\[ I \da \int_\RR {xe^{2ix} \over x^2-1}\dx = i\pi \cos(2) .\]
Solution
Let \(f(z)={ze^{2iz}\over z^2-1}\). The denominator is \((z-1)(z+1)\), so \(f\) has simple poles at \(\pm1\) on \(\RR\), and \(I\) is a principal value there. For \(0<\eps<1<R\), define the contour
- \(C_1: [-R, -1-\eps]\), \([-1+\eps, 1-\eps]\) and \([1+\eps, R]\) on the real axis
- \(C_2: -1 + \eps e^{it}\), \(t\) from \(\pi\) to \(0\) (clockwise, above \(-1\))
- \(C_4: 1 + \eps e^{it}\), \(t\) from \(\pi\) to \(0\) (clockwise, above \(1\))
- \(C_R: Re^{it}, t\in [0, \pi]\)
- \(\Gamma = C_1 + C_2 + C_4 + C_R\)
The figure shows the same indentations on a rectangular contour; here the outer arc is the semicircle \(C_R\):
\(\int_\Gamma f = 0\) since \(\Gamma\) encloses no singularities. By Jordan’s lemma, \(\int_{C_R}f\to 0\), since \(\abs{z/(z^2-1)}\le R/(R^2-1)\) on \(C_R\). A clockwise half-circle about a simple pole contributes \(-\pi i\) times the residue in the limit \(\eps\to0\). With \[ \Res_{z=-1} f(z) = \lim_{z\to -1} {ze^{2iz} \over z-1} = {e^{-2i} \over 2}, \qquad \Res_{z=1} f(z) = \lim_{z\to 1} {ze^{2iz} \over z+1} = {e^{2i} \over 2} ,\] letting \(\eps\to0\) and \(R\to\infty\) gives \[ 0 = I - \pi i\qty{{e^{-2i} \over 2} + {e^{2i}\over 2}} ,\] so \[ I = \pi i \cos(2) .\]