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Prove that if $a_1,a_2,\cdots, a_n$ are real numbers then $$\sum_{i,j=1}^n f(a_i - a_j) = \int_{-\infty}^\infty \left(\sum_{i=1}^n \frac{1}{1+(x-a_i)^2}\right)^2 dx,$$ where $f(y) = \int_{-\infty}^\infty \frac{dx}{(1+x^2)(1+(x+y)^2)}$.

I know that $\int_{-\infty}^\infty \big(\sum_{i=1}^n \frac{1}{1+(x-a_i)^2}\big)^2 dx = \int_{-\infty}^\infty \sum_{i,j=1}^n\frac{1}{1+(x-a_i)^2}\frac{1}{1+(x-a_j)^2}dx$. Also, $\sum_{i,j=1}^n f(a_i-a_j) = \int_{-\infty}^\infty \sum_{i=1}^n \sum_{j=1}^n \frac{1}{(1+x^2)(1+(x+(a_i-a_j)^2)}dx.$ But I can't seem to show that for all $i,j$, $\frac{1}{(1+x^2)(1+(x+(a_i-a_j)^2)} =\frac{1}{1+(x-a_i)^2}\frac{1}{1+(x-a_j)^2}$. If this isn't true, then how do I show the given sums are equal?

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    $\begingroup$ If someone answers this are you going to delete it right away? That's what just happened with another recent question of yours... $\endgroup$
    – Stephen
    Commented Aug 10, 2022 at 0:35
  • $\begingroup$ @Stephen sorry about that. I thought the question I asked was really stupid. But my question was based off of something that's likely an error in a solution, which is discussed in this post. $\endgroup$
    – Gord452
    Commented Aug 10, 2022 at 2:23
  • $\begingroup$ $\frac{1}{(1+x^2)(1+(x+(a_i-a_j)^2)}\neq\frac{1}{1+(x-a_i)^2}\frac{1}{1+(x-a_j)^2}$ but still $$\int_{-\infty}^\infty\frac{1}{(1+x^2)(1+(x+(a_i-a_j)^2)}dx=\int_{-\infty}^\infty\frac{1}{1+(x-a_i)^2}\frac{1}{1+(x-a_j)^2}dx$$ (because of a simple substitution). $\endgroup$
    – metamorphy
    Commented Aug 10, 2022 at 9:46
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    $\begingroup$ Your original question wasn't really stupid; much simpler questions are asked and answered here every day. I would not say, however, that there is an error in the solution you mention. That solution does not (implicitly or otherwise) need the answer to your question to be yes. It only needs that the denominator of $(an+1)/bn$ not be $b$ for some sequence of $n$'s tending towards $\infty$. $\endgroup$
    – Stephen
    Commented Aug 10, 2022 at 15:28

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