📂Complex Anaylsis

Singularities on the Real Axis and Improper Integrals via Jordan's Lemma

Buildup

The overall flow is similar to Jordan’s Lemma for Improper Integrals. For two polynomials $p(z) , q(z)$, let’s say $\displaystyle f(z) = {{q(z)} \over {p(z)}}$.

If there exists a real solution $a$ that satisfies $p(z) = 0$, then $f$ has a real singularity at $a$. The reason we haven’t dealt with such cases until now was to use the Residue Theorem. Of course, just because a singularity is added on the real axis, we don’t strictly have to give up on the residue theorem; we use a trick of bending the integration path.

Considering a simple closed path $\mathscr{C}$ as shown above, we can bypass the singularity with a small semicircle like $\gamma$. The semicircle does not necessarily have to be centered on $0$, and it’s fine if it exists finitely and complexly on the real axis. By choosing $r \to 0$ to tighten the small semicircle and taking $R \to \infty$, we can use the residue theorem as we originally did. To use this method, we need to prepare by proving the following auxiliary lemma first. Though it sounds complicated, one can easily see it resembles the Residue Theorem.

Auxiliary Lemma ¹

Let’s suppose the function $f : \mathbb{C} \to \mathbb{C}$ is analytic near the vicinity $0 < |z - a | < r_{0}$ of the singularity $a$. If a semicircle $z( \theta) - a = r e^{i \theta}, 0 \le \theta \le \pi$ with radius $r$, which is a small positive number smaller than $r_{0}$, centered at $a$ $$ \lim_{r \to 0} \int_{\gamma} f(z) dz = - i \pi \text{Res}_{a} f(z) $$

Proof

According to the assumption, $$ f(z) = \sum_{m=0}^{\infty} {{b_{2m+1}} \over {(z-a)^{2m+1} }} + \sum_{n=0}^{\infty} a_{n} (z-a)^{n} $$ $(z-a) = r e^{i \theta}$, so when $n \ne -1$, $$ \int_{\gamma} (z-a)^{n} dz = -i r^{n+1} \int_{0}^{\pi} e^{(n+1) i \theta} d \theta = {{r^{n+1}} \over {n+1}} (1 + \cos {n \pi}) $$ Therefore, either $n \ge 0$ or $n<-1$, but if $n$ is odd, $$ \lim_{r \to 0} \int_{\gamma} (z - a)^{n} dz = 0 $$ On the other hand, if considering the case of $n = -1$, $$ \lim_{r \to 0} \int_{\gamma} (z-a)^{-1} dz = \lim_{r \to 0} \int_{0}^{\pi} - i d \theta = - i \pi $$ Therefore, $\displaystyle \lim_{r \to 0} \int_{\gamma} f(z) dz = - i \pi b_{1} = - i \pi \text{Res}_{\alpha} f(z)$

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Example

As an example, let’s calculate the improper integral of the sinc function $\displaystyle \int_{0}^{\infty} {{\sin x} \over {x} } dx$.

Solution

Since $e^{i z} = \cos z + i \sin z$, first consider $\displaystyle \int_{\mathscr{C}} {{e^{iz} } \over {z}} dz$, and let’s assume $\mathscr{C}$ as given in the figure above. Then $$ \int_{\mathscr{C}} {{e^{iz} } \over {z}} dz = \color{red} { \int_{ \Gamma} {{e^{iz} } \over {z}} dz } + \int_{-R}^{-r} {{e^{ix} } \over {x}} dx + \color{blue} { \int_{ \gamma} {{e^{iz} } \over {z}} dz } + \int_{r}^{R} {{e^{ix} } \over {x}} dx $$

Residue Theorem: Suppose an analytic function $f: A \subset \mathbb{C} \to \mathbb{C}$ has a finite number of singularities inside a Simple closed contour $\mathscr{C}$. Then $\displaystyle \int_{\mathscr{C}} f(z) dz = 2 \pi i \sum_{k=1}^{m} \text{Res}_{z_{k}} f(z)$

Since the singularity of the function $\displaystyle f(z) := {{e^{iz} } \over {z}}$ is only $z = 0$ and there are no singularities inside $\mathscr{C}$, by the Residue Theorem $$ \int_{\mathscr{C}} {{e^{iz} } \over {z}} dz = 0 $$ That is, $$ \color{red} { \int_{ \Gamma} {{e^{iz} } \over {z}} dz } + \int_{-R}^{-r} {{e^{ix} } \over {x}} dx + \color{blue} { \int_{ \gamma} {{e^{iz} } \over {z}} dz } + \int_{r}^{R} {{e^{ix} } \over {x}} dx = 0 $$

Residue at Simple Poles: Suppose function $f$ can be represented as $\displaystyle f(z) = {{g(z)} \over {h(z)}}$. Here $g$ and $h$ are analytic at $\alpha$, and if $g(\alpha) \ne 0 , h(\alpha) = 0, h ' (\alpha) \ne 0$ then $\alpha$ is a simple pole of $f$ and $$` \text{Res}_{\alpha} f(z) = {{g(\alpha)} \over {h ' (\alpha)}}$$

Therefore, since $\displaystyle \text{Res}_{0} {{e^{iz} } \over {z}} = {{e^{i \cdot 0} } \over {1}} = 1$, $$ \lim_{r \to 0} \color{blue} { \int_{\gamma} {{e^{iz} } \over {z}} dz } = - i \pi \text{Res}_{0} {{e^{iz} } \over {z}} = - i \pi $$ When $r \to 0$, $$ \color{red} { \int_{ \Gamma} {{e^{iz} } \over {z}} dz } + \int_{-R}^{0} {{e^{ix} } \over {x}} dx \color{blue} { - i \pi } + \int_{0}^{R} {{e^{ix} } \over {x}} dx = 0 $$ Summarizing, $$ \color{red} { \int_{ \Gamma} {{e^{iz} } \over {z}} dz } + \int_{-R}^{R} {{e^{ix} } \over {x}} dx = i \pi $$

Jordan’s Lemma: When a semicircle $\Gamma$ is represented as $z(\theta) = R e^{i \theta} , 0 \le \theta \le \pi$, if a function $f$ is continuous on $\Gamma$ and $\displaystyle \lim_{z \to \infty} f(z) = 0$, then for a positive $m \in \mathbb{R}^{+}$, $$\lim_{R \to \infty} \int_{\Gamma} e^{m i z } f(z) dz = 0$$

By Jordan’s Lemma, since $\displaystyle \lim_{R \to \infty} \color{red} { \int_{ \Gamma} {{e^{iz} } \over {z}} dz } = 0$, $$ \int_{- \infty }^{ \infty } {{e^{ix} } \over {x}} dx = i \pi $$ Once again, since $e^{i z} = \cos z + i \sin z$, $$ \int_{- \infty }^{ \infty } {{ \cos x } \over {x}} dx + i \int_{- \infty }^{ \infty } {{ \sin x } \over {x}} dx = i \pi $$ Taking only the imaginary part gives $$ \int_{- \infty}^{\infty} {{\sin x} \over {x} } dx = \pi $$ And since the sinc function is an even function, $$ \int_{0}^{\infty} {{\sin x} \over {x} } dx = {{\pi} \over {2} } $$

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Corollary

It’s also easy to see that the following equation holds for $a\in \mathbb{R}$ through variable substitution. $$ \int_{0}^{\infty} \frac{\sin x}{x}dx=\int_{0}^{\infty} \frac{\sin (ax)}{x}dx=\frac{\pi}{2} $$ Although it may seem long and complicated at first glance, if you examine the solution closely, there really isn’t much to the calculation. Using complex analysis seems simpler and more straightforward.