## How can this contour integral be non-zero? [closed]

Hi,

I have a contour integral whose integrand $\rightarrow 0$ as $|z| \rightarrow \infty$ (integrand goes as $\frac{1}{z^2}$). It is analytic everywhere except for 2 poles ($\omega \pm i\gamma/2$).

I came to this problem by wishing to integrate it along the real axis. However, as far as I can tell I can complete the contour in either half plane since the integrals around large semicircles in either half plane will vanish.

My question is this: If we have an integrand (mine is an example, but if I've made a mistake make it a hypothetical one) the integrals of which vanish when integrating over large semicircles in either half plane then by integrating over one semicircle and then the other we would make a large circular contour enclosing the entire plane. By our reasoning so far this should be zero, since both component integrals that we're summing are. But that cannot be possible if the integrand has poles anywhere at all due to the residue theorem.

My specific integral is this: $\int_{-\infty}^{+\infty} \frac{\gamma}{(z - \omega_0)^2 + \gamma^2/4} \frac{\mathrm{d} z}{2\pi}$

Could anybody please explain this to me?

JMA

-
The sentence "But that cannot be possible if the integrand has poles anywhere at all due to the residue theorem." is wrong. The two poles have residues $\pm i/(2\pi)$, so they sum to $0$. If $\gamma>0$, then the pole in the upper half-plane is at $\omega_0 + i \gamma/2$, and it has residue $-i/(2\pi)$. If you deform the contour from the real line to infinity, you pick up $2\pi i \cdot (-i/(2\pi)) = 1$, so the integral is $1$. If $\gamma<0$, you get $-1$ instead. – Henry Cohn Jan 16 2012 at 17:45
First, this problem is so simple that the indefinite integral has a closed form: $\frac{1}{\pi}arctan(\frac{1}{4} - 8\frac{z-\omega_{0}}{\gamma})$. Taking limits finishes the problem. Second, the definite integral itself is also so easy, that Maple immediately returns the answer (signum(g) where g=$\gamma$ is assumed real), as per Henry's comment. – Jacques Carette Jan 16 2012 at 18:22