Integration over contours.

Question

Imagine we wanted to calculate the following integral: $$ \int_{-\infty}^{\infty} \frac{e^{i\omega t}}{\omega^2 +1} d\omega $$ We know, by residue theory that: $$ \int_{-\infty}^{\infty} \frac{e^{i\omega t}}{\omega^2 +1} d\omega = 2\pi i\sum_i^n \text{Res} (f, z_i) - \int_{arc} f(\omega) d\omega $$ Then we finally have (for $t > 0$): $$ \int_{-\infty}^{\infty} \frac{e^{i\omega t}}{\omega^2 +1} = \pi e^{-t} $$ But, now, what I don't exactly understand is, why you need to integrate over a contour that encloses $-i$, to find the integral but for $t < 0$. Same goes for $t > 0$, why integrating along a contour that encloses $+i$, gives me a result for $t > 0$. Any hint?

Is there any relation between integrating along a contour that encloses singularities with $\mathrm {Im} (z) > 0$ and with $t > 0$ and vice versa.

Your hint is look at the integral on the arc in more detail, which you seem to be overlooking — Ninad Munshi, Jun 05 '23 at 17:12
I wrote an explanation in this post. https://math.stackexchange.com/questions/4710787/deriving-the-characteristic-function-of-the-cauchy-distribution-by-residue-theor/4710802#4710802 — MathFail, Jun 05 '23 at 17:12

score 2 · Accepted Answer · answered Jun 05 '23 at 18:29

While the arcs that enclose $i$ and $-i$ can both close the loop, you need to choose the one where $\mathrm{Re}[i\omega t] < 0$ if you want its integral to vanish instead of blow up as you take the radius to infinity. For $t < 0$, this requires $\mathrm{Im}[\omega] < 0$, so we use the arc that encloses $-i$. Similarly, for $t > 0$, we require $\mathrm{Im}[\omega] > 0$, and thus the arc enclosing $i$ is the one to use.

Integration over contours.

1 Answers1