Proof of an identity involving $\int \exp(-|x-s|)dx$ over an even sphere 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*}
 A: 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.
A: 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.
A: 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).
