Calculate $\int_0^\infty \frac{\ln x}{1 + x^4} \mathrm{d}x$ using residue calculus I need to evaluate this integral using calculus of residues: 
$$\int_0^\infty\frac{\ln(x)}{1+x^4}\mathrm{d}x$$
I know I need to consider $\displaystyle \int_0^\infty$$\frac{\ln(z)}{1+z^4}\mathrm{d}z$.
Then the integrand has singularities at at $-e^{i\pi/4}$, $e^{i\pi/4}$, $-e^{3i\pi/4}$, and $e^{3i\pi/4}$. Then I believe I should find the residues of the integrand, however I am not sure how to proceed.
Could anyone point me in the right direction?
 A: There are a number of ways to attack this using the residue theorem.  One way is to consider the following contour integral:
$$\oint_C dz \frac{\log{z}}{1+z^4} $$
where $C$ is a quarter circle of radius $R$ in the upper-right quadrant, modified by a small quarter-circle of radius $\epsilon$ so as to avoid the branch point at $z=0$.  Thus the contour integral is equal to
$$\int_{\epsilon}^R dx \frac{\log{x}}{1+x^4} + i R \int_0^{\pi/2} d\theta \, e^{i \theta} \frac{\log{(R e^{i \theta})}}{1+R^4 e^{i 4 \theta}} \\ + i \int_R^{\epsilon} dy \, \frac{\log{y} + i \pi/2}{1+y^4} + i \epsilon \int_{\pi/2}^0 d\phi \, e^{i \phi} \frac{\log{(\epsilon e^{i \phi})}}{1+\epsilon^4 e^{i 4 \phi}}$$
As $R\to\infty$, the second integral vanishes as $\log{R}/R^3$; as $\epsilon \to 0$, the fourth integra. vanishes as $\epsilon \log{\epsilon}$.  In these limits, the contour integral becomes
$$(1-i) \int_0^{\infty} dx  \frac{\log{x}}{1+x^4} + \frac{\pi}{2} \int_0^{\infty}  \frac{dx}{1+x^4}$$
By the residue theorem, the contour integral is $i 2 \pi$ times the residue of the pole inside $C$, namely $e^{i \pi/4}$, so that we have
$$(1-i) \int_0^{\infty} dx  \frac{\log{x}}{1+x^4} + \frac{\pi}{2} \int_0^{\infty}  \frac{dx}{1+x^4} = i 2 \pi \frac{i \pi/4}{4 e^{i 3 \pi/4}} = \frac{\pi^2}{8} e^{i \pi/4} $$
By equating imaginary parts, we find that
$$\int_0^{\infty} dx  \frac{\log{x}}{1+x^4} = -\frac{\sqrt{2}}{16} \pi^2 $$
We can also find the other integral as well from the real part:
$$\int_0^{\infty}  \frac{dx}{1+x^4} = \frac{\sqrt{2}}{4} \pi $$
A: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
$\ds{\int_{0}^{\infty}{\ln\pars{x} \over 1 + x^{4}}\,\dd x:\ {\large ?}}$

The following method 'avoids' the four poles of the original integral.
Instead, we just have to consider a
  $\color{#c00000}{\ds{\it\underline{\mbox{one pole-integral}}}}$:

Note that
\begin{align}&\color{#c00000}{\int_{0}^{\infty}{\ln\pars{x} \over 1 + x^{4}}
\,\dd x}
=\int_{0}^{\infty}{\ln\pars{x^{1/4}} \over 1 + x}\,{1 \over 4}\,x^{-3/4}\,\dd x
={1 \over 16}\int_{0}^{\infty}{x^{-3/4}\ln\pars{x} \over 1 + x}\,\dd x
\\[3mm]&={1 \over 16}\lim_{\mu\ \to\ -3/4}\,\,\,\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_{\infty}^{0}
{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}
\\[3mm]&\imp\quad
\color{#00f}{\int_{0}^{\infty}{x^{\mu} \over 1 + x}\,\dd x}
={2\pi\ic\expo{\pi\mu\ic} \over 1 - \expo{2\pi\mu\ic}}
=-\pi\,{2\ic \over \expo{\pi\mu\ic} - \expo{-\pi\mu\ic}}
=\color{#00f}{-\,{\pi \over \sin\pars{\pi\mu}}}
\end{align}

Replacing in $\pars{1}$:
\begin{align}&\color{#66f}{\large\int_{0}^{\infty}{\ln\pars{x} \over 1 + x^{4}}
\,\dd x}
={1 \over 16}\bracks{\pi^{2}\cot\pars{\pi\mu}\csc\pars{\pi\mu}}_{\mu\ =\ -3/4}
\,\,\,=\color{#66f}{\large -\,{\root{2} \over 16}\,\pi^{2}}
\approx -0.8724
\end{align}
A: From the link Calculating the integral $\int_0^{\infty}{\frac{\ln x}{1+x^n}}$ using complex analysis, we have for $n=4$
$$ \int_0^\infty\frac{\ln(x)}{1+x^4}\mathrm{d}x=-\frac{\pi^2\cos\frac{\pi}{4}}{4^2\sin^2\frac{\pi}{4}}=-\frac{\pi^2\sqrt2}{16}. $$
