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\begin{document}
\noindent Author: Jack Ebersole \\
Institution: Northern Illinois University\\
Date: February 6, 2025\\
\noindent\textbf{Problem statement:} Compute the following integral over the surface of a sphere of radius $R$:
\begin{align*}
    I = \oint_S \frac{\cos(\theta^\prime)}{|\vec{x} - \vec{x}^\prime|}dS
\end{align*}
\textbf{Answer:}
\begin{align*}
    I(r \leq R, \theta) &= \frac{4\pi r\cos\theta}{3}\\
    I(r \geq R, \theta) &= \frac{4\pi R^3\cos\theta}{3r^2}
\end{align*}
\textbf{Proof:}
First expand the relative distance $|\vec{x} - \vec{x}^\prime|$ in the denominator:
\begin{align*}
    \vec{x} &= r\hat{r}\\
    \vec{x}^\prime &= R \hat{r}^\prime \text{    because the source points are on the sphere of radius R}\\
    I &= \oint_S \frac{\cos(\theta^\prime)}{|\vec{x} - \vec{x}^\prime|}dS\\
    &= \int_0^{2\pi} \int_0^\pi \frac{\cos(\theta^\prime)}{|\vec{x} - \vec{x}^\prime|} R^2\sin(\theta^\prime)d\theta^\prime d\phi^\prime\\
    &= \int_0^{2\pi}\int_0^\pi \frac{\cos(\theta^\prime)}{|\vec{x} - \vec{x}^\prime|} R^2\sin(\theta^\prime)d\theta^\prime d\phi^\prime \\
    |\vec{x} - \vec{x}^\prime| &= \sqrt{\vec{x}\cdot \vec{x} + \vec{x}^\prime \cdot \vec{x}^\prime - 2\vec{x}\cdot \vec{x}^\prime}\\
    &= \sqrt{r^2 + R^2 - 2rR\left(\hat{r}\cdot\hat{r}^\prime\right)}\\
    \therefore I &= \int_0^{2\pi}\int_0^\pi\frac{\cos(\theta^\prime)}{\sqrt{r^2 + R^2 - 2rR\left(\hat{r}\cdot\hat{r}^\prime\right)}}R^2\sin(\theta^\prime)d\theta^\prime d\phi^\prime
\end{align*}

\noindent Since $I$ has to be finite everywhere, it cannot diverge as $r \to 0$ or $r \to \infty$. The only way to achieve this is to break $I$ into two piecewise continuous functions for $r \leq R$ and $r \geq R$. We'll do $I(r \leq R)$ first. It needs to not diverge as $r \to 0$, so pull $R$ out of the denominator so that the integrand in terms of Legendre polynomials is a sum of powers of $\frac{r}{R}$, all of which vanish at $r=0$ and are continuous everywhere else for $r \leq R$:

\begin{align*}
    I(r \leq R) &= \int_0^{2\pi}\int_0^\pi\frac{\cos(\theta^\prime)}{R\sqrt{1 + \frac{r^2}{R^2} - 2\frac{r}{R}\left(\hat{r}\cdot\hat{r}^\prime\right)}}R^2\sin(\theta^\prime)d\theta^\prime d\phi^\prime\\
    &= R\int_0^{2\pi}\int_0^\pi\cos(\theta^\prime)\frac{1}{\sqrt{1 + \frac{r^2}{R^2} - 2\frac{r}{R}\left(\hat{r}\cdot\hat{r}^\prime\right)}}\sin(\theta^\prime)d\theta^\prime d\phi^\prime\\
