For \(a>0\), compute and justify \[ \int_0^\infty\frac{\cos(ax)}{(1+x^2)^2}\,dx =\frac{\pi(a+1)e^{-a}}4. \]
Solution
1. The exponential integrand has one double pole in the upper half-plane.
Set \(F(z)=e^{iaz}/(1+z^2)^2\). Its only upper-half-plane pole is \(i\), of order two. Differentiating the holomorphic factor gives its residue [@SS03]: \[ \begin{aligned} \operatorname{Res}_{z=i}F &=\left.\frac{d}{dz}\frac{e^{iaz}}{(z+i)^2}\right|_{z=i}\\ &=e^{-a}\left(\frac{ia}{(2i)^2}-\frac{2}{(2i)^3}\right) =-\frac{i(a+1)e^{-a}}4. \end{aligned} \] On the positively oriented contour formed by \([-R,R]\) and the upper semicircle \(\Gamma_R\), for \(R>1\), the residue theorem gives \[ \int_{-R}^R\frac{e^{iax}}{(1+x^2)^2}\,dx +\int_{\Gamma_R}F(z)\,dz =\frac{\pi(a+1)e^{-a}}2. \]
2. The semicircle vanishes and symmetry gives the requested value.
On \(\Gamma_R\), the assumption \(a>0\) implies \(|e^{iaz}|=e^{-a\operatorname{Im}z}\leq1\). Also \(|1+z^2|\geq R^2-1\). Hence \[ \left|\int_{\Gamma_R}F(z)\,dz\right| \leq\frac{\pi R}{(R^2-1)^2}\longrightarrow0. \] The integral on the real line converges absolutely: its absolute integrand is \((1+x^2)^{-2}\), bounded on \([-1,1]\) and at most \(|x|^{-4}\) outside that interval. Letting \(R\to\infty\) in step 1 is therefore justified. Its real part is \[ \int_{-\infty}^{\infty}\frac{\cos(ax)}{(1+x^2)^2}\,dx =\frac{\pi(a+1)e^{-a}}2. \] The integrand is even, so the half-line integral is \(\pi(a+1)e^{-a}/4\), as asserted.