Evaluate \[ \int_{-\infty}^{\infty}\frac{e^{-ix}}{x^2-2x+4}\,dx. \]
Solution
Let \[ f(z)\coloneqq\frac{e^{-iz}}{z^2-2z+4}. \]
1. The poles of \(f\) are\[ z=1\pm i\sqrt{3}, \]and the only pole in the lower half-plane is\[ z_0\coloneqq1-i\sqrt{3}. \]
One has \[ z^2-2z+4 = (z-1)^2+3 = (z-1-i\sqrt{3})(z-1+i\sqrt{3}). \] Both poles are simple, and only \(1-i\sqrt{3}\) has negative imaginary part.
2. If the real interval \([-R,R]\) is closed by the lower semicircle, thenthe integral over that semicircle tends to \(0\) as \(R\to\infty\).
For \(z=x+iy\) in the lower half-plane, \[ \abs{e^{-iz}}=e^y\leq1. \] On the semicircle \(\abs z=R\), \[ \abs{z^2-2z+4} \geq R^2-2R-4. \] Its length is \(\pi R\), so the \(ML\) estimate gives \[ \abs{\int_{\text{lower arc}}f(z)\,dz} \leq \frac{\pi R}{R^2-2R-4} \longrightarrow0. \]
3. The residue at \(z_0=1-i\sqrt{3}\) is\[ \Res_{z=z_0}f(z) = \frac{e^{-\sqrt{3}-i}}{-2i\sqrt{3}}. \]
Since the pole is simple, \[ \Res_{z=z_0}f(z) = \frac{e^{-iz_0}}{2z_0-2}. \] Now \[ 2z_0-2=-2i\sqrt{3}, \] and \[ e^{-iz_0} = e^{-i(1-i\sqrt{3})} = e^{-\sqrt{3}-i}. \] Substitution gives the displayed residue.
4. One has\[ \int_{-\infty}^{\infty}\frac{e^{-ix}}{x^2-2x+4}\,dx = \boxed{\frac{\pi}{\sqrt{3}}e^{-\sqrt{3}-i}}. \]
The lower semicircular contour is clockwise. By the residue theorem, \[ \int_{-R}^{R}f(x)\,dx + \int_{\text{lower arc}}f(z)\,dz = -2\pi i\Res_{z=z_0}f(z). \] Letting \(R\to\infty\) and using step 2 gives \[ \int_{-\infty}^{\infty}f(x)\,dx = -2\pi i \frac{e^{-\sqrt{3}-i}}{-2i\sqrt{3}} = \frac{\pi}{\sqrt{3}}e^{-\sqrt{3}-i}. \]
5. Q.E.D.
Step 4 gives the requested value.