Calculating the Dirichlet Integral using Complex Analysis

In this post i will prove, that the Dirichlet Integral is equal to $\pi/2$ (see corollary below) by using complex analysis techniques. The following integral is more amenable to complex analysis techniques:

Proposition: \begin{equation} \lim_{R \to \infty} \int_{-R}^R \frac{\sin (x)}{x} dx = \pi \end{equation} Proof: Note: The integrand can be continuously extended to $0$ by $1$, because $\lim_{x \to 0} \sin (x) /x =1$. Let $R \in [1,\infty)$. Then \begin{equation} \int_{-R}^R \frac{\sin (x)}{x} dx = \lim_{(0,1) \ni \varepsilon \to 0} \bigg( \int_{-R}^{-\varepsilon} \frac{\sin (x)}{x} dx + \int_{\varepsilon}^{R} \frac{\sin (x)}{x} dx \bigg) \end{equation} Let $\varepsilon \in (0,1)$. Then (using $\forall z \in \mathbb{C}: \exp ( iz) = \cos(z) + i \sin (z)$) \begin{equation} \int_{-R}^{-\varepsilon} \frac{\sin (x)}{x} dx + \int_{\varepsilon}^{R} \frac{\sin (x)}{x} dx = \Im \bigg ( \int_{[-R,-\varepsilon]} \frac{e^{iz}}{z} dz + \int_{[\varepsilon,R]} \frac{e^{iz}}{z} dz \bigg). \end{equation} For $\varepsilon \in (0,1)$ let $\gamma_\varepsilon: [0, \pi]\to \mathbb{C}$, $\gamma_\varepsilon (t) := -\varepsilon e^{-it}$. For $R \in [1,\infty)$ let $\beta_R :[0,\pi] \to \mathbb{C}$, $\beta_R (t):= R e^{it}$. By Cauchys integral theorem: $\forall R \in [1,\infty), \varepsilon \in (0,1)$: \begin{equation} \int_{[\varepsilon,R]\beta_R [-R,-\varepsilon] \gamma_\varepsilon} \frac{e^{iz}}{z}dz=0. \end{equation} Therefore \begin{equation} \int_{[-R,-\varepsilon]} \frac{e^{iz}}{z} dz + \int_{[\varepsilon,R]} \frac{e^{iz}}{z} dz = -\int_{\gamma_\varepsilon} \frac{e^{iz}}{z} dz - \int_{\beta_R} \frac{e^{iz}}{z} dz . \end{equation} By using the power series of the exponential function it follows that $\mathbb{C}\setminus \{0\} \ni z \mapsto \frac{e^{iz}}{z} - \frac{1}{z}$ can be holomorphically extended to $\mathbb{C}$. By the fundamental estimate \begin{equation} \bigg|\int_{\gamma_\varepsilon} \frac{e^{iz}}{z} - \frac{1}{z}dz\bigg| \leq \pi \varepsilon \max_{z \in \bar D_1(0)} \bigg|\frac{e^{iz}}{z}- \frac{1}{z}\bigg| \to 0 \quad (\varepsilon \to 0). \end{equation} Now \begin{equation} \int_{\gamma_\varepsilon} \frac{1}{z} dz = \int_0^\pi \frac{1}{-\varepsilon e^{-it} } i \varepsilon e^{-it} dt = -i \pi. \end{equation} By combining Eq. (6) and (7): \begin{equation} \lim_{(0,1) \ni \varepsilon \to 0} \int_{\gamma_\varepsilon} \frac{e^{iz}}{z} dz = -i \pi. \end{equation} Using Jordan’s lemma (see my previous blog post): \begin{equation} \bigg|\int_{\beta_R} \frac{e^{iz}}{z} dz \bigg| \leq \frac{\pi}{R} \to 0 \quad ( R\to \infty). \end{equation} And so in total, combining Eq. (5), (8) and (9): \begin{equation} \lim_{R \to \infty} \lim_{(0,1) \ni\varepsilon \to 0} \bigg( \int_{[-R,-\varepsilon]} \frac{e^{iz}}{z} dz + \int_{[\varepsilon,R]} \frac{e^{iz}}{z} dz \bigg) = i \pi. \end{equation} Since $\Im$ is continuous we obtain from Eq. (3) and (2) \begin{equation} \lim_{R \to \infty} \int_{-R}^R \frac{\sin(x)}{x}dx = \pi. \end{equation} This ends the proof. $\square$

Corollary (Dirichlet Integral): \begin{equation} \lim_{R \to \infty} \int_0^R \frac{\sin (x)}{x} dx = \frac{\pi}{2}. \end{equation} Proof: This follows from the proposition, because for all $x \in \mathbb{R} \setminus \{0\}$: \begin{equation} \frac{\sin (x)}{x} = \frac{\sin (-x)}{-x}, \end{equation} since $\sin$ is an odd function. Now define $\varphi :\mathbb{R}\to \mathbb{R}$, $\varphi (x) := -x$. Then $\varphi^\prime =-1$. Let $R \in [0,\infty)$: By the substitution rule \begin{equation} \int_{-R}^0 \frac{\sin (x)}{x}dx = -\int_{-R}^0 \varphi^\prime (y) \frac{\sin (\varphi(y))}{\varphi(y)}dy =-\int_{\varphi(-R)}^{\varphi (0)} \frac{\sin (x)}{x}dx = \int_0^R \frac{\sin (x)}{x}dx . \end{equation} Therefore \begin{equation} \int_{-R}^R \frac{\sin (x)}{x}dx = \int_{-R}^0 \frac{\sin (x)}{x}dx +\int_{0}^R \frac{\sin (x)}{x}dx = 2 \int_{0}^R \frac{\sin (x)}{x}dx \end{equation} and so the proposition implies the corollary. $\square$