How to calculate $\int\frac{e^{iz}}{z}\, dz$ on the semi-circle given by $re^{i\theta}$ where $\theta:\,0\to\pi$

Question

I was reading this post and in the comments someone said that the difficult in calculating the limit as $r$ goes to $0$ is a lot different than calculating the limit. I tried to calculate the integral and don't seems that difficult but a lot of algebraic work, are there a trick to easily calculate the integral?.

Answer 1

There's no closed form. Take a counterclockwise contour comprising a large semicircle $\{r e^{i \theta} | 0 \leq \theta \leq \pi\}$, a small semicircle $\{\epsilon e^{i \theta} | 0 \leq \theta \leq \pi\}$, and real intervals $[-r, -\epsilon]$ and $[\epsilon, r]$. This looks like a semicircle with a dimple cut out around the origin. The integrand has no poles in this region, so the integral around the whole contour is zero. If $x$ is real, then $$\frac{e^{ix}}{x} = \frac{i \sin x}{x} = \frac{i \sin (-x)}{-x} = \frac{e^{-ix}}{-x}$$ so the integrals over the real intervals are the same and equal $i \int_\epsilon^r \frac{\sin x}{x}\, dx$, which has no closed form; the special function $\operatorname{Si}(x) = \int_0^x \frac{\sin t}{t}\, dt$ is sometimes used.

Finally, as shown in the question you linked, at the $\epsilon \to 0$ limit, the integral over the small semicircle is $-\pi i$ (with the negative sign because the contour traverses this small semicircle in the wrong direction, left-to-right). Thus the integral over the outer semicircle is $$\pi i - 2i \operatorname{Si}(r).$$

It's possible to show that the $r \to \infty$ limit of this expression is zero.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

By closing the contour along the real axis while an indent $\ds{\braces{z\ \mid\ z = \epsilon\expo{\ic\theta},\ r > \epsilon > 0, \ \theta \in \bracks{0,\pi} }}$ is provided, I'll find:

\begin{align} &\bbox[10px,#ffd]{\int_{\large z\ \in\ r\exp\pars{\ic\bracks{0,\pi}} \atop \large r\ >\ 0}{\expo{\ic z} \over z}\,\dd z} \\[5mm] = &\ \lim_{\epsilon \to 0^{\large +}}\pars{-\int_{-r}^{-\epsilon}{\expo{\ic x} \over x}\,\dd x - \int_{\pi}^{0}\ic\,\dd\theta -\int_{\epsilon}^{r}{\expo{\ic x} \over x}\,\dd x} \\[5mm] = &\ \lim_{\epsilon \to 0^{\large +}}\pars{\int_{\epsilon}^{r}{\expo{-\ic x} \over x}\,\dd x + \pi\ic -\int_{\epsilon}^{r}{\expo{\ic x} \over x}\,\dd x} \\[5mm] = &\ \pi\ic - 2\ic\int_{0}^{r}{\sin\pars{x} \over x} \,\dd x = \bbx{\pi\ic - 2\ic\,\mrm{Si}\pars{r}} \end{align}

$\ds{\mrm{Si}}$ is the Sine Integral Function.