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?