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.
 A: Let us define, for $r>0$,
$$\Gamma_{r}=\{r e^{i\alpha}\ :\ 0\leq\alpha\leq\theta\}$$
The circular arcs of your contour are then $\Gamma_{\epsilon}$ and $\Gamma_{R}$. 
We note that, for $z\in\Gamma_{\epsilon}$
$$\left|\frac{\ln z}{1+z^n}\right|\leq|\ln z|\leq 2\ln\epsilon$$
if $\epsilon$ is small enough.
So
$$\left|\int_{\Gamma_\epsilon} f(z)\right|\leq 2\theta\epsilon\ln\epsilon$$
which goes to $0$ with $\epsilon$.
Also, for $R$ large enough,
$$|\ln z|\leq 2\ln R$$
when $z\in \Gamma_R$. So
$$\left|\frac{\ln z}{1+z^n}\right|\leq\frac{2\ln R}{R^n-1}$$
hence
$$\left|\int_{\Gamma_R} f(z)\right|\leq\theta R\frac{2\ln R}{R^n-1}$$
which goes to $0$ when $R\to\infty$, if $n>1$.
(So, yes, these two pieces go to zero)
Now take $\Gamma_1=\{r\ :\ \epsilon\leq r\leq R\}$ and $\Gamma_2=e^{i\theta}\Gamma_1$; define the contour $\Gamma$ by walking along $\Gamma_1$, then $\Gamma_R$, then backwards along $\Gamma_2$ and $\Gamma_\epsilon$.
The roots of $1+z^n=0$ are the $n$-th roots of $-1$, that is $$z_k=e^{i\frac{(2k+1)\pi}{n}}$$
so, if we take $\theta\in (\frac{\pi}{n},\frac{3\pi}{n})$, $\Gamma$ surrounds exactly one pole of $f$.
Now, following the hint of Christopher A. Wong, we set $\theta=\frac{2\pi}{n}$; therefore, if $z\in\Gamma_2$, we get $z^n=|z|^ne^{in\theta}=|z|^n$ and $\ln z=\ln|z| + i\frac{2\pi}{n}$.
So
$$\int_{\Gamma_2}f(z)=\int_{\epsilon}^R\frac{\ln x + i\frac{2\pi}{n}}{1+x^n}e^{i\frac{2\pi}{n}}dx$$
We recall that
$$\int_0^\infty\frac{1}{1+x^n}dx=\frac{\pi/n}{\sin(\pi/n)}$$
with the same method we used now (integrating on $\Gamma$!).
Call $J=\int_{0}^\infty \ln x/(1+x^n) dx$ and $L$ the residue of $f(z)=\ln z /(1+z^n)$ at $z=e^{i\frac{\pi}{n}}$, then
$$2i\pi L=J-e^{i\frac{2\pi}{n}}\left(J+i\frac{2\pi}{n}\frac{\pi/n}{\sin(\pi/n)}\right)$$
that is
$$J=\frac{2i\pi L+ie^{i\frac{2\pi}{n}}\frac{2\pi^2/n^2}{\sin(\pi/n)}}{1-e^{i\frac{2\pi}{n}}}$$
Now it is enough to perform the computations.
A: Here's an alternative solution. Consider a standard keyhole contour:

and let
$$f(z) = \frac{(\log z)^2}{1+z^n}$$
where $\log$ denotes the natural branch of the complex logarithm. The usual estimates over $\gamma$ and $\Gamma$ show that
$$\int_\gamma f(z)\,dz \to 0 \qquad\text{and}\qquad \int_\Gamma f(z)\,dz \to 0$$
as $r \to 0$ and $R \to \infty$. In the limit, we will be left with
$$\int_0^\infty \frac{(\ln x)^2}{1+x^n}\,dx$$
from the ''upper'' part of the real axis and
$$-\int_0^\infty \frac{(\ln x + 2\pi i)^2}{1+x^n}\,dx$$
from the ''lower'' part. Adding these together, the $(\ln x)^2$-terms will cancel, and we end up with
$$4\pi i \int_0^\infty \frac{\ln x}{1+x^n} - 4\pi^2 \int_0^\infty \frac{1}{1+x^n}\,dx.$$
By the residue theorem, this sum will be equal to
$$2\pi i \sum \operatorname{Res}(f; \alpha_k)$$
where the sum is taken over all poles of $f$. These poles are given by the solutions to $z^n = -1$, i.e. $\alpha_k = \exp\Big( \frac{i\pi}{n}(2k+1) \Big)$ for $0 \le j \le n-1$. Also,
\begin{align}
\operatorname{Res}(f; \alpha_k) &= \frac{(\log \alpha_k)^2}{n\alpha_k^{n-1}} = -\frac{\alpha_k(\log \alpha_k)^2}{n} \\
&= -\frac{\exp\Big( \frac{i\pi}{n}(2k+1) \Big)\Big(\frac{i\pi}{n}(2k+1) \Big)^2}{n} \\
&= \frac{\pi^2}{n^3} \exp\Big( \frac{i\pi}{n}(2k+1) \Big)\Big((2k+1) \Big)^2.
\end{align}
Summing up (while keeping track of all the $i$:s), we end up with
\begin{align}
\int_0^\infty \frac{\ln x}{1+x^n} &= -\frac12 \operatorname{Re}
\left( \sum \operatorname{Res}(f; \alpha_k) \right) \\
&= -\frac{\pi^2}{2n^3} \sum_{j=0}^{n-1} \exp\Big( \frac{i\pi}{n}(2k+1) \Big)\Big((2k+1) \Big)^2.
\end{align}
(I'll leave the algebra that shows that this answer is the same as the other one given as an exercise.)
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{\int_{0}^{\infty}{\ln\pars{x} \over 1 + x^{n}}\,\dd x:\ {\large ?}.\qquad
     n \geq 2.}$

