I am trying to show the following:

enter image description here

I can get to the following result:

enter image description here

I would appreciate any help with solving this problem. Thanks in advance.


First thing to do is replace for $$x=\frac{s}{2}$$ $$\implies \frac{-q^2}{8\pi e} \left( \frac{1}{2\frac{s}{2}} + \sum_{n=1}^{\infty} \left(\frac{ns}{(ns)^2-(\frac{s}{2})^2} - \frac{1}{ns} \right) \right)$$ $$\frac{-q^2}{8\pi e} \left( \frac{1}{s} + \sum_{n=1}^{\infty} \left( \left(\frac{s}{s^2}\right)\frac{n}{(n)^2-(\frac{1}{2})^2} - \frac{1}{ns} \right) \right)$$ $$\frac{-q^2}{8\pi es}\left( 1 + \sum_{n=1}^{\infty} \left( \frac{n}{(n)^2-(\frac{1}{2})^2} - \frac{1}{n} \right) \right)$$ $$\frac{-q^2}{8\pi es}\left( 1 + \sum_{n=1}^{\infty} \frac{1}{n(2n-1)(2n+1)} \right)$$

The sum can be rewritten as:

$$\sum_{n=1}^{\infty} \frac{1}{n(2n-1)(2n+1)}= 2 \sum_{n=1}^{\infty} \frac{1}{(2n)(2n-1)(2n+1)}$$

Method 1:

If we let,

$$f(x)=\sum_{n=1}^{\infty} \frac{x^{2n+1}}{(2n)(2n-1)(2n+1)}$$ $$\implies f^{(3)}(x)=\sum_{n=1}^{\infty}x^{2n-2}=\sum_{n=0}^{\infty}x^{2n}=\frac{1}{1-x^2}$$ Now you must find $f(1)$

Since $$f(0)=f'(0)=f''(0)=0$$ $$\implies \int_0^{1} \int_0^{x} \int_0^{y}f'''(z)dzdydx=f(1)$$ $$\therefore 2f(1)=\int_0^1 (1-z)^2 f'''(z)dz=\int_0^1 (1-z)^2\frac{1}{1-z^2}dz=\int_0^1 \frac{1-z}{1+z}dz=2\ln(2)-1$$

So, $f(1)=\ln(2)-\frac{1}{2}$

$$\sum_{n=1}^{\infty} \frac{1}{n(2n-1)(2n+1)}= 2 \sum_{n=1}^{\infty} \frac{1}{(2n)(2n-1)(2n+1)}=2\ln(2)-1$$

So, the final sum is

$$E_i(x=s/2)=\frac{-q^2}{8\pi es} (1 + 2 \ln(2)-1)=\frac{-q^2}{8\pi es} \ln(4)$$

Method 2 Use partial fractions:


Solve for $A,B,C$

Proof: On how to prove $\int_0^x \int_0^y \int_0^z f(t) \ dt \ dz \ dy=\frac{1}{2} \int_0^x (x-t)^2 f(t) \ dt$

First imagine the triple integral as the integral of a region A,

$$\iiint_A f(t) dV$$

where $A=\{(t,z,y) \ | \ 0<t<z, \ 0<z<y, \ 0<y<x\}$

The objective is to change the order of integration to $dt$ being last because this avoids the complexity of having to integrate $f(t)$. So, now take $B$ as the projection on the $t\text{-} y$ plane then

$$B=\{(t,y) \ | \ 0<t<x, \ t<y<x \}$$

So, what is the new region $A$ with an integration order of $dz \ dy \ dt$?

Once you find the new region $A$ (which is best found by drawing it); it is simply:

$$A=\{(t,z,y)\ | \ t\leq z \leq y, \ t\leq y \leq x , \ 0\leq t \leq x \} $$ In this case, you can skip the drawing because the inequalities of region $A$ are simple and its just a matter of changing the function according to the order of integration. More specifically, $z$ is the first order, so its limits, from the first region of $A$, are $t<z$ and $z<y$ which is $t<z<y$. The other limits come from the projection, $B$ $$\therefore \int_0^x \int_0^y \int_0^z f(t) \ dt \ dz \ dy = \int_0^x \int_t^x \int_t^y f(t) \ dz \ dy \ dt=\int_0^x \left[\int_t^x (y-t)f(t) \ dy \right] \ dt$$ $$=\int_0^x \left(\frac{1}{2}x^2-tx-\frac{1}{2}t^2+t^2 \right) f(t) \ dt= \frac{1}{2} \int_0^x \left( x^2-2tx+t^2 \right) f(t) \ dt$$ $$=\frac{1}{2}\int_0^x \left(x-t\right)^2 f(t) \ dt$$

  • $\begingroup$ Just had chance to check the question. Thank you very much. $\endgroup$ – user118670 Jan 2 '14 at 21:15
  • $\begingroup$ @user118670 Any questions or concerns? Or Have I shocked you with triple integral that you haven't taken yet :\ $\endgroup$ – John Jan 2 '14 at 21:16
  • $\begingroup$ How did you go between the triple integral of x,y,z to single integral of z? $\endgroup$ – user118670 Jan 3 '14 at 7:42
  • $\begingroup$ @user118670 Its a known formula that $$\int_0^x \int_0^y \int_0^z f(t) \ dt \ dz \ dy=\frac{1}{2} \int_0^x (x-t)^2 f(t) \ dt$$ The proof isn't very difficult. $\endgroup$ – John Jan 3 '14 at 9:03
  • 1
    $\begingroup$ thanks again. Appreciate it. $\endgroup$ – user118670 Jan 3 '14 at 12:19

First, I suggest you decompose the element in your sum using partial fractions. I suppose that you will not have any problem from here.


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