S_{d-2}y^{d-2}dy$. $\endgroup$ Commented Mar 15, 2020 at 17:14
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    $\begingroup$ @Indeterminate: Looks like something was broken in my head that day ;) Your remark on $y^{d-2}$ is quite right. Fixed. $\endgroup$
    – metamorphy
    Commented Mar 22, 2020 at 5:10

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