\begin{align}
&\color{#c00000}{\int_{0}^{\infty}{\ln\pars{x} \over 1 + x^{n}}\,\dd x}
=\int_{0}^{\infty}{\ln\pars{x^{1/n}} \over 1 + x}\,{1 \over n}\,x^{1/n - 1}\dd x
={1 \over n^{2}}\int_{0}^{\infty}{x^{1/n - 1}\ln\pars{x} \over 1 + x}\,\dd x
\\[3mm]&={1 \over n^{2}}\lim_{\mu \to 1/n - 1}\partiald{}{\mu}
\color{#00f}{\int_{0}^{\infty}{x^{\mu} \over 1 + x}\,\dd x}\tag{1}
\end{align}

\begin{align}
&\color{#00f}{\int_{0}^{\infty}{x^{\mu} \over 1 + x}\,\dd x}
=2\pi\ic\expo{\pi\mu\ic}
-\int^{0}_{\infty}{x^{\mu}\expo{2\pi\mu\ic} \over 1 + x}\,\dd x 
=2\pi\ic\expo{\pi\mu\ic}
+\expo{2\pi\mu\ic}\color{#00f}{\int_{0}^{\infty}{x^{\mu} \over 1 + x}\,\dd x}
\end{align}

$$
\color{#00f}{\int_{0}^{\infty}{x^{\mu} \over 1 + x}\,\dd x}
=2\pi\ic\expo{\pi\mu\ic}\,{1 \over 1 - \expo{2\pi\mu\ic}}
=\pi\,{2\ic \over \expo{-\pi\mu\ic} - \expo{\pi\mu\ic}}
=-\pi\csc\pars{\pi\mu}
$$

Replacing in $\pars{1}$:

\begin{align}
&\color{#66f}{\large\int_{0}^{\infty}{\ln\pars{x} \over 1 + x^{n}}\,\dd x}
={1 \over n^{2}}
\lim_{\mu \to 1/n - 1}\partiald{\bracks{-\pi\csc\pars{\pi\mu}}}{\mu}
\\[3mm]&={1 \over n^{2}}\,
\lim_{\mu \to 1/n - 1}\bracks{\pi^{2}\cot\pars{\pi\mu}\csc\pars{\pi\mu}}
={\pi^{2} \over n^{2}}\,\cot\pars{{\pi \over n} - \pi}
\csc\pars{{\pi \over n} - \pi}
\\[3mm]&=\color{#66f}{\large-\,{\pi^{2} \over n^{2}}\,\cot\pars{{\pi \over n}}
\csc\pars{{\pi \over n}}}\,,\qquad n > 1
\end{align}

