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?
Best Answer
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).