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

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?

• 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. – Fabian Nov 22 '18 at 7:06
• Oh, I see the residues. I believe the second and fourth residue should be minus from what you write. – Fabian Nov 22 '18 at 7:09
• 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 – James Liu Nov 22 '18 at 7:24
• We can also get the other answers by the same procedure. – James Liu Nov 22 '18 at 7:31
• @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. – spaceisdarkgreen Nov 22 '18 at 7:46