📂Complex Anaylsis

Evaluation of Improper Integrals through the Jordan Lemma

Description ¹

First, similar to the divergent semicircular complex path integration for rational functions’ improper integrals, let’s start with two polynomials $p(z) , q(z)$, assuming $\displaystyle f(z) = {{q(z)} \over {p(z)}}$.

If a real solution does not exist satisfying $p(z) = 0$, then $f$ would not have a real singularity. Considering an integral in the form of $\displaystyle \int_{- \infty}^{\infty} \sin{mx}f(x) dx$ or $\displaystyle \int_{- \infty}^{\infty} \cos{mx}f(x) dx$ for a positive $m \in \mathbb{R}^{+}$,

The condition for the existence of the improper integral is relaxed down from $\displaystyle f(z) \sim {{1} \over {z^{p}}}$ to $p > 0$, significantly. Thinking in terms of series, this is similar to how the harmonic series $\displaystyle \sum_{n=1}^{\infty} {{1 }\over {n}}$ diverges, but the alternating harmonic series $\displaystyle \sum_{n=1}^{\infty} (-1)^{n} {{1 }\over {n}}$ converges.

Thinking about such a simple closed semicircle $\mathscr{C} = {\color{red}\Gamma} \cup [-R,R]$, $$\int_{\mathscr{C}} e^{m i z} f(z) dz = \color{red} {\int_{\Gamma} e^{m i z} f(z) dz } + \int_{-R}^{R} \cos m z f(z) dz + i \int_{-R}^{R} \sin m z f(z) dz$$ it can be considered in parts.

Jordan’s lemma: For a function $f$ that is continuous in $\Gamma$ and if $\displaystyle \lim_{z \to \infty} f(z) = 0$, then for a positive $m \in \mathbb{R}^{+}$, $$\lim_{R \to \infty} \int_{\Gamma} e^{m i z } f(z) dz = 0$$

Since $R \to \infty$ when $\displaystyle \color{red} {\int_{\Gamma} e^{m i z} f(z) dz } \to 0$, utilizing the residue theorem to calculate $\displaystyle \int_{\mathscr{C}} e^{m i z}f(z) dz$, the real part is found to be $\displaystyle \int_{-\infty}^{\infty} \cos m z f(z) dz$, and the imaginary part, $\displaystyle \int_{-\infty}^{\infty} \sin m z f(z) dz$.