\[ \int_\RR {1 \over (1+x^2)^2} \dx .\]
Solution
- Factor \((1+z^2)^2 = ((z-i)(z+i))^2\), so \(f\) has poles at \(\pm i\) of order 2.
- Take a semicircular contour \(\gamma \da I_R \union D_R\), then \(f(z) \approx 1/z^4 \to 0\) for large \(R\) and \(\int_{D_R} f \to 0\).
- Note \(\int_{I_R} f \to \int_\RR f\), so \(\int_\gamma f \to \int_\RR f\).
- \(\int_\gamma f = 2\pi i \sum_{z_0} \Res_{z=z_0} f\), and \(z_0 = i\) is the only pole in this region.
- Compute \[ \Res_{z=i} f &= \lim_{z\to i} {1\over (2-1)!} \dd{}{z} (z-i)^2 f(z) \\ &= \lim_{z\to i} \dd{}{z} {1\over (z+i)^2 } \\ &= \lim_{z\to i} {-2 \over (z+i)^3 } \\ &= -{2 \over (2i)^3 } \\ &= {1\over 4i} \\ \\ \implies \int_\gamma f &= {2\pi i \over 4i} = \pi/2 ,\]