This integral comes from a physics book when calculating a field of an uniformly charged sphere (without Gauss' Law).
It says that it can be done by partial fractions, but I cannot imagine how.
This integral comes from a physics book when calculating a field of an uniformly charged sphere (without Gauss' Law).
It says that it can be done by partial fractions, but I cannot imagine how.
As GEdgar pointed out in a comment, this is
\frac{1}{2}\int\frac{f'(z)}{f(z)^{3/2}}\mathrm du\;,
with $f(z)=R^2 + z^2 - 2Rzu$, so perhaps the easiest way to do this is
\begin{eqnarray} \frac{1}{2}\int\frac{f'(z)}{f(z)^{3/2}}\mathrm du &=& -\int\frac{\mathrm d}{\mathrm dz}f(z)^{-1/2}\mathrm du \\ &=& -\frac{\mathrm d}{\mathrm dz}\int f(z)^{-1/2}\mathrm du \\ &=& -\frac{\mathrm d}{\mathrm dz}\int(R^2 + z^2 - 2Rzu)^{-1/2}\mathrm du \\ &=& \frac{\mathrm d}{\mathrm dz}\frac{(R^2 + z^2 - 2Rzu)^{1/2}}{Rz}\;. \end{eqnarray}
If it is indeed the integral with respect to $u$ that you want, and $z$ is an oddly named constant, make the substitution $w=R^2+z^2 -2Rzu$. Despite appearances, this is a simple linear substitution.
Go through the process, not forgetting that $dw=-2Rz du$.
You end up with an integral of the shape $\int \frac{A+Bw}{w^{3/2}} dw.$ where $A$ and $B$ are somewhat messy constants. Split the integrand into two parts, $A/w^{3/2}$ and $B/w^{1/2}$.
If in your integral $du$ is a typo for $dz$, life is even simpler, make the same substitution, and you end up with $dw/2$ on top.
Let $\sqrt{R^2+z^2-2Rzu}=t$ and then differentiating both sides gives
$\dfrac{-2Rz}{2\sqrt{R^2+z^2-2Rzu}} du = dt$, which is equivalent to $\dfrac{-Rz}{t}du=dt$. Substituting this,
$\displaystyle \int{ \frac{z-\frac{R^2+z^2-t^2}{2z}}{t^3} \frac{-t}{Rz} dt}=\int{ \frac{-R^2+z^2+t^2}{2z} \frac{-1}{Rzt^2} dt} = \frac{1}{2Rz^2} \int{ (\frac{R^2-z^2}{t^2} -1) dt}$.
I guess you can do it now.
$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\color{#66f}{\large% \int_{-1}^{1}{z - Ru \over \pars{R^{2} + z^{2} - 2Rzu}^{3/2}}\,\dd u} =\int_{u\ =\ -1}^{u\ =\ 1}\pars{z - Ru}\, \dd\pars{1/\pars{Rz} \over \root{R^{2} + z^{2} - 2Rzu}} \\[5mm]&=\left.{1 \over Rz}\, {z - Ru \over \root{R^{2} + z^{2} - 2Rzu}}\,\right\vert_{\,u\ =\ -1}^{\,u\ =\ 1} -\int_{-1}^{1}{1 \over Rz}\,{-R \over \root{R^{2} + z^{2} - 2Rzu}}\,\dd u \\[1cm]&={1 \over Rz}\pars{{z - R \over \root{R^{2} + z^{2} - 2Rz}} -{z + R \over \root{R^{2} + z^{2} + 2Rz}}} \\[5mm]&\phantom{=}+{1 \over z\,d_{>}} \int_{-1}^{1}{\dd u \over \root{1 - 2\pars{d_{<}/d_{>}}u + \pars{d_{<}/d_{>}}^{2}}} \quad\mbox{where}\quad \left\{\begin{array}{lcl} d_{<} & \equiv & \min\pars{\verts{R},\verts{z}} \\[2mm] d_{>} & \equiv & \max\pars{\verts{R},\verts{z}} \end{array}\right. \end{align}
In the last integral, the integrand can be expanded in Legendre Polynomias $\ds{\,{\rm P}_{\ell}:\bracks{-1,1} \to {\mathbb R}}$, with $\ds{\ell = 0,1,2,3,\ldots}$, which satisfy $\ds{\int_{-1}^{1}\,{\rm P}\ell\pars{x}\,{\rm P}\ell'\pars{x} ={2\,\delta_{\ell\ell'} \over 2\ell + 1}}$. Then,
\begin{align}&\color{#66f}{\large% \int_{-1}^{1}{z - Ru \over \pars{R^{2} + z^{2} - 2Rzu}^{3/2}}\,\dd u} \quad\pars{~\mbox{Note that}\ \,{\rm P}_{0}\pars{x} = 1\,,\ \forall\ x \in \bracks{-1,1}~} \\[5mm]&={\sgn\pars{z - R} - \sgn\pars{z + R} \over Rz} +{1 \over z\,d_{>}} \sum_{\ell\ =\ 0}^{\infty}\pars{d_{<} \over d_{>}}^{\ell}\ \overbrace{% \int_{-1}^{1}\,{\rm P}_{\ell}\pars{u}\,\dd u} ^{\ds{=\ \dsc{2\,\delta_{\ell,0}}}} \\[5mm]&=\color{#66f}{\large% {1 \over z}\bracks{{\sgn\pars{z - R} - \sgn\pars{z + R} \over R} +{2 \over d_{>}}}} \end{align}
where $\ds{\quad\left\{\begin{array}{lcl} d_{<} & \equiv & \min\pars{\verts{R},\verts{z}} \\[2mm] d_{>} & \equiv & \max\pars{\verts{R},\verts{z}} \end{array}\right.}$