Use a semicircular contour and the residue theorem to evaluate \[ I = \int_\RR {\cos(x) \over x^2 + 1 }\dx .\]
Solution
Write \[ I = \Re \int_\RR f(x) \dx && f(z) \da {e^{iz} \over z^2 + 1} .\] Define contours \(C_1 = [-R, R]\), \(C_2 = \ts{Re^{it} \st t\in [0, \pi]}\), and \(\Gamma = C_1 + C_2\). Then noting that \(z_0 = i\) is the only pole of \(f\) in \(\HH\), by the residue theorem \[ \int_\Gamma f(z) \dz = 2\pi i \Res_{z=i} f(z) = \qty{\int_{C_1} + \int_{C_2}} f .\] Note also that \[ I = \Re \lim_{R\to\infty} \int_{C_1} f .\]
By the ML estimate, \[ \abs{\int_{C_2} f(z)\dz } &\leq \pi R \sup_{z\in C_2} \abs{f(z)} \\ &= \pi R \sup_{z\in C_2} \abs{e^{iz}\over z^2 + 1} \\ &\leq \pi R \sup_{z\in C_2} {1 \over \abs{ z^2 + 1} } \text{ since } \abs{e^{iz}}=e^{-\Im z}\le1 \text{ on } \HH\\ &\leq \pi R\sup_{z\in C_2} {1 \over \abs{z}^2 - 1 } \text{ by the reverse triangle ineq.}\\ &= {\pi R\over R^2-1} \\ &= \bigo(R^{-1}) \\ &\convergesto{R\to\infty}0 .\]
For the residue, \(z=i\) is at worst an order 1 pole, and remains order 1 since \(i\) is not a zero of the numerator. \[ \Res_{z=i} f(z) &= \Res_{z=i} {e^{iz} \over (z+i)(z-i)} \\ &= { e^{iz} \over z+i}\evalfrom_{z=i} \\ &= {1\over 2ei} .\] Thus \[ \lim_{R\to\infty}\int_{C_1} f = 2\pi i \Res_{z=i} f = {2\pi i \over 2ei} = {\pi \over e} \\ \implies I = \Re\qty{\pi \over e} = {\pi \over e} .\]
Note that this also shows that \[ \int_\RR { \sin(z) \over z^2 + 1} \dz = 0 .\]