📂Complex Anaylsis

Divergent Semicircle Complex Path Integral for the Improper Integral of Rational Functions

Buildup

Let’s assume for two polynomial functions $p(z) , q(z)$ that $\displaystyle f(z) = {{q(z)} \over {p(z)}}$.

If there are no real solutions satisfying $p(z) = 0$, then $f$ will not have a real singularity. The condition for the existence of an improper integral $\displaystyle \int_{-\infty}^{\infty} f(z) dz$ of such a rational function is from $\displaystyle f(z) \sim {{1} \over {z^{p}}}$ to $p > 1$. Thinking in terms of the concept of infinite series, it can be related to the necessary and sufficient condition for $\displaystyle \sum_{n=0}^{\infty} {{{1} \over {n^{p}}} }$ to converge being $p>1$. This condition can be expressed more concisely as $\displaystyle \lim_{z \to \infty } z f(z) = 0$.

Meanwhile, consider the simple closed semicircular path $\mathscr{C} = {\color{red}\Gamma} \cup [-R,R]$ as follows: Then, it can be considered as split into $\displaystyle \int_{\mathscr{C}} f(z) dz = { \color{red} \int_{\Gamma} f(z) dz } + \int_{-R}^{R} f(z) dz$, and if $R \to \infty$, the original equation regarding the improper integral we intend to solve can be obtained.

Theorem ¹

If the function $f$ is continuous on the semicircle $\Gamma$ with radius $R$ and center at $0$ and if $\displaystyle \lim_{z \to \infty } z f(z) = 0$, then $$\lim_{R \to \infty} \int_{\Gamma} f(z) dz = 0$$

Description

With such a lemma as mentioned above, when $R \to \infty$, $\displaystyle {\color{red} \int_{\Gamma} f(z) dz } \to 0$ hence $\displaystyle \int_{\mathscr{C}} f(z) dz = \int_{-\infty}^{\infty} f(z) dz$.

On the other hand, $\displaystyle \int_{\mathscr{C}} f(z) dz$ can be easily calculated using the residue theorem, and the improper integral falls into $\displaystyle \int_{-\infty}^{\infty} f(z) dz = 2 \pi i \sum \text{Res} f(z)$. This method will continue to appear in techniques using the Jordan’s lemma later on. The proof of the lemma is not particularly difficult.

Proof

Given $\displaystyle \lim_{z \to \infty} z f(z) = 0$, there will exist $\delta > 0$ for any $\varepsilon >0$ satisfying $\displaystyle |{{1} \over {z}}| < \delta \implies |zf(z)| < \varepsilon$.

On $\Gamma$, since $|z|=R$, when sorting out regarding $R$, $${{1} \over {R}} < \delta \implies |f(z)| < { {\varepsilon} \over {R}}$$ By the ML lemma, $${{1} \over {R}} < \delta \implies |\int_{\mathscr{C}} f(z) dz | < \left( R \cdot \pi + 2 R \right) \cdot { {\varepsilon} \over {R}} = ( \pi + 2 ) \varepsilon$$ Therefore, $\displaystyle \lim_{R \to \infty} \int_{\Gamma} f(z) dz = 0$

■

Generalization

As can be seen by examining the proof, there’s no need for $\Gamma$ to specifically be a semicircle, hence the following generalization is possible:

Diverging arc’s complex path integration: If the function $f$ is continuous on the arc $\Gamma : z(\theta) = R e^{i \theta} , \alpha \le \theta \le \beta$ with radius $R$ and center $0$, and if $\displaystyle \lim_{z \to \infty } z f(z) = l$, then $$\lim_{R \to \infty} \int_{\Gamma} f(z) dz = ( \beta - \alpha) l i$$

The lemma introduced in this post is a corollary case of the above general theorem for $\alpha = 0, \beta = \pi, l = 0$. Essentially using the same method, so proof is omitted.

Contracting semicircle’s complex path integration: If the function $f$ is continuous on the semicircle $\gamma$ with radius $r$ and center $0$, and if $\displaystyle \lim_{z \to 0 } z f(z) = 0$, then $$\lim_{r \to 0} \int_{\gamma} f(z) dz = 0$$

Additionally, such a lemma can also be induced without much difficulty.

Of course, it doesn’t matter if it’s not a semicircle but an arc.