A: For what is worth, one can exhibit the solution as $$\sum\limits_{k \in \Bbb Z} {{{{{\left( { - 1} \right)}^{k+1}}} \over {{{\left( {kn + 1} \right)}^2}}}}  = \sum\limits_{ - \infty }^{ + \infty } {{{{{\left( { - 1} \right)}^{k+1}}} \over {{{\left( {kn + 1} \right)}^2}}}} $$
by making a change of varibles $x\mapsto e^u$ and then integrating over the positive and negative sides of $\Bbb R$ by expanding $$\frac 1{1+e^{nu}}$$ as $$\sum\limits_{k = 0}^\infty  {{{\left( { - 1} \right)}^k}} {e^{nku}}$$ or $$\sum\limits_{k = 0}^\infty  {{{\left( { - 1} \right)}^k}} {e^{-n(k+1)u}}$$ to assure convergence. Then one uses $$ - \int\limits_{ - \infty }^0 {u{e^{au}}du}  = \int\limits_0^\infty  {u{e^{ - au}}du}  = {1 \over {{\alpha ^2}}}$$ to obtain a split series 
$$\sum\limits_{k = 0}^\infty  {{{\left( { - 1} \right)}^k}} {{ - 1} \over {{{\left( {nk + 1} \right)}^2}}} + \sum\limits_{k = 0}^\infty  {{{\left( { - 1} \right)}^k}} {1 \over {{{\left( {n\left( {k + 1} \right) - 1} \right)}^2}}}$$
which one can merge by $$\eqalign{
  & \sum\limits_{k = 0}^\infty  {{{\left( { - 1} \right)}^k}} {{ - 1} \over {{{\left( {nk + 1} \right)}^2}}} - \sum\limits_{k = 0}^\infty  {{{\left( { - 1} \right)}^{k + 1}}} {1 \over {{{\left( {n\left( {k + 1} \right) - 1} \right)}^2}}} =   \cr 
  &  - \sum\limits_{k = 0}^\infty  {{{\left( { - 1} \right)}^k}} {1 \over {{{\left( {nk + 1} \right)}^2}}} - \sum\limits_{k = 1}^\infty  {{{\left( { - 1} \right)}^k}} {1 \over {{{\left( {nk - 1} \right)}^2}}} =   \cr 
  &  - \sum\limits_{k = 0}^\infty  {{{\left( { - 1} \right)}^k}} {1 \over {{{\left( {nk + 1} \right)}^2}}} - \sum\limits_{k = 1}^\infty  {{{\left( { - 1} \right)}^k}} {1 \over {{{\left( { - nk + 1} \right)}^2}}} =   \cr 
  &  - \sum\limits_{ - \infty }^\infty  {{{\left( { - 1} \right)}^k}} {1 \over {{{\left( {nk + 1} \right)}^2}}} =   \cr 
  & \sum\limits_{ - \infty }^\infty  {{{\left( { - 1} \right)}^{k + 1}}} {1 \over {{{\left( {nk + 1} \right)}^2}}} \cr} $$
Most probably, one can confirm the series equals what robjohn wrote by using the identities relating polygamma-1 to the cotangent. In fact $$\int\limits_0^\infty  {{{\log x} \over {1 + {x^n}}}dx}  = 1 - {1 \over {{n^2}}}\left[ {\varphi \left( {{1 \over n}} \right) + \varphi \left( { - {1 \over n}} \right)} \right]$$ where $$\sum\limits_{k = 0}^\infty  {{{\left( { - 1} \right)}^k}} {1 \over {{{\left( {k + z} \right)}^2}}} = \varphi \left( z \right)$$ if I'm not getting any sign wrong.
ADD Then, we have that $$I(n) = 1 - {1 \over {{n^2}}}\left[ {\varphi \left( {{1 \over n}} \right) + \varphi \left( { - {1 \over n}} \right)} \right] = 1 - {1 \over {{{\left( {2n} \right)}^2}}}\left[ {{\psi ^{\left( 1 \right)}}\left( {{1 \over {2n}}} \right) - {\psi ^{\left( 1 \right)}}\left( {{1 \over {2n}} + {1 \over 2}} \right) + {\psi ^{\left( 1 \right)}}\left( { - {1 \over {2n}}} \right) - {\psi ^{\left( 1 \right)}}\left( { - {1 \over {2n}} + {1 \over 2}} \right)} \right]$$
and W|A confirms that $$\psi^{(1)} \left( {{1 \over {2n}}} \right) - \psi^{(1)} \left( {{1 \over {2n}} + {1 \over 2}} \right) + \psi^{(1)} \left( { - {1 \over {2n}}} \right) - \psi^{(1)} \left( { - {1 \over {2n}} + {1 \over 2}} \right) ={4n}^2 + {\pi ^2}{\csc ^2}{\pi  \over {2n}} - {\pi ^2}{\sec ^2}{\pi  \over {2n}}$$ which proves what robjohn wrote, but I still can't find a source that proves this last identity.
A: 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
$$ \psi^{(1)} \left( \frac{1}{2} - \frac{1}{2n} \right) + \psi^{(1)} \left( \frac{1}{2} + \frac{1}{2n} \right) = \pi^2 \csc^2 \left(\frac{\pi}{2} - \frac{\pi}{2n} \right) = \pi^2 \sec^2 \left( \frac{\pi}{2n} \right).  $$
A: Using the contour in the question:
$\hspace{4cm}$
contour integration gives
$$
\begin{align}
&2\pi i\,\mathrm{Res}\left(\frac{\log(z)}{1+z^n},z=e^{i\pi/n}\right)\\
&=\frac{2\pi^2e^{i\pi/n}}{n^2}\\
&=\color{#00A000}{\lim_{R\to\infty}\int_0^R\frac{\log(x)}{1+x^n}\,\mathrm{d}x}
\color{#C00000}{+\int_0^{2\pi/n}\frac{\log(R)+ix}{1+R^ne^{inx}}iRe^{ix}\mathrm{d}x}
\color{#0000FF}{-e^{i2\pi/n}\int_0^R\frac{\log(x)+i2\pi/n}{1+x^n}\,\mathrm{d}x}\\
&=\left(1-e^{i2\pi/n}\right)\int_0^\infty\frac{\log(x)}{1+x^n}\,\mathrm{d}x
-e^{i2\pi/n}\int_0^\infty\frac{i2\pi/n}{1+x^n}\,\mathrm{d}x\tag{1}
\end{align}
$$
where the green integral is the outgoing line, the red integral is the circular arc, and the blue integral is the returning line. For $n>1$, the integral in red vanishes as $R\to\infty$.
Dividing $(1)$ by $-2ie^{i\pi/n}$, we get
$$
i\frac{\pi^2}{n^2}=\sin(\pi/n)\int_0^\infty\frac{\log(x)}{1+x^n}\,\mathrm{d}x
+e^{i\pi/n}\int_0^\infty\frac{\pi/n}{1+x^n}\,\mathrm{d}x\tag{2}
$$
Taking the imaginary part of $(2)$, we get
$$
\frac{\pi^2}{n^2}=\sin(\pi/n)\int_0^\infty\frac{\pi/n}{1+x^n}\,\mathrm{d}x\tag{3}
$$
and taking the real part of $(2)$, we get
$$
\sin(\pi/n)\int_0^\infty\frac{\log(x)}{1+x^n}\,\mathrm{d}x
+\cos(\pi/n)\int_0^\infty\frac{\pi/n}{1+x^n}\,\mathrm{d}x=0\tag{4}
$$
Combining $(3)$ and $(4)$ yields
$$
\int_0^\infty\frac{\log(x)}{1+x^n}\,\mathrm{d}x=-\frac{\pi^2}{n^2}\csc(\pi/n)\cot(\pi/n)\tag{5}
$$

