# Integral in d-dimensional spherical coordinates over $r_{x_1}>0$

I have the following $$d-$$dimensional integral $$\int_{r_{x_1}>0, \lvert r\rvert

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.

• What is the meaning of $e^x$ with a vector $x$? Mar 15, 2020 at 12:28
• I meant $e^{\lvert x\rvert}$, since $r$ is the radial function so I assumed it was understood. I will edit the question anyway. Mar 15, 2020 at 13:19
• @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. Mar 15, 2020 at 13:50

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.
• 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 $z^{d+1}$? Mar 15, 2020 at 16:32
• 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? Mar 15, 2020 at 16:40
• 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 $0<r<r_0$?). Mar 15, 2020 at 17:03
• 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 $d^{d-1}\mathbf{y} = S_{d-2}y^{d-2}dy$. Mar 15, 2020 at 17:14
• @Indeterminate: Looks like something was broken in my head that day ;) Your remark on $y^{d-2}$ is quite right. Fixed. Mar 22, 2020 at 5:10