I want to prove the following identity calculating the integral of an exponential over an even dimensional sphere in terms of functions $\chi_i(R)$ and $\tilde\psi_i(s)$ (described below) which are essentially modified spherical Bessel functions.

First set up some notation. Let $p>0$ be an integer and $R>0$. Let $S_R^{2p}\subset \mathbb{R}^{2p+1}$ denote the radius $R$ sphere. Let $\mathrm{s}\in \mathbb{R}^{2p+1}$ be a point inside the sphere and let $s=|\mathrm{s}|$ so $0\le s<R$. Finally, let $\omega_n$ be the volume of the unit $n$-ball. I want to show the following. $$ \frac{1}{(2p+1)!\,\omega_{2p+1}} \int_{\mathrm{x}\in S_R^{2p}} e^{-\left|\mathrm{x}-\mathrm{s}\right|}\,\mathrm{d}\mathrm{x} = \frac{(-1)^p e^{-R}}{2^p p!}\sum_{i=0}^p \binom{p}{i}\chi_{p+i}(R)\tilde\psi_i(s). $$

I can get Sage to check this is true up to, say $p=18$, using the integral form of the left hand side given below. I had previously asked for a reference for the integral, but as none was forthcoming I've posted the whole statement in the hope of a proof!

Here $(\chi_i)_{i=0}^\infty$ denote the sequence of 'reverse Bessel polynomials', so that $\chi_i(R)$ is a degree $i$ integer polynomial in $R$. The sequence begins as follows: \begin{align*} \chi_0(R)&=1;\\ \chi_1(R)&=R;\\ \chi_2(R)&=R^2+R;\\ \chi_3(R)&=R^3+3R^2+3R. %;\\ %\chi_4(R)&=R^{4} + 6 R^{3} + 15 R^{2} + 15 R. %\chi_5(R)&=R^{5} + 10 R^{4} + 45 R^{3} + 105 R^{2} + 105 R \end{align*} There are many ways to define this sequence, but we can take the recursion relation as a definition: $$ \chi_{i+2}(R)=R^2\chi_i(R)+(2i+1)\chi_{i+1}(R). $$

These functions are related to the modified spherical Bessel functions by $\chi_i(R)=\frac{2}{\pi} e^R R^{i+1}k_{i-1}(R)$. I will describe an integral form at the bottom.

The sequence $(\tilde\psi_i)_{i=0}^\infty$ denotes the sequence of functions $\mathbb{R}\to \mathbb{R}$ which begins in the following way: \begin{align*} \tilde\psi_0(s)&=\cosh(s);\\ \tilde\psi_1(s) &=-\frac{\sinh\left(s\right)}{s};\\ \tilde\psi_2(s) &=\frac{\cosh\left(s\right)}{s^{2}} - \frac{\sinh\left(s\right)}{s^{3}};\\ \tilde\psi_3(s) &= -\frac{\sinh\left(s\right)}{s^{3}} + \frac{3 \, \cosh\left(s\right)}{s^{4}} - \frac{3 \, \sinh\left(s\right)}{s^{5}}. %;\\ %\tilde\psi_4(s) %&= %\frac{\cosh\left(s\right)}{s^{4}} - \frac{6 \, \sinh\left(s\right)}{s^{5}} + \frac{15 \, \cosh\left(s\right)}{s^{6}} - %\frac{15 \, \sinh\left(s\right)}{s^{7}}. \end{align*} You can take the following recursion relation as a definition. $$ \tilde\psi_{i+1}(s)=-\frac{1}{s}\frac{\mathrm d \tilde\psi_i(s)}{\mathrm{d} s}. $$ These functions are related to the modified spherical Bessel functions by $\tilde\psi_i(s)=(-1)^i i^{(1)}_{i-1}(s)/s^{i-1}$. Again there are integral forms which I will give at the bottom.

The left hand side of the identity I want to prove can be expressed as a line integral as follows which makes it more tractable. \begin{multline*} \frac{1}{(2p+1)!\,\omega_{2p+1}} \int_{\mathrm{x}\in S^{2p}} e^{-\left|\mathrm{x}-\mathrm{s}\right|}\,\mathrm{d}\mathrm{x}\\ =\frac{ R^{2p}}{p!(p-1)!(2)^{2p}}\int_{\theta=0}^{\pi}e^{-\sqrt{R^2+s^2-2Rs\cos \theta}}\sin^{2p-1}\theta\,\mathrm{d}\theta\\ =\frac{ R}{p!(p-1)!(4s)^{2p-1}}\int_{\rho=R-s}^{R+s}e^{-\rho}((R+s)^2-\rho^2)^{p-1} (\rho^2-(R-s)^2)^{p-1}\rho\,\mathrm{d}\rho. \end{multline*}