    &= R\int_0^{2\pi}\int_0^\pi\cos(\theta^\prime)\left(\sum_{\ell = 0}^\infty P_\ell\left(\hat{r}\cdot\hat{r}^\prime\right)\left(\frac{r}{R}\right)^\ell\right)\sin(\theta^\prime)d\theta^\prime d\phi^\prime\\
    &= \sum_{\ell = 0}^\infty\frac{r^\ell}{R^{\ell - 1}} \int_0^{2\pi}\int_0^\pi \cos(\theta^\prime)P_\ell\left(\hat{r}\cdot\hat{r}^\prime\right) \sin(\theta^\prime)d\theta^\prime d\phi^\prime\\
    &= \sum_{\ell = 0}^\infty\frac{r^\ell}{R^{\ell - 1}} \int_0^{2\pi}\int_0^\pi \cos(\theta^\prime)\left(\frac{4\pi}{2\ell + 1}\sum_{m = -\ell}^\ell Y_{\ell, m}(\theta, \phi)Y^*_{\ell, m}(\theta^\prime, \phi^\prime)\right)\sin(\theta^\prime)d\theta^\prime d\phi^\prime\\
    &= \sum_{\ell = 0}^\infty\sum_{m = -\ell}^\ell\frac{4\pi}{2\ell + 1}\frac{r^\ell}{R^{\ell - 1}} Y_{\ell, m}(\theta, \phi)\int_0^{2\pi}\int_0^\pi \cos(\theta^\prime)Y^*_{\ell, m}(\theta^\prime, \phi^\prime)\sin(\theta^\prime)d\theta^\prime d\phi^\prime
\end{align*}
There are different conventions for defining the spherical harmonics in terms of the Legendre polynomials and sometimes the factor of $\frac{4\pi}{2\ell + 1}$ in the expansion of $P_\ell\left(\hat{r}\cdot\hat{r^\prime}\right)$ is absorbed into the spherical harmonics as a factor of $\sqrt{4\pi/(2\ell + 1)}$ rather than written out explicitly. This will not effect the final result because it will be canceled out by the convention that's chosen for the normalization condition. Your book may use a different convention.\\

Now use the fact that $\cos(\theta^\prime)$ is itself a spherical harmonic, specifically $\cos(\theta^\prime) = \sqrt{\frac{4\pi}{3}}Y_{1,0}(\theta^\prime, \phi^\prime)$:
\begin{align*}
    I(r \leq R) &= \sqrt{\frac{4\pi}{3}}\sum_{\ell = 0}^\infty\sum_{m = -\ell}^\ell\frac{4\pi}{2\ell + 1}\frac{r^\ell}{R^{\ell - 1}} Y_{\ell, m}(\theta, \phi)\int_0^{2\pi}\int_0^\pi Y_{1,0}(\theta^\prime, \phi^\prime)Y^*_{\ell, m}(\theta^\prime, \phi^\prime)\sin(\theta^\prime)d\theta^\prime d\phi^\prime
\end{align*}
Spherical harmonic functions are orthonormal. This means that:
\begin{align*}
    \int_0^{2\pi}\int_0^\pi Y_{a,b}(\theta^\prime, \phi^\prime)Y^*_{c, d}(\theta^\prime, \phi^\prime)\sin(\theta^\prime)d\theta^\prime d\phi^\prime = \delta_{ac}\delta_{bd},
\end{align*}
where $\delta_{ij}$ is the Kronecker delta symbol:
\begin{align*}
    \delta_{ij} = \begin{cases}
        1 \text{ if } i &= j\\
        0 \text{ if } i &\neq j
    \end{cases}
\end{align*}
As previously noted, there exist other conventions for orthonormality.  Your book should tell you which one it is assuming. In any case, the result is:

\begin{align*}
