I have the following $d$-dimensional integral:
$$\displaystyle \int_{\mathbb{R^d}} |x|^{-d/2}|b-x|^{-d/2}J_{d/2}(\rho |x|)J_{d/2}(\rho |b-x|)\mathrm{d}x,$$
where $|\cdot|$ denotes the Euclidean norm on $\mathbb{R}^d$, and $x, b \in \mathbb{R}^d$, $\rho > 0$ does not depend on $x$, and $J_{d/2}$ denotes the Bessel function of the first kind. My supervisor has suggested the following substitution, which appears to be a generalisation of polar co-ordinates, but I'm not entirely sure it makes sense. He has suggested letting
$$\displaystyle r = |(0,x_2,x_3, ... ,x_d)|, \ \ |x| = \sqrt{x_1^2 + r^2},$$ $$\displaystyle \mathrm{d}x = Cr^{d-2}\mathrm{d}x_1 \mathrm{d}r\mathrm{d}\varphi,$$
but I don't understand how this works. This does not seem to be hyperspherical co-ordinates, but I'm mainly confused how that expression for $\mathrm{d}x$ was derived. I also don't know where this $\varphi$ comes from, and what $\mathrm{d}\varphi$ actually means here. Can anyone tell me how this is derived? I have a strong suspicion that there is something left out here, and would greatly appreciate it if someone could tell me what it could be. What is especially confusing is that this seems to jump from an integral involving $d$ variables to one involving just three, unless $\varphi \in \mathbb{R}^{d-2}.$
For the integral itself, I'm mainly interested in proving its convergence, and estimating its value. But I'd like to understand this co-ordinate transformation first.
Let us consider the following integral \begin{align} \int_{\mathbb{R}^d} \frac{dx}{|x|^{(d+1)/2}|x-b|^{(d+1)/2}}. \end{align}
Let us first cut up $\mathbb{R}^d$ into three overlapping pieces.
The first piece is $\frac{1}{2}|b|<|x|<2|b|$.
The second piece is $\frac{1}{2}|b|<|x-b|<2|b|$.
And the last piece is $|x-b|>2|b|$ or $|x|>2|b| $.
For the first piece, we see that $|x-b|<3|b|$ since $\frac{1}{2}|b|<|x|<2|b|$ which means \begin{align} \int_{\frac{1}{2}|b|<|x|<2|b|} \frac{dx}{|x|^{(d+1)/2}|x-b|^{(d+1)/2}} \leq&\ C\int_{|x-b|<3|b|} \frac{dx}{|b|^{(d+1)/2}|x-b|^{(d+1)/2}}\\ =&\ \frac{C}{|b|^{(d+1)/2}} \int_{|x-b|<3|b|} \frac{dx}{|x-b|^{(d+1)/2}}\\ =&\ \frac{C}{|b|^{(d+1)/2}} \int^{3|b|}_0 \int_{|x-b|=r} \frac{dS}{|x-b|^{(d+1)/2}}\ dr\\ =&\ \frac{C}{|b|^{(d+1)/2}} \int^{3|b|}_0 \frac{r^{d-1}}{r^{(d+1)/2}}\ dr <\infty \end{align} if $d\geq 3$.
For the second piece, we see that $|x|<3|b|$ since $\frac{1}{2}|b|<|x-b|<2|b|$. Same calculation as above.
For the third piece, observe since $|x-b|>2|b|$ and $|x|>2|b|$, then it follows $|x|>|b|$ and $2|x-b|\geq |x-b|+|b| \geq |x|$. Hence it follows \begin{align} \int_{|x|>2|b| \text{ or } |x-b|>2|b|} \frac{dx}{|x|^{(d+1)/2}|x-b|^{(d+1)/2}} \leq&\ C\int_{|x|\geq |b|} \frac{dx}{|x|^{d+1}} =C \int^\infty_{|b|} \int_{|x|=r} \frac{dS}{r^{d+1}}\ dr\\ =&\ C\int^\infty_{|b|}\frac{r^{d-1}}{r^{d+1}}\ dr<\infty. \end{align}
Hence your integral converges.