$\displaystyle f=\int_V \dfrac{x-x'}{|\mathbf{r}-\mathbf{r'}|^3}\ dV'$

where $V'$ is a finite volume in space

$\mathbf{r}=(x,y,z)$ are coordinates of all space

$\mathbf{r'}=(x',y',z')$ are coordinates of $V'$


How to prove that:

$\lim\limits_{\Delta x \to 0} \dfrac{f(x+\Delta x,y,z)-f(x,y,z)}{\Delta x}$ exist

$\text{ }$


I am not sure whether this method would work. If it doesn't please suggest another method to reach my goal.

\begin{align} &\lim\limits_{\Delta x \to 0} \dfrac{f(x+\Delta x,y,z)-f(x,y,z)}{\Delta x}\\ =&\lim\limits_{\Delta x \to 0}\dfrac{\displaystyle\int_{V'} \dfrac{(x+\Delta x)-x'}{|\mathbf{r}(x+\Delta x,y,z)-\mathbf{r'}|^3}\ dV' - \int_{V'} \dfrac{x-x'}{|\mathbf{r}(x,y,z)-\mathbf{r'}|^3}\ dV'}{\Delta x}\\ =&\lim\limits_{\Delta x \to 0}\displaystyle\int_{V'} \dfrac{\left( \dfrac{(x+\Delta x)-x'}{|\mathbf{r}(x+\Delta x,y,z)-\mathbf{r'}|^3} -\dfrac{x-x'}{|\mathbf{r}(x,y,z)-\mathbf{r'}|^3} \right)}{\Delta x}dV' \end{align}

Now if only I could take the limit inside the integral (with respect to $V′$),I can proceed to show the limit exists.

If we can't do that and this method doesn't work, please suggest another method to show that the limit exists.

  • $\begingroup$ You wrote $f(x)$ and $f(x+\Delta x)$ -- what about the $y$ and $z$? $\endgroup$ – user10354138 May 7 at 15:38
  • $\begingroup$ $y$ and $z$ are constants. Let me edit $\endgroup$ – Joe May 7 at 16:01
  • $\begingroup$ You want $\int_{V'}\dots$ not $\int_V\dots$ in the first formula so you are only integrating over a finite region. Show that for fixed $x,y,z$, $g(x',y',z')=\frac{\partial}{\partial x}\frac{x-x'}{\lvert\mathbf{r}-\mathbf{r}'\rvert^3}$ is absolutely integrable on $V'$. $\endgroup$ – user10354138 May 8 at 1:21
  • $\begingroup$ @user10354138 Then what shall we do to reach our conclusion that the limit exists. Can you please elaborate in an answer. $\endgroup$ – Joe May 8 at 8:24

Assume that $C$ is a $3$-dimensional body with a smooth boundary and ${\rm vol}\ C<\infty$. Define $f(\textbf{r}) =\int_C\ \frac{x-x'}{|\textbf{r}-\textbf{r}'|^3}\ d{\rm vol}\ (\textbf{r}' ) $ Prove that $f$ is finite.

Proof : $\int_{B_\epsilon (0)}\ \frac{1}{|{\bf r}|^2} \ d{\rm vol}\ ({\bf r}) \leq C\epsilon$ for some $l>0$ and note that $ \frac{|x-x'|}{|\textbf{r}-\textbf{r}'|^3}\leq \frac{1}{| \textbf{r}-\textbf{r}'|^2}$

So $$ |f( \textbf{r} )| \leq \int_{B_\epsilon ({\bf r} )}\ \frac{|x-x'|}{|\textbf{r}-\textbf{r}'|^3}\ d{\rm vol}\ (\textbf{r}') +\int_{C-B_\epsilon(\textbf{r}) }\ \frac{1}{|\textbf{r}- \textbf{r}'|^2}\ d{\rm vol}\ (\textbf{r} ') $$

$$ \leq l\epsilon +\int_{C-B_\epsilon(\textbf{r}) }\ \frac{1}{\epsilon^2} \ d{\rm vol}\ (\textbf{r} ') \leq l\epsilon + \frac{1}{\epsilon^2}{\rm vol}\ C $$


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