Is the integration of the arc in contour integration always zero?

Question

Is the integration of the arc in contour integration always zero or is it just a most common coincidence?

By arc I mean the arc $|z|=R$ and $\Im(z)\ge0$, and by integration I mean the contour integration used to evaluate many improper integrals.

Answer 1

If $\deg P(z) \le \deg Q(z) - 2$, then for large $|z|$, we have $\left|\dfrac{P(z)}{Q(z)}\right| \le \dfrac{C}{|z|^2}$ for some constant $C$.

Then, as the radius $R$ of the arc tends to $\infty$, we have

$\displaystyle\left|\int_{|z| = R}\dfrac{P(z)}{Q(z)}\,dz\right| \le \int_{|z| = R}\left|\dfrac{P(z)}{Q(z)}\right|\,|dz| \le \dfrac{C}{R^2}\int_{|z| = R}|dz| = \dfrac{C}{R^2} \cdot 2\pi R = \dfrac{2\pi C}{R} \to 0$.

However, if $\deg P(z) \ge \deg Q(z) - 1$, then this doesn't hold.

For instance $\displaystyle\int_{|z| = R}\dfrac{dz}{z} = 2\pi i$ regardless of how big you make $R$.