\[ I\da \int_0^\infty {\log(x) \over 1+x^a}\dx &= - \qty{\pi \over a}^2\cos\qty{\pi\over a}\csc^2\qty{\pi \over a} \\ &= - {\pi^2\over a^2} {\cos\qty{\pi\over a} \over \sin^2\qty{\pi\over a}} .\]
Solution
Assume \(a>1\), so that the integral converges. The pole of \(f(z)\da{\log(z)\over1+z^a}\) nearest the positive axis is \(\omega_a \da e^{i\pi\over a}\). Take the sector \(0\le\arg z\le 2\pi/a\) indented at \(0\), which contains only this pole. By the ML estimate, the integrals over its arcs of radius \(R\) and \(\eps\) are \(O\qty{R\log R/R^a}\) and \(O\qty{\eps\abs{\log\eps}}\), so both tend to zero:
Set \(\zeta_a \da e^{2\pi i \over a}\). Contributions from the contours: let \(\gamma_1\) be the contour along \(\RR\) and \(\gamma_2\) along \(\zeta_a \RR\), oriented so the overall contour is counterclockwise. Then \(\int_{\gamma_1}f(z)\dz \to I\) for \(f(z) \da {\log(z) \over 1+z^a}\), so compute the monodromy term: parameterize \(\gamma_2 \da \ts{\zeta_a t \st t\in [\eps, R]}\), so \[ \int_{\gamma_2}f(z) \dz &=\int_R^\eps f(\zeta_a t) \zeta_a \dt \\ &= -\zeta_a \int_\eps^R {\log(\zeta_a t) \over (\zeta_a t)^a + 1}\dt \\ &= -\zeta_a \int_{\eps}^R {\log(t) + {2\pi i \over a} \over t^a+1}\dt \\ &\to -\zeta_a I - \zeta_a {2\pi i \over a} \int_0^\infty {1\over t^a + 1 }\dt \\ &\da -\zeta_a I - {2\pi i\over a}\zeta_a I' .\]
\[ I' = {\pi\over a}\csc\qty{\pi\over a} .\]
Computing the auxiliary integral \(I'\): the integrand has the same pole at \(\omega_a\), so apply the same technique. Write \(g(z) \da {1\over z^a+1}\).
The contributions from the contours: \[ \qty{ \int_{\gamma_1} + \int_{\gamma_2}} g(z) \to (1-\zeta_a)I' = -2i\sin\qty{\pi\over a}e^{i\pi\over a}I' ,\] using \(1-e^{2i\theta}=-2i\sin(\theta)e^{i\theta}\).
Computing the residue: \[ \Res_{z=e^{i\pi\over a}}g(z) &= {1\over az^{a-1}}\evalfrom_{z=e^{i\pi\over a}} = {1\over ae^{i\pi\qty{a-1\over a}}} = {1\over a e^{i\pi} e^{-i\pi\over a}} \\ \implies 2\pi i \Res_{z=e^{i\pi\over a}}g(z) &= -{2\pi i \over a}e^{i \pi \over a} .\]
Combining and solving: \[ I' = { - {2\pi i\over a} e^{i\pi \over a} \over -2i\sin\qty{\pi\over a}e^{i\pi\over a}} = {\pi \over a}\csc\qty{\pi \over a} .\]
Given this, the RHS of the residue theorem limits to \[ (1-\zeta_a) I - {2\pi i\over a}\zeta_a \qty{ {\pi \over a}\csc\qty{\pi\over a} } = (1-\zeta_a) I - {2\pi^2\over a^2}i\zeta_a \csc\qty{\pi\over a} .\]
For the LHS, we compute the residue at \(\omega_a\): \[ \Res_{z=\omega_a} f(z) &= \lim_{z\to \omega_a} {(z-\omega_a) \log(z) \over z^a + 1} \\ &\eqLH \lim_{z\to \omega_k} {\log(z) \over az^{a-1}}\\ &= {\log\qty{e^{i\pi \over a}} \over ae^{i\pi \qty{a-1\over a}} } \\ &= -{i\pi /a \over ae^{- i\pi \over a} } \\ &= -{i\pi \over a^2} e^{i\pi\over a} ,\] so \[ 2\pi i \Res_{z=\omega_a} f(z) = -2\pi i \qty{i\pi\over a^2}e^{i\pi \over a} = {2\pi^2\over a^2}e^{i\pi\over a} .\]
Combining the residue and contour contributions gives: \[ {2\pi^2\over a^2}e^{i\pi\over a} &= (1-\zeta_a) I - {2\pi^2\over a^2}i\zeta_a \csc\qty{\pi\over a} \\ \\ \implies {2\pi^2\over a^2}e^{i\pi\over a} &= (1-e^{2\pi i\over a}) I - {2\pi^2\over a^2}ie^{2\pi i \over a} \csc\qty{\pi\over a} \\ \\ \implies {2\pi^2\over a^2} &= (e^{-{i\pi\over a}}-e^{\pi i\over a}) I - {2\pi^2\over a^2}ie^{\pi i \over a} \csc\qty{\pi\over a} \\ \\ &= -2i\sin\qty{\pi\over a}I - {2\pi^2\over a^2}ie^{\pi i \over a} \csc\qty{\pi\over a} \\ \\ \implies I &= { {2\pi^2\over a^2} + {2\pi^2\over a^2}ie^{\pi i \over a} \csc\qty{\pi\over a} \over -2i\sin\qty{\pi \over a} } \\\\ &= {2\pi^2\over a^2} \qty{ 1 + ie^{\pi i \over a} \csc\qty{\pi\over a} \over -2i\sin\qty{\pi \over a} } \\ \\ &= -{\pi^2\over a^2}\csc\qty{\pi\over a} \cdot (-i)\qty{1 + ie^{i\pi\over a}\csc\qty{\pi\over a}} \\ \\ &= -{\pi^2\over a^2}\csc\qty{\pi\over a} \qty{ -i + \qty{\cos\qty{\pi\over a} +i\sin\qty{\pi\over a}}\csc\qty{\pi\over a}} \\ \\ &= -{\pi^2\over a^2}\csc\qty{\pi\over a} \qty{-i + \cot\qty{\pi\over a} + i } \\ \\ &= -{\pi^2\over a^2}\csc\qty{\pi\over a} \cot\qty{\pi\over a} \\ &= -{\pi^2\over a^2}\cos\qty{\pi\over a} \csc^2\qty{\pi\over a} \\ .\]