- (10 points) Use residues to verify that
\[ \int _ { 0 } ^ { \infty } \frac { \ln x } { ( x ^ { 2 } + 4 ) ^ { 2 } } d x = \frac { \pi } { 3 2 } ( \ln 2 - 1 ) . \]
Solution
1. The real integral is absolutely convergent.
For \(0<x\leq1\), its absolute integrand is at most \(|\log x|/16\), whose integral is finite. For \(x\geq1\), the inequalities \(\log x\leq x\) and \((x^2+4)^2\geq x^4\) bound it by \(x^{-3}\). Both endpoint integrals therefore converge absolutely. Denote the requested integral by \(I\).
2. An indented upper semicircle determines \(2I\) by a double-pole residue.
Choose the logarithm with \(-\pi/2<\arg z<3\pi/2\) and set \[ F(z)=\frac{\operatorname{Log}z}{(z^2+4)^2}. \] This branch is holomorphic off the nonpositive imaginary axis. For \(0<\varepsilon<1\) and \(R>3\), integrate along \([-R,-\varepsilon]\), the clockwise upper semicircle of radius \(\varepsilon\), \([\varepsilon,R]\), and the counterclockwise upper semicircle of radius \(R\). The only enclosed pole is \(2i\), of order two. With \(L=\operatorname{Log}(2i)=\log2+i\pi/2\), its residue is \[ \begin{aligned} \operatorname{Res}_{2i}F &=\left.\frac{d}{dz}\frac{\operatorname{Log}z}{(z+2i)^2} \right|_{z=2i}\\ &=\frac{1}{(2i)(4i)^2}-\frac{2L}{(4i)^3} =\frac{i(1-L)}{32}. \end{aligned} \] The residue theorem gives total contour integral \(\pi(L-1)/16\) [@SS03]. Since \(\operatorname{Log}(-x)=\log x+i\pi\) for \(x>0\), the sum of the two straight segments is \[ \int_\varepsilon^R\frac{2\log x+i\pi}{(x^2+4)^2}\,dx. \] Its real part is twice the corresponding truncated integral.
3. Both arcs vanish and yield the asserted value.
On the outer semicircle the integral has modulus at most \[ \frac{\pi R(\log R+\pi)}{(R^2-4)^2}\longrightarrow0. \] On the inner semicircle the analogous bound is \[ \frac{\pi\varepsilon(|\log\varepsilon|+\pi)} {(4-\varepsilon^2)^2}\longrightarrow0. \] These follow from the arc lengths, \(|\arg z|\leq\pi\) on the upper half-plane, and the reverse triangle inequality for \(z^2+4\). Taking real parts in step 2 and then letting \(\varepsilon\downarrow0\), \(R\to\infty\) is justified by these estimates and absolute convergence. It gives \[ 2I=\frac{\pi}{16}(\log2-1), \qquad \boxed{I=\frac{\pi}{32}(\log2-1)}. \]