The title states the integral in question.

I have tried arguing that there is no pole in $z=0$ because $$(1-\cos(z))/z^2 \approx (1-1+z^2)/z^2 = 1.$$ Then using the contour of a half circle in the upper half plane and rewriting cosine in terms of complex exponentials and use the residue thm. to get: $-\pi (1-\exp(-at))/a^3$. But this does not seem to be the right answer.

If $z=0$ is a pole of order 2, then I should avoid it in my contour, but I do not know what the contribution to the integral is then.

If it was a simple pole it would contribute with half the residue, but what about a pole of order 2?

  • $\begingroup$ I don't see a connection between que question at the end of your post and the rest of the post. As you wrote, the function that you are trying to integrate has no pole at $0$: $\endgroup$ – José Carlos Santos Jun 2 at 8:46

$\dfrac{1-\cos\omega}{\omega^2(a^2+\omega^2)}$ is indeed regular at $\omega=0$, but the plan to put $\cos\omega=(e^{i\omega}+e^{-i\omega})/2$ breaks when trying to let the radius of the half-circle tend to $\infty$, because $e^{-i\omega}$ "explodes". Instead, you can drive this way with integrand $\dfrac{1-e^{i\omega}}{\omega^2(a^2+\omega^2)}$ (which has a simple pole at $\omega=0$ so your last note applies).

  • $\begingroup$ Thanks, that seemed to work! $\endgroup$ – Nikolaj Jun 2 at 10:08

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