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Recently I had to use the fact that the Dirichlet integral evaluates as

$\int_{0}^{\infty} \frac{\sin{x}}{x} \ dx = \frac{\pi}{2}$ a couple of times.

There already is a question that specifically ask for methods to show this result $\textbf{not}$ using complex integration. In this question I am interested in seeing the derivation via contour integration. ( I am aware of the wikipedia entry, but am looking for more detail )

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We need to use $f(z) = (e^{iz} - 1)/z$ because it has a removable singularity at $z = 0$. Consider a contour $C = [-R, R] \cup C_R$ for $R > 0$. Then $I \equiv \int_{-R}^R f(z)dz + \int_{C_R} f(z)dz = 0$ by Cauchy Theorem, i.e., $\int_{-R}^R f(z)dz = \int_{C_R} \frac{1}{z}dz - \int_{C_R} \frac{e^{iz}}{z}dz$ but $\int_{C_R} \frac{1}{z}dz = \pi i$ and we can show that the other integral goes to zero as $R \to \infty$. Therefore, because $\int_{-R}^R \frac{\sin x}{x}dx = \operatorname{Im}I,$ we see that $\int_{-\infty}^\infty \frac{\sin x}{x}dx = \pi$ or $\int_0^\infty \frac{\sin x}{x}dx = \frac{\pi}{2}.$

Hope this helps.

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    @N3buchadnezzar $C_R$ is the upper half-circle with radius R > 0 traversed once in the counterclockwise direction, i.e., $C_R = \{z : z = Re^{it}, t \in [0, \pi]\}$.2013-07-31