- Compute the integral
\[ \int _ { 0 } ^ { \infty } { \frac { ( \log x ) ^ { 2 } } { 1 + x ^ { 2 } } } d x . \]
Solution
The value is \(\boxed{\pi^3/8}\).
1. The improper integral converges.
On \((0,1]\) the integrand is at most \((\log x)^2\), and on \([1,\infty)\) it is at most \((\log x)^2/x^2\). Substituting \(x=e^{-t}\) in the first comparison integral and \(x=e^t\) in the second makes each equal to \(\int_0^\infty t^2e^{-t}\,dt=2\), by two integrations by parts. Thus the requested nonnegative integral is finite.
2. An indented upper semicircle gives a residue identity.
Choose \(\operatorname{Log}z=\log|z|+i\arg z\) with \(-\pi/2<\arg z<3\pi/2\). Its cut is the nonpositive imaginary axis, so it is holomorphic near the upper half-plane away from zero [@SS03]. For \(0<\varepsilon<1<R\), use the contour consisting of \([-R,-\varepsilon]\), the clockwise upper arc of radius \(\varepsilon\), \([\varepsilon,R]\), and the counterclockwise upper arc of radius \(R\).
For \(F(z)=(\operatorname{Log}z)^2/(1+z^2)\) the only pole inside is \(i\), with \[ \operatorname{Res}_{z=i}F=\frac{(i\pi/2)^2}{2i} =\frac{i\pi^2}{8}. \] The residue theorem therefore gives total contour integral \(-\pi^3/4\) [@SS03]. Since \(\operatorname{Log}(-x)=\log x+i\pi\) for \(x>0\), the two real segments contribute \[ \int_\varepsilon^R \frac{(\log x+i\pi)^2+(\log x)^2}{1+x^2}\,dx. \] Its real part is \[ 2\int_\varepsilon^R\frac{(\log x)^2}{1+x^2}\,dx -\pi^2\int_\varepsilon^R\frac{dx}{1+x^2}. \]
3. Both arcs vanish, determining the integral.
1. Since\(\int_0^\infty dx/(1+x^2)=\pi/2\) by the arctangent primitive,the resulting equation is\[ 2I-\frac{\pi^3}{2}=-\frac{\pi^3}{4}. \]Hence \(I=\pi^3/8\).