2
$\begingroup$

I've these two series, and I would like a closed form:

$$ \sum_{k=-\infty}^{\infty} \frac{x+kx_0-h}{|x+kx_0|^3}$$

$$ 3\sum_{k=-\infty}^{\infty} \frac{(x+kx_0-h)(x+kx_0)^2}{|x+kx_0|^5} $$

Mathematica gives me these closed forms ($A=x_0$, $B=h$):

Sum[(x + k*A - B)/((abs (x + k*A))^3), {k, -Infinity, Infinity}] 
  =   (1/(2 A^3 abs^3))(2 A PolyGamma[1, x/A] + 2 A PolyGamma[1, 1 - x/A] +       B PolyGamma[2, x/A] - B PolyGamma[2, 1 - x/A]) 
Sum[((x + k*A - B)*(x + k*A)^2)/(abs (x + k*A)^5), {k, -Infinity, Infinity }] 
=  (1/(2 A^3 abs))(2 A PolyGamma[1, x/A] + 2 A PolyGamma[1, 1 - x/A] + B      PolyGamma[2, x/A] - B PolyGamma[2, 1 - x/A]) 

Now, first of all, when it writes $abs^3$ this should mean $|x|^3$ or $|x-kx_0|^3$? And at last, does this make sense? Is there an analytical way to reach this (or a better) conclusion?

Thanks!

  • 0
    You may want to post your question here as well: http://mathematica.stackexchange.com/2012-06-22
  • 0
    Good idea! I didn't know about that!2012-06-22
  • 1
    Are A and B the same as $x_0$ and $h$? I suspect you mistakenly used () instead of [] and Mathematica thinks abs is the name of a variable.2012-06-22
  • 0
    Added input code and yes $A=x_0$, $B=h$!2012-06-22
  • 1
    Yes, that's your mistake. It should be Abs[...], not abs (...).2012-06-22
  • 0
    Ups, thank you so much!2012-06-22

1 Answers 1