An alternate method
Using this result
$$
\frac{\pi}{n}\csc\left(\pi\frac{m+1}{n}\right)=\int_0^\infty\frac{x^m}{1+x^n}\,\mathrm{d}x\tag{6}
$$
Differentiating in $m$ yields
$$
-\frac{\pi^2}{n^2}\csc\left(\pi\frac{m+1}{n}\right)\cot\left(\pi\frac{m+1}{n}\right)=\int_0^\infty\frac{\log(x)x^m}{1+x^n}\,\mathrm{d}x\tag{7}
$$
Therefore, setting $m=0$ gives
$$
-\frac{\pi^2}{n^2}\csc\left(\frac\pi n\right)\cot\left(\frac\pi n\right)=\int_0^\infty\frac{\log(x)}{1+x^n}\,\mathrm{d}x\tag{7}
$$
A: We can use Integration under Differentiation to solve this problem instead of residues or special functions. It is much easier. Let
$$ J(a)=\int_0^\infty \frac{x^a}{1+x^n}dx. $$
Note $J'(0)=I$. First we calculate
\begin{eqnarray}
J(a)&=&\int_0^1\frac{x^a+x^{n-a-2}}{1+x^n}dx\\
&=&\int_0^1\sum_{k=0}^\infty(-1)^n(x^a+x^{n-a-2})x^{kn}dx\\
&=&\sum_{k=0}^\infty(-1)^k\left(\frac{1}{nk+a+1}+\frac{1}{nk+n-a-1}\right)\\
&=&\sum_{k=0}^\infty(-1)^k\frac{1}{nk+a+1}\\
&=&\sum_{k=-\infty}^\infty(-1)^k\frac{1}{nk+a+1}\\
&=&\frac{\pi}{n\sin(\frac{a+1}{n}\pi)}.
\end{eqnarray}
So
$$ J'(0)=-\frac{\pi^2\cos(\frac{\pi}{n})}{n^2\sin^2(\frac{\pi}{n})}. $$
Here we used the following result
$$ \sum_{k=-\infty}^\infty(-1)^k\frac{1}{ak+b}=\frac{\pi}{a\sin(\frac{b}{a}\pi)}.$$
