Calculating the integral $\int_0^{\infty}{\frac{\ln x}{1+x^n}}$ using complex analysis I need to calculate $\int_0^{\infty}{\frac{\ln x}{1+x^n}}$ $,n\geq2$ using complex analysis. I probably need to use the Residue Theorem.
I use the function $f(z)={\frac{\ln z}{1+z^n}}$ in the normal branch.
I've tried to use this contour.

Where $\theta$ is an angle so that only $z_{0}$ will be in the domain (I hope this is clear from the drawing)
I estimated $|\int_{\Gamma_{R}}|$ and $|\int_{\Gamma_{\epsilon}}|$ and showed that they must tend to $0$ when $\epsilon \rightarrow0$ and $R \rightarrow\infty$. (Is it true?)
However I'm having trouble calculating $\int_{\Gamma_{2}}$ . Does it have something to do with choosing the "right" $\theta$?
Any ideas?
Thanks!
UPDATE
After Christopher's comment I chose $\theta=\frac{2\pi}{n}$ which gives, after the paramatrization $\Gamma (t) = te^{\frac{2\pi i}{n}}$, $t\in(\epsilon,R)$:
$$\int_{\Gamma_{2}}{\frac{\ln z}{1+z^n}dz} = \int_{\epsilon}^{R}{\frac{\ln (te^\frac{2\pi i}{n})}{1+t^n}e^\frac{2\pi i}{n}dt} = 
e^\frac{2\pi i}{n}\int_{\epsilon}^{R}{\frac{\ln t}{1+t^n}dt} +
ie^\frac{2\pi i}{n}\int_{\epsilon}^{R}{\frac{\frac{2\pi}{n}}{1+t^n}dt}  =$$
$$
=
e^\frac{2\pi i}{n}\int_{0}^{\infty}{\frac{\ln t}{1+t^n}dt} +
ie^\frac{2\pi i}{n}\int_{0}^{\infty}{\frac{\frac{2\pi}{n}}{1+t^n}dt}
$$
But I have no idea how to deal with the second integral.

This is NOT an answer to the original question. This is a justification of the final identity in Peter's answer. This was requested in chat, but too long to put there or in a comment box.
@Peter, all you need is recurrence and reflection for the trigamma function. Taking $z = 1/2n,$ first we find
$$ \psi^{(1)} \left( 1 + \frac{1}{2n} \right) = \psi^{(1)} \left(  \frac{1}{2n} \right) - \frac{1}{z^2} = \psi^{(1)} \left(  \frac{1}{2n} \right) -4 n^2.  $$ 
Next
$$  \psi^{(1)} \left( 1 - \left( 1 + \frac{1}{2n} \right) \right)  + \psi^{(1)} \left( 1 + \frac{1}{2n} \right) = \pi^2 \csc^2 \frac{\pi}{2n},  $$ so
$$  \psi^{(1)} \left(  -  \frac{1}{2n}  \right)  + \psi^{(1)} \left(  \frac{1}{2n} \right) - 4 n^2 = \pi^2 \csc^2 \frac{\pi}{2n},  $$ or
$$  \psi^{(1)} \left(  -  \frac{1}{2n}  \right)  + \psi^{(1)} \left(  \frac{1}{2n} \right) = 4 n^2 + \pi^2 \csc^2 \frac{\pi}{2n}.  $$
Then
$$ \psi^{(1)} \left( \frac{1}{2} - \frac{1}{2n} \right) + \psi^{(1)} \left( \frac{1}{2} + \frac{1}{2n} \right) = \pi^2 \csc^2 \left(\frac{\pi}{2} - \frac{\pi}{2n} \right) = \pi^2 \sec^2 \left( \frac{\pi}{2n} \right).  $$
