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An exercise in the second chapter of Stein-Shakarchi's Complex Analysis, asks us to evaluate the famous integral

$$\int^{\infty}_{0}\frac{\sin(x)}{x}dx=\frac{\pi}{2}$$

We are expected to do this via Cauchy's theorem by evaluating a contour integral about an indented semi-circle. The text hints that this integral equals $\frac{1}{2i} \int^{\infty}_{-\infty} \frac{e^{ix}-1}{x}$, so I think the complex function being integrated is

$$f(z)=\frac{e^{iz}-1}{z}$$

I am struggling with the integral about the indention, that is let $\gamma_{\epsilon}(\theta)=\epsilon e^{i\theta}$ parametrized clockwise so that the integral I am having issue with is

$$\int_{\gamma_{\epsilon}}\frac{e^{iz}-1}{z}dz.$$

If I consider the Taylor expansion of $e^{iz}-1=iz + E(z)$ where $E(z) \to 0$ as $z\to 0$ and then try to evaluate this integral, I obtain that

$$\int_{\gamma_{\epsilon}}\frac{e^{iz}-1}{z}dz \approx -\int_{\pi}^{0} \epsilon e^{i\theta}d\theta $$ Where I am confused is that it looks like the later integral goes to $0$ as $\epsilon \to 0$, which is not what I am looking for.

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marked as duplicate by Guy Fsone, Arnaud D., kingW3, José Carlos Santos, user99914 Nov 23 '17 at 1:53

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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You are correct. But recall that

$$\begin{align} \oint_C \frac{e^{iz}-1}{z}\,dz&=\int_{-R}^{-\epsilon}\frac{e^{ix}-1}{x}\,dx+\int_{\epsilon}^{R}\frac{e^{ix}-1}{x}\,dx\\\\ &+\int_\pi^0 \frac{e^{\epsilon e^{i\phi}}-1}{\epsilon e^{i\phi}}\,i\epsilon e^{i\phi}\,d\phi+\int_0^\pi \frac{e^{R e^{i\phi}}-1}{R e^{i\phi}}\,iR e^{i\phi}\,d\phi \tag 1 \end{align}$$

Evaluating the limit as $R\to \infty$ of the last integral on the right-hand side of $(1)$ yields

$$\begin{align} \lim_{R\to \infty}\int_0^\pi \frac{e^{iRe^{i\phi}}-1}{Re^{i\phi}}\,iRe^{i\phi}\,d\phi&=i\lim_{R\to \infty}\int_0^\pi (e^{iRe^{i\phi}}-1)\,d\phi\\\\ &=-i\pi \tag 2 \end{align}$$

Note that in arriving at $(2)$, we used the following estimates

$$\begin{align} \left|\int_0^\pi e^{iRe^{i\phi}}\,d\phi\right|&\le \int_0^\pi e^{-R\sin(\phi)}\,d\phi\\\\ &= 2\int_0^{\pi/2} e^{-R\sin(\phi)}\,d\phi\\\\ &\le 2\int_0^{\pi/2}e^{-2R\phi/\pi}\,d\phi\\\\ &=\pi\left(\frac{1-e^{-R}}{R}\right)\to 0\,\,\text{as}\,\,R\to \infty \end{align}$$

Hence, we have

$$\int_0^\infty \frac{\sin(x)}{x}\,dx=\text{Re}\left(\frac{1}{2i}(i\pi )\right)=\pi/2$$

as was to be shown!

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  • $\begingroup$ Oh thank you, for some reason I thought that since $\vert \frac{e^{iz}-1}{z} \vert \leq \frac{2}{\vert z \vert}$, that the integral you described should go to $0$ as $R \to \infty$. $\endgroup$ – user135520 Dec 30 '16 at 18:15
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    $\begingroup$ You're welcome! My pleasure. Note that $\frac{1}{z}\,dz=i\,d\phi$. And so while the integrand does approach zero, the path length approaches infinity such that the integral approaches the stated limit. $\endgroup$ – Mark Viola Dec 30 '16 at 18:30

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