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$$r>1$$

The following is applying gauss' law by explicit integration.

$$\int_0^{\pi } \frac{\sin (\theta )}{\sqrt{r^2-2 r \cos (\theta )+1}} \, d\theta=\frac{2}{r}$$

The following is finding the potential of a charged spherical shell by explicit integration. Mathematica takes forever to solve this and I cannot figure out how it would be done.

$$\int _0^{2 \pi }\int _0^{\pi }\frac{\sin (\theta ) \cos (\theta )}{\sqrt{r^2-2 r (\cos (\psi )\cos (\theta ) +\sin (\psi )\sin (\theta ) \cos (\phi ))+1}}d\theta d\phi=\frac{4 \pi }{3}\frac{\cos(\psi)}{r^2}$$

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  • $\begingroup$ Actually, your first integral's evaluation is incomplete: see my question here. $\endgroup$ – Chappers Oct 1 '15 at 16:46
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    $\begingroup$ The divergence theorem (en.wikipedia.org/wiki/Divergence_theorem) may give a great help. $\endgroup$ – Jack D'Aurizio Oct 1 '15 at 17:07
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    $\begingroup$ Why is there a $\cos \theta$ term in the numerator? $\endgroup$ – Mark Viola Oct 1 '15 at 18:01
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    $\begingroup$ The integrand; why is the term $\cos \theta$ implicated here? $\endgroup$ – Mark Viola Oct 1 '15 at 18:17
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    $\begingroup$ I believe that the final answer has a typographical error. The exponent on $r$ is $2$, not $1$. $\endgroup$ – Mark Viola Oct 1 '15 at 18:48
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We begin with the integral $I(r,\psi)$ given by

$$I(r,\psi)=\int_{0}^{2\pi}\int_{0}^{\pi}\frac{\sin \theta \cos \theta}{\sqrt{r^2-2r\left(\cos \theta \cos \psi+\sin \theta \sin \psi \cos \phi\right)+1}}\,d\theta\,d\phi$$

Using spherical harmonics, we can expand the denominator of the integrand as

$$\frac{1}{\sqrt{r^2-2r\left(\cos \theta \cos \psi+\sin \theta \sin \psi \cos \phi\right)+1}}=4\pi\sum_{\ell=0}^{\infty}\sum_{m=-\ell}^{\ell}\frac{1}{2\ell +1}\frac{r_{<}^{\ell}}{r_{>}^{\ell+1}}Y_{\ell m}(\theta,\phi)Y_{\ell m}(\psi,0)$$

The integration over $\phi$ is trivial and reveals

$$\begin{align} I(r,\psi)&=2\pi \int_{0}^{\pi}\sin \theta \cos \theta \sum_{\ell=0}^{\infty}\frac{r_{<}^{\ell}}{r_{>}^{\ell+1}}P_{\ell}(\cos \theta)P_{\ell}(\cos \psi)\,d\theta\\\\&=2\pi \sum_{\ell=0}^{\infty}\frac{r_{<}^{\ell}}{r_{>}^{\ell+1}}P_{\ell}(\cos \psi)\int_{0}^{\pi}\sin \theta \cos \theta P_{\ell}(\cos \theta)\,d\theta \tag 1 \end{align}$$

Noting that $P_1(\cos \theta)=\cos \theta$ and exploiting the orthogonality of the Legendre Polynomials, we obtain

$$\begin{align} I(r,\psi)&=2\pi\sum_{\ell=0}^{\infty}\frac{r_{<}^{\ell}}{r_{>}^{\ell+1}}P_{\ell}(\cos \psi)\left(\frac{2}{2\ell +1}\delta_{\ell,1}\right)\\\\ &=\frac{4\pi}{3}\frac{r_{<}}{r_{>}^{2}}P_{1}(\cos \psi)\\\\ &=\frac{4\pi}{3}\frac{r_{<}}{r_{>}^{2}}\cos \psi\\\\ \end{align}$$

For $r>1$, $r_{<}=1$ and $r_{>}=r$. Finally, we have

$$\bbox[5px,border:2px solid #C0A000]{I(r,\psi)=\frac{4\pi}{3} \frac{\cos \psi}{r^2}}$$

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    $\begingroup$ It's always a bit disappointing when we have to resort to orthogonal polynomials. But I can't see a better way, so (+1). $\endgroup$ – Chappers Oct 1 '15 at 19:04
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    $\begingroup$ @rongordon In my former life, I was a scientist also. And you're correct. The use herein is a bit mind-numbing. But, as I stated earlier, brute force works well when it works. Mark $\endgroup$ – Mark Viola Oct 1 '15 at 19:29
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    $\begingroup$ @mickep Well, pleased that this warmed you. ;-))... Mark $\endgroup$ – Mark Viola Oct 1 '15 at 19:40
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    $\begingroup$ @Chappers Well, I confess .... I took the easy way here. There might be much more creative approaches, but brute force works when it works. And thank you for the up vote! Much appreciative. $\endgroup$ – Mark Viola Oct 1 '15 at 21:05
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    $\begingroup$ @grdgfgr The square root term is a Green Function (or Green's Function, depending on whom you wish to believe). It is a solution to Laplace's Equation for $r\ne 1$, $\theta \ne \psi$, $\phi \ne 0$. It can be decomposed, therefore, in terms of spherical harmonics, which one can show by using separation of variables (for example). $\endgroup$ – Mark Viola Oct 1 '15 at 21:37
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Substitution Method

Let $r^2-2r\cos \theta+1=u^2\implies 2r\sin\theta d\theta=2udu$ or $\sin\theta d\theta=\frac{udu}{r}$ $$\int_{0}^{\pi}\frac{\sin\theta}{\sqrt{r^2-2r\cos \theta+1}}d\theta=\int_{r-1}^{r+1}\frac{1}{u}\frac{udu}{r}$$ $$=\frac{1}{r}\int_{r-1}^{r+1}du=\frac{2}{r}$$

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  • $\begingroup$ Given already ? $\endgroup$ – Narasimham Oct 1 '15 at 17:27
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    $\begingroup$ Unfortunately, this doesn't begin to answer the question. $\endgroup$ – Mark Viola Oct 1 '15 at 18:49

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