$\int_\RR\frac{x\sin x}{1+x^2}\,dx$ by Jordan's lemma

Write \(f(z) = {ze^{iz} \over 1+z^2}\). Since \(f\in \bigo\qty{1\over z}\) on the semicircle of radius \(R\), whose length is \(\pi R\), the ML estimate bounds the arc integral only by a constant; Jordan’s lemma gives decay:

Write \(f(z) = e^{iz}g(z)\) where \(g(z) \da {z\over 1 + z^2}\). Write \(C_1 = [-R, R]\) and \(C_R = \ts{Re^{it} \st t\in [0, \pi]}\). By Jordan’s lemma, \[ \abs{\int_{C_R} e^{iz} g(z)\dz }\leq \pi M_R,\, \qquad M_R \da \sup_{z\in C_R}\abs{z\over 1+z^2}\leq {R\over R^2-1} ,\] so the arc integral tends to zero. By the residue theorem, \[ 2\pi i \sum_{z_k\in \HH}\Res_{z=z_k}f(z) = \qty{\int_{C_1} + \int_{C_R}}f \converges{R\to\infty} \int_\RR {xe^{ix}\over 1+x^2}\dx ,\] and \(I\) is the imaginary part of the right side. Since \(1+z^2 = (z+i)(z-i)\), the only pole in \(\HH\) is the simple pole \(z = i\), and \[ 2\pi i \Res_{z=i} f(z) &= 2\pi i \lim_{z\to i} {ze^{iz} \over z+i} \\ &= 2\pi i {ie^{-1}\over 2i} = {\pi i \over e} ,\] so \[ I = \Im{\pi i\over e} = {\pi \over e} .\]