So we know that if $f(z) = \frac{P(z)}{Q(z)}$, whereas $deg(Q) \geq deg(P) + 2$, then $\int_{-\infty}^{\infty} f(x) dx = 2 \pi i \sum_{ Im k > 0} \text{Res}_k f(z)$, but now the exercise is as follows:
Let f(z) be a rational function function which does not have any poles on the real axis and which has a zero of order 1 at $\infty$ (i.e $deg(Q) =deg(P) + 1$), argue why the limit $\lim_{R \to \infty} \int_{-R}^{R} f(x) dx$ still exists and determine the difference $\lim_{R \to \infty} \int_{-R}^{R} f(x) dx$ - $ 2 \pi i \sum_{ \text{Im k} > 0} \text{Res}_k f(z)$.
(Im k = imaginary part of the k-th pole) So I assume that when the zero at $\infty$ is of order 1, then the integral won't be absolutely convergent anymore? but what exactly guarantees its existence and how could I proceed calculating the difference?