# Contour integration with 2 simple poles on contour

Ok for this one I would appreciate if someone could give me a conceptual answer first. I am supposed to integrate $\int_{-\infty}^{\infty} \frac{e^{-i q t}}{p^2 - q^2} dq$ along a half circle C (whose radius goes to infinity), which comprises a horizontal path along the real line circumventing the two real poles -p & +p with a semicircle in the upper half plane, just as in part (a)

I am to prove that the result is 0 if t<0 and $2 \pi i (-\frac{i}{p} ) \sin{pt}$ if t>0.

Now I don't quite understand why the sign of t would change anything. Can someone enlighten me?

• You add to your initial contour of integration something which was not there (a semi circle). Now for this to make sense, one must prove that the corresponding semicircle integral goes to zero as the radius increases. That is usually done with Jordan's lemma. The sign of $t$ will really matter for how one will apply it. – Start wearing purple Oct 14 '13 at 1:38

The sign of $t$ determines which contour you use. The idea is that the integral over the large circular arc shall tend to $0$ when the radius tends to $\infty$, and for that, the integrand must become small, in particular the exponential factor $e^{-iqt}$. We have $\lvert e^z\rvert = e^{\operatorname{Re} z}$, and $\operatorname{Re} (-iqt) = t\cdot \operatorname{Im} q$. For $t > 0$, that becomes negative, and hence the integrand small if $q$ is in the lower half plane, for $t < 0$, the integrand becomes small when $q$ is in the upper half plane.

The contour in the upper half plane does not enclose a pole, hence by the Cauchy integral theorem, the integral is $0$. The contour with the large semicircle in the lower half plane encloses the two poles, and by the residue theorem, it is then $-2\pi i$ times the sum of the residues in the poles ($-2\pi i$ because the contour is negatively oriented).