    I(r \leq R) &= \sqrt{\frac{4\pi}{3}}\sum_{\ell = 0}^\infty\sum_{m = -\ell}^\ell\frac{4\pi}{2\ell + 1}\frac{r^\ell}{R^{\ell - 1}}Y_{\ell, m}(\theta, \phi)\delta_{1, \ell}\delta_{0, m}\\
    &= \sqrt{\frac{4\pi}{3}}\frac{4\pi r}{3}Y_{1, 0}(\theta, \phi)\\
    &= \sqrt{\frac{4\pi}{3}}\frac{4\pi r}{3}\sqrt{\frac{3}{4\pi}}\cos\theta\\
    &= \frac{4\pi r\cos\theta}{3}
\end{align*}
Now use the same procedure for $r \geq R$, but this time pull $r$ out of the square root in the denominator:
\begin{align*}
    I(r\geq R) &= \int_0^{2\pi}\int_0^\pi\frac{\cos(\theta^\prime)}{\sqrt{r^2 + R^2 - 2rR\left(\hat{r}\cdot\hat{r}^\prime\right)}}R^2\sin(\theta^\prime)d\theta^\prime d\phi^\prime\\
    &= \frac{R^2}{r}\int_0^{2\pi}\int_0^\pi\cos(\theta^\prime)\frac{1}{\sqrt{1 + \frac{R^2}{r^2} - 2\frac{R}{r}\left(\hat{r}\cdot\hat{r}^\prime\right)}}\sin(\theta^\prime)d\theta^\prime d\phi ^\prime\\
    &= \frac{R^2}{r}\int_0^{2\pi}\int_0^\pi\cos(\theta^\prime)\left(\sum_{\ell = 0}^\infty P_\ell\left(\hat{r}\cdot \hat{r}^\prime\right)\left(\frac{R}{r}\right)^\ell\right)\sin(\theta^\prime)d\theta^\prime d\phi^\prime \\
    &= \sum_{\ell = 0}^\infty \frac{R^{\ell + 2}}{r^{\ell + 1}}\int_0^{2\pi}\int_0^\pi\cos(\theta^\prime)P_\ell\left(\hat{r}\cdot\hat{r}^\prime\right)\sin(\theta^\prime)d\theta^\prime d\phi^\prime\\
    &= \sum_{\ell = 0}^\infty \frac{R^{\ell + 2}}{r^{\ell + 1}}\int_0^{2\pi}\int_0^\pi\cos(\theta^\prime)\left(\frac{4\pi}{2\ell + 1}\sum_{m = -\ell}^\ell Y_{\ell, m}(\theta, \phi)Y^*_{\ell, m}(\theta^\prime, \phi^\prime)\right)\sin(\theta^\prime)d\theta^\prime d\phi^\prime\\
    &= \sum_{\ell = 0}^\infty\sum_{m = -\ell}^\ell \frac{4\pi}{2\ell + 1}\frac{R^{\ell + 2}}{r^{\ell + 1}}Y_{\ell, m}(\theta, \phi)\int_0^{2\pi}\int_0^\pi \cos(\theta^\prime)Y^*_{\ell,,m}(\theta^\prime, \phi^\prime)d\theta^\prime \sin(\theta^\prime)d \phi^\prime \\
    &= \sqrt{\frac{4\pi}{3}}\sum_{\ell = 0}^\infty\sum_{m = -\ell}^\ell \frac{4\pi}{2\ell + 1}\frac{R^{\ell + 2}}{r^{\ell + 1}}Y_{\ell, m}(\theta, \phi)\int_0^{2\pi}\int_0^\pi Y_{1,0}(\theta^\prime, \phi^\prime)Y^*_{\ell,,m}(\theta^\prime, \phi^\prime)\sin(\theta^\prime)d\theta^\prime d \phi^\prime \\
    &= \sqrt{\frac{4\pi}{3}}\sum_{\ell = 0}^\infty\sum_{m = -\ell}^\ell \frac{4\pi}{2\ell + 1}\frac{R^{\ell + 2}}{r^{\ell + 1}}Y_{\ell, m}(\theta, \phi)\delta_{1, \ell}\delta_{0, m}\\
    &= \sqrt{\frac{4\pi}{3}}\frac{4\pi}{3}\frac{R^3}{r^2}\sqrt{\frac{3}{4\pi}}\cos\theta\\
    \therefore I(r \geq R) &= \frac{4\pi R^3\cos\theta}{3r^2}
\end{align*}
\end{document}