For $i\ge 1$ and $R, s>0$ we have the following integral forms for the functions. \begin{align*} \chi_i(R) &= \frac{e^R R^{2i}}{2^{i-1} (i-1)!}\int_{t=0}^\infty e^{-R\cosh t}\sinh^{2i-1}t\, \mathrm d t\\ &= \frac{e^R R}{2^{i-1} (i-1)!}\int_{y=R}^\infty e^{-y}(y^2-R^2)^{i-1}\, \mathrm d y. \end{align*} \begin{align*} \tilde\psi_i(s) &= \frac{(-1)^{i}}{2^{i} (i-1)!}\int_{\theta=0}^\pi e^{s\cos \theta}\sin^{2i-1}\theta\, \mathrm d \theta\\ &= \frac{(-1)^{i}}{2^{i} (i-1)!s^{2i-1}}\int_{x=-s}^s e^{x}(s^2-x^2)^{i-1}\, \mathrm d x. \end{align*}


3 Answers 3


Following on from the answer of Sam Dolan, I generalized my conjecture to the following, which I prove as Theorem 4 in my paper The magnitude of odd balls via Hankel determinants of reverse Bessel polynomials. [The statement in the question is case where $j=0$.]

Theorem. For $0\le j \le p$, $\mathbf{s}\in \mathbb{R}^{2p+1}$, with $s=|\mathbf{s}|$ and $R>s\ge 0$ \begin{equation*} \int_{\mathbf{x}\in S^{2p}_R} \psi_j(\left | \mathbf{x}-\mathbf{s}\right|)\,\mathrm{d}\mathbf{x} = (-2\pi)^p 2 e^{-R}\sum_{i=0}^{p-j} \binom{p-j}{i}\chi_{i+p}(R) \tilde\psi_{i+j}(s). %\label{eq:GeneralizedKeyIntegral} \end{equation*}

Here $\psi_i$ (as opposed to $\tilde\psi_i$) can be defined by $\psi_i(r)=r^{-2i}e^{-r}\chi_i(r)$; it is related to a modified spherical Bessel function of the second kind.

The sequence of such functions begins in the following way: \begin{align*} \psi_0 (r) &= e^{-r}\\[0.5em] \psi_1 (r) &= e^{-r} \left( \frac{1}{r} \right)\\ \psi_2 (r) &= e^{-r} \left( \frac{1}{r^{2}} + \frac{1}{r^{3}} \right)\\ \psi_3 (r) &= e^{-r} \left( \frac{1}{r^{3}} + \frac{3}{r^{4}} + \frac{3}{r^{5}} \right)\\ \psi_4 (r) &= e^{-r} \left( \frac{1}{r^{4}} + \frac{6}{r^{5}} + \frac{15}{r^{6}} + \frac{15}{r^{7}} \right) . \end{align*}

This theorem is stated in terms of Bessel functions in another question.


Let $L_p$ and $R_p$ be the left and right hand sides. Put $L(z)=\sum_{p=0}^\infty L_p .(z/R)^{2p}$ and similarly for $R(z)$; it will suffice to show that $L(z)=R(z)$. Using the first integral form for $L_p$ one can show that $$ L(z)= \int_{\theta=0}^\pi\frac{z}{2} I_1(\sin(\theta)z) e^{-\sqrt{R^2+s^2-2Rs\cos(\theta)}} \,d\theta, $$ where $I_1$ is a Bessel function. That's not a huge step forwards, but perhaps it is worth something.


My suggestion would be to try replacing the exponential term in the line integral (second line after "... makes it more tractable") with the identity 10.2.35 in Abram&Stegun (http://people.math.sfu.ca/~cbm/aands/page_445.htm). The RHS of that identity is an infinite sum of the product of modified spherical Bessel functions i_n and k_n, and Legendre polynomials in z=cos(theta). My hunch is that the line integral over z could then be evaluated (after swapping the order of integral and sum).

  • $\begingroup$ Thanks! So it seems like 'addition formulas' are the relevant identities. It looks like there are also addition formulas that involve the shift in one index from i to p+i as I have in the right hand side of the thing I'm trying to prove. This is probably worth looking further into. $\endgroup$ Oct 5, 2016 at 9:38
  • $\begingroup$ Hmm. After initial optimism, this approach doesn't seem to help. $\endgroup$ Oct 8, 2016 at 12:27

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