\[ I \da \int_{0}^{\infty} \frac{\cos (x)}{x^{2}+b^{2}} d x=\frac{\pi \mathrm{e}^{-b}}{2 b} .\]
Solution
The integrand is even, so \[ I = \Re{1\over 2} \tilde I \da {1\over 2} \int_\RR {e^{iz} \over (z+ib)(z-ib)} \]
Since \(f \sim 1/x^2\), the ML estimate on a semicircular contour works:
Then \(\int_{C_R} f\to 0\) and \(\int_{C_1} f\to \tilde I\). Thus \[ \tilde I &= 2\pi i \Res_{z=ib} \\ &= 2\pi i \lim_{z\to ib}{e^{iz}\over (z+ib)} \\ &= 2\pi i {e^{-b} \over 2i b} \\ &= {\pi e^{-b} \over b} \] and \[ I = \Re {1\over 2} \tilde I = {\pi e^{-b} \over 2b} .\]