Compute \[ \int_\RR {\sin(x) \over x}\dx .\]
Solution
Note that the ML bound is not sufficient to bound a semicircular contour: \[ \abs{\int_{C_R} { e^{iz} \over z}\dz } \leq \pi R \sup_{z\in C_R} \abs{1\over z} = \pi \not\to 0 .\] Jordan’s lemma on this contour yields \[ \abs{\int_{C_R} {e^{iz} \over z} \dz } \leq \pi \sup_{z\in C_R}\abs{1\over z} = {\pi \over R} \to 0 .\]
To compute the full integral, use an indented semicircular contour:
- \(C_+ \da [\eps, R]\)
- \(C_- \da [-R, -\eps]\)
- \(C_\eps \da \eps e^{it}\) with \(t\in [0, \pi]\)
- \(C_R \da R e^{it}\) with \(t\in [0, \pi]\)
- \(\Gamma \da C_+ + C_R + C_- - C_\eps\), where \(C_\eps\) is taken with reversed orientation.
Write \(I\) for the original integral, and \[ f(z) \da { e^{iz} \over z} \implies I = \Im \lim_{\eps\to 0}\lim_{R\to \infty} \qty{\int_{C_-} + \int_{C_+}}f .\] By the residue theorem \[ \int_\gamma f(z) \dz &= 2\pi i \sum_{z_k \in \HH} \Res_{z=z_k} f(z) = 0 \\ &= \qty{\int_{C_+} + \int_{C_R} + \int_{C_-} - \int_{C_\eps} }f \\ &= \tilde I + \qty{\int_{C_R} - \int_{C_\eps}}f ,\] where \(\Im(\tilde I) = I\) is the original integral, so \[ \tilde I = -\qty{\int_{C_R} - \int_{C_\eps}}f .\]
Since \(\int_{C_R}f \to 0\), it just remains to compute \(\int_{C_\eps}\). By the fractional residue formula, \[ \lim_{\eps \to 0} \int_{C_\eps}{e^{iz} \over z}\dz = i\pi \Res_{z=0} {e^{iz} \over z} = i\pi .\] Thus \[ I = \Im(i\pi) = \pi .\]