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](https://cloud.sagemath.com/projects/49e1082f-34a4-4d89-8776-af032ad7caeb/files/Verifying%20an%20Identity%20Involving%20an%20Integral%20over%20an%20Even%20Sphere%20-%20mathoverflow.sagews) 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](https://mathoverflow.net/questions/239531/the-integral-of-exp-x-a-over-an-even-dimensional-sphere) 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](http://dlmf.nist.gov/10.49) 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](http://dlmf.nist.gov/10.49) 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\*}



Proof of an identity involving $\int \exp(-|x-s|)dx$ over an even sphere
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](https://arxiv.org/abs/1708.03227). [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](https://mathoverflow.net/questions/253902/a-bessel-function-integral-identity-involving-int-0-pi-frack-j-1-2ww).


