Use residues to calculate the integral \(\int_0^\infty \frac{1}{(1 + x^2)^2} \, dx\).
Solution
1. The upper-half-plane contour integral is \(\pi/2\).
Let \(F(z)=(1+z^2)^{-2}\). For \(R>1\), use the contour formed by \([-R,R]\) followed by the counterclockwise upper semicircle \(\Gamma_R\) of radius \(R\). The only enclosed pole is the double pole at \(i\). Factoring \(F(z)=(z-i)^{-2}(z+i)^{-2}\) gives \[ \operatorname{Res}_{i}F =\left.\frac{d}{dz}(z+i)^{-2}\right|_{z=i} =-\frac{2}{(2i)^3}=\frac{1}{4i}. \] The residue theorem therefore gives \[ \int_{-R}^R\frac{dx}{(1+x^2)^2} +\int_{\Gamma_R}F(z)\,dz =2\pi i\frac{1}{4i}=\frac\pi2 \]
2. The requested integral is\[ \boxed{\int_0^\infty \frac{dx}{(1+x^2)^2}=\frac{\pi}{4}}. \]
For \(\abs{z}=R\), the inequality \(\abs{1+z^2}\geq R^2-1\) yields \[ \abs{\int_{\Gamma_R}F(z)\,dz} \leq\frac{\pi R}{(R^2-1)^2}\longrightarrow0. \] The real integral converges absolutely, since its integrand is bounded by one on \([-1,1]\) and by \(\abs{x}^{-4}\) outside that interval. Letting \(R\to\infty\) in step 1 thus gives \(\int_{-\infty}^\infty(1+x^2)^{-2}\,dx=\pi/2\). The integrand is even, so the requested half-line integral is one half of this value, namely \(\pi/4\).
3. Q.E.D.
Step 2 gives the requested value.