# Showing $\int_{0}^{\infty} \frac{\sin{x}}{x} \ dx = \frac{\pi}{2}$ using complex integration

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 so $$\int_{-R}^R f(z)dz = \int_{C_R} \frac{1}{z}dz - \int_{C_R} \frac{e^{iz}}{z}dz$$ but $$\int_{C_R} \frac{1}{z}dz = \pi i$$ and we can show that the other integral goes to zero as $R \to \infty$. Since $$\int_{-R}^R \frac{\sin x}{x}dx = \operatorname{Im}I$$ we have $$\int_{-\infty}^\infty \frac{\sin x}{x}dx = \pi$$ and $$\int_0^\infty \frac{\sin x}{x}dx = \frac{\pi}{2}.$$
 thanks, very helpful ! – Beltrame Oct 24 '12 at 11:48 @glebovg: can you show us how to prove that the other integral goes to zero ? – aziiri Apr 5 at 22:35 @aziiri Recall that $e^{iz} = \cos z + i\sin z$ and use the Squeeze Theorem. – glebovg May 6 at 16:55