Evaluate \(\int_0^\infty \frac{\log x}{x^2+2}\,dx\) by contour integration using the positively oriented contour from \(-R\) to \(R\) on the real axis, indented at the origin, and the positively oriented semicircle \(|z|=R\), \(\operatorname{Im}z>0\). Choose an appropriate branch of logarithm.
Solution
The value is \(\boxed{\pi\log2/(4\sqrt2)}\).
1. The chosen branch gives an identity for the truncated integral.
Use \(\operatorname{Log}z=\log|z|+i\arg z\) with \(-\pi/2<\arg z<3\pi/2\), whose cut is the nonpositive imaginary axis. This branch is holomorphic on a neighborhood of the closed upper half-plane away from zero [@SS03]. Let \(0<\varepsilon<1<\sqrt2<R\), and take the contour consisting, in order, of \([-R,-\varepsilon]\), the clockwise upper semicircle of radius \(\varepsilon\), \([\varepsilon,R]\), and the counterclockwise upper semicircle of radius \(R\). It bounds the upper half-annulus positively.
The function \(F(z)=\operatorname{Log}z/(z^2+2)\) has only the simple pole \(i\sqrt2\) inside this contour, with \[ \operatorname{Res}_{i\sqrt2}F =\frac{\log\sqrt2+i\pi/2}{2i\sqrt2}. \] For \(x>0\) the boundary values are \(\operatorname{Log}x=\log x\) and \(\operatorname{Log}(-x)=\log x+i\pi\). Substituting \(z=-x\) on the negative real segment therefore makes the sum of the two real-segment integrals equal to \[ 2\int_\varepsilon^R\frac{\log x}{x^2+2}\,dx +i\pi\int_\varepsilon^R\frac{dx}{x^2+2}. \] Adding the two arc integrals, the residue theorem says that this expression equals \(\pi(\log\sqrt2+i\pi/2)/\sqrt2\) [@SS03].
2. Both arcs vanish in the limit, and taking real partsdetermines the required integral.
On either upper arc, \(|\arg z|\leq\pi\). On the large arc \(\Gamma_R\), the reverse triangle inequality gives \(|z^2+2|\geq R^2-2\), so \[ \left|\int_{\Gamma_R}F(z)\,dz\right| \leq\frac{\pi R(\log R+\pi)}{R^2-2}\longrightarrow0. \] On the small arc \(\gamma_\varepsilon\) one similarly has \[ \left|\int_{\gamma_\varepsilon}F(z)\,dz\right| \leq\frac{\pi\varepsilon(|\log\varepsilon|+\pi)} {2-\varepsilon^2}\longrightarrow0. \] The real logarithmic integral converges absolutely: near zero its absolute integrand is at most \(|\log x|/2\), and for \(x\geq1\) it is at most \(\log x/x^2\). Both comparison integrals are finite. The integral of \(1/(x^2+2)\) also converges.
Letting \(R\to\infty\) and \(\varepsilon\to0\) in step 1 and taking real parts gives \[ 2\int_0^\infty\frac{\log x}{x^2+2}\,dx =\frac{\pi}{\sqrt2}\log\sqrt2 =\frac{\pi\log2}{2\sqrt2}. \] Dividing by two proves the displayed answer.