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Why is the contour integral in upper plane different from the lower plane in this case?

$\int_{-\infty}^{\infty}\mathrm{d}k\frac{1}{(k+a)(k-a)(p-k-b)(p-k+b)}$

where Im[a] and Im[b] are negative and p is real. Besides, Re[a] and Re[b], and p are positive.

The the poles in complex plane are shown below: enter image description here

$\frac{1}{2a[(p-a)^2-b^2]},\frac{1}{2b[(p-b)^2-a^2]},-\frac{1}{2a[(p+a)^2-b^2]},-\frac{1}{2a[(p+b)^2-a^2]}$

I get these residues by just putting $a$ into the other three denominators $(k+a)$, $(p-k-b)$, $(p-k+b)$. and I also use the same procedure to the other three poles $p+b, -a, p-b$ (since each pole is a simple pole). I think the summation of these four poles should give zero. What's the problem?

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  • $\begingroup$ What are the residues that you obtain. The sum of the residues has to be zero in this case. Otherwise, you have a mistake in the calculation. $\endgroup$ – Fabian Nov 22 '18 at 7:06
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    $\begingroup$ Oh, I see the residues. I believe the second and fourth residue should be minus from what you write. $\endgroup$ – Fabian Nov 22 '18 at 7:09
  • $\begingroup$ I get one of the residue $Res_{k=p+b}=\frac{1}{(p+b+a)(p+b-a)(-2b)}$ by put p+b into k+a, k-a, p-k-b $\endgroup$ – James Liu Nov 22 '18 at 7:24
  • $\begingroup$ We can also get the other answers by the same procedure. $\endgroup$ – James Liu Nov 22 '18 at 7:31
  • $\begingroup$ @JamesLiu That is not how you compute a residue. In order for this heuristic to work, you need to change the initial expression it so all signs of $k$ are positive. (i.e. have the last two factors in the denominator be $(k-(p-b))(k-(p+b)),$ otherwise you will be off by a minus sign for two of them. $\endgroup$ – spaceisdarkgreen Nov 22 '18 at 7:46

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