Contour integration of a meromorphic function

Question

Given a meromorphic function $f$ which is uniformly bounded on the upper half plane. Assume that $\int_{-\infty}^{+\infty} f(x)dx$ is absolutely integrable. Then Cauchy's integral theorem suggests $\int_{-\infty}^{+\infty}f(x)dx=0$ except there is some tricky business around infinity. Can someone give me a counterexample or a supporting argument?

Answer 1

If $f$ is bounded on the upper half-plane, it has no poles there or on the real axis.

For $\varepsilon > 0$, consider the functions

$$g_\varepsilon(z) = \frac{f(z)}{1 - i\varepsilon z},\quad h_\varepsilon(z) = g_\varepsilon(z)\cdot e^{i\varepsilon z}.$$

$g_\varepsilon(z)$ satisfies the same assumptions as $f$, and additionally, we have $g_\varepsilon(z) \to 0$ for $z\to\infty$ in the upper half plane (or on the real axis). By Jordan's lemma,

$$\int_{\gamma_R} h_\varepsilon(z)\,dz \xrightarrow{R\to+\infty} 0,$$

where $\gamma_R$ is the semicircle with radius $R$ and centre $0$ in the upper half plane. By Cauchy's integral theorem, we deduce

$$\int_{-\infty}^\infty h_{\varepsilon}(x)\,dx = 0.$$

By the dominated convergence theorem (we have $\lvert h_\varepsilon(z)\rvert \leqslant \lvert f(z)\rvert$), we have

$$\int_{-\infty}^\infty f(x)\,dx = \lim_{\varepsilon\searrow 0} \int_{-\infty}^\infty h_\varepsilon(x)\,dx = 0.$$