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I have an exercise (from physics) where I am supposed to show

$$\sum_{k'<k_f} \frac{1}{|k-k'|^2} = C \left( \frac{1}{2} + \frac{1-(\frac{k}{k_f})^2}{4 \left( \frac{k}{k_f} \right) } ln |\frac{1 + \frac{k}{k_f} }{1-\frac{k}{k_f} }| \right)$$

by approximating this sum by an integral, unfortunately this notation is very sloppy. I guess the sum starts at $1$. I also noticed that this logarithm in this expression is just a the artanh function and this is nice, cause this is the antiderivative to $(artanh)'(x) = \frac{1}{1-x^2}.$ Now I was hoping that we can also make some sense out of the term $\frac{1-(\frac{k}{k_f})^2}{4 \left( \frac{k}{k_f} \right) }$ and try to understand how this could be related to the sum.

I know this question is ill-defined, but I would like to get a few ideas, how these two expressions could be related to each other?

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  • $\begingroup$ sorry, typo in the question $\endgroup$ – user159356 Nov 6 '14 at 18:55
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This may or may not help, but it's at least some thought/direction. Notice that $$ \frac{1}{|\, k-k'\,|^2} = \frac{1}{k_f^2} \frac{1}{\big| \frac{k}{k_f} - \frac{k'}{k_f}\big|^2} $$ and so $$ \sum_k\sum_{k'} \frac{1}{|\, k-k'\,|^2} = \sum_k \frac{1}{k_f} \Big(\sum_{k'} \frac{1}{k_f} \frac{1}{\big| \frac{k}{k_f} - \frac{k'}{k_f}\big|^2}\Big) =:\sum_k \Delta y \Big[ \sum_{k'} \Delta x \Big( \frac{1}{|\,x_k - y_k\,|^2} \Big)\Big] \\ \sim \iint \frac{1}{|x-y|^2} \,dxdy $$ where the bounds on the integral will depend on what $k,k'<k_f$ actually mean. According to WolframAlpha $$ \iint \frac{1}{|x-y|^2} \,dxdy = (x-y)\log(x-y) - x +\text{constant} $$

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