Show that \[ \int_0^{\infty} {\cos(x) \over x^2 + b^2}\dx = {\pi e^{-b} \over 2b} .\]
Solution
Let \(I\) be the integral over \(\RR\). Since \(f(x)\) is even, the original integral is \({1\over 2}I\).
Write \(f(z) = e^{iz} / (z^2 + b^2)\). Take a semicircular contour \(\Gamma \da \gamma_1 + \gamma_2\) where \(\gamma_1\) is \([-R, R]\) on \(\RR\) and \(\gamma_2\) is the usual half-circle of radius \(R\).
Claim: \(\int_{\gamma_2} f \converges{R\to\infty}\too 0\), so \(\int_\Gamma \to \int_\RR f(z)\).
- Estimate the exponential factor: \[ \abs{f} = { \abs{e^{iz} } \over \abs{z^2 + b^2} } = {e^{-\Re z} \over \abs{z^2 + b^2} } \leq {1\over \abs{z^2 + b^2}} \converges{R\to\infty}\too 0 .\]
Compute \(\int_\Gamma f\) by residues: factor \(z^2 + b^2 = (z+ib)(z-ib)\), so the contour only contains the order 1 pole \(z_0 = ib\).
Compute the residue: \[ \Res_{z=ib}f = \lim_{z\to ib} (z-ib) {e^{iz} \over (z+ib)(z-ib) } = { e^{iz} \over z+ib} \evalfrom_{z=ib} = {e^{i(ib)} \over 2ib} = {e^{-b} \over 2ib} .\]
So the intermediate integral is \(I\) is \(2\pi i\) times this, i.e. \(I = \pi e^{-b} / b\).
And the original integral is \({1\over 2}I = \pi e^{-b} \over 2b\).