Confusion over integral substitution

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.

• Sorry, I've changed the integral -- it now varies over $\mathbb{R}^d$, and the integrand contains Bessel functions. I would like to show that it converges, and also, if possible, estimate it. – user363087 Oct 31 '16 at 20:44
• Could you tell me how you know that it converges? I tried to bound the Bessel functions using a bound like $|J_{\nu}(z)| \leqslant C|z|^{-1/2},$ but I think that leads to a divergent integral (bounding the Bessel functions means that we cannot take advantage of any positive-negative cancellation that occurs). – user363087 Oct 31 '16 at 20:47
• In that case, how do you know it converges? – user363087 Oct 31 '16 at 20:51
• It's an asymptotic bound for Bessel functions of the first kind, with large argument. (In this example I'm interested in $\rho \rightarrow \infty.$) – user363087 Oct 31 '16 at 20:54
• So basically, you are trying to estimate this integral\begin{align} \int_{\mathbb{R}^d} \frac{dx}{|x|^{(d+1)/2}|x-b|^{(d+1)/2}}. \end{align} – Jacky Chong Oct 31 '16 at 20:55

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.

• Nice answer! And it seems that the case for $0 \leqslant |x| \leqslant \frac{1}{2}|b|$ is implied by your second piece, yes? – user363087 Oct 31 '16 at 23:07
• @user363087 Yes. If you draw out the picture in $\mathbb{R}^2$, you will see it does cover that region. – Jacky Chong Nov 1 '16 at 0:15
• Also, I notice that this proof is valid for all $d \geqslant 3.$ Does the integral still converge if $d = 2$? – user363087 Nov 1 '16 at 7:07
• Nope. The proof doesn't work for $d=2$ unless you have better bounds for your Bessel functions. – Jacky Chong Nov 1 '16 at 7:10