\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\newcommand{\undb}{\underbrace}
\newcommand{\pl}{\partial}
\parindent=0pt
\begin{document}

{\bf Question}

Verify the following integral.

$\ds\int_0^{\infty}\,dx \ds\frac{\log
x}{x^4+1}=-\ds\frac{\pi^2\sqrt{2}}{16}$
\medskip

{\bf Answer}

Consider $J=\ds\oint_C \ds\frac {dz \log z}{(z^4+1)}$ where $C$ is

PICTURE \vspace{2in}

simple poles at $z^4+1=0 \Rightarrow z=e^{\frac{i\pi}{4}},\
e^{\frac{3i\pi}{4}},\ e^{\frac{5i\pi}{4}},\ e^{\frac{7i\pi}{4}}$

Branch points at $z=0$.

Set cut along negative imaginary axis only $e^{\frac{i\pi}{4}}$
and $e^{\frac{3i\pi}{4}}$ count.

\begin{eqnarray*} J & = & 2\pi i \times
[\rm{res}(e^{\frac{i\pi}{4}})+\rm{res}(e^{\frac{3i\pi}{4}})]\\ & =
& 2\pi i \times \left\{ \lim_{z \to
e^{\frac{i\pi}{4}}}\left[\ds\frac{(z-e^{\frac{i\pi}{4}})}{(z^4+1)}
\log z\right]+\lim_{z \to
e^{\frac{3i\pi}{4}}}\left[\ds\frac{(z-e^{\frac{3i\pi}{4}}) \log
z}{(z^4+1)} \right] \right\}\\ & & \rm{use\ l'Hopital}\\ & = & 2
\pi i \left\{
\ds\frac{1}{4e^{\frac{3i\pi}{4}}}\log(e^{\frac{3i\pi}{4}})
+\ds\frac{1}{4e^{\frac{9i\pi}{4}}}\log(e^{\frac{3i\pi}{4}})\right\}\\
& = & \ds\frac{2\pi
i}{4}e^{\frac{-3i\pi}{4}}\left\{\ds\frac{i\pi}{4}+\ds\frac{3i\pi}{4}
e^{\frac{-6i\pi}{4}}\right\}\\ & & \rm{since}\
e^{\frac{-3i\pi}{2}}=+i\\ & = &
\ds\frac{-\pi^2}{8}\ds\frac{(-1-i)}{\sqrt{2}}(1+3i)\\ & = &
\ds\frac{-\pi^2}{8}\ds\frac{(2-4i)}{\sqrt{2}}\\ & = &
\ds\frac{-\pi^2\sqrt{2}}{8}+\ds\frac{\pi^2i}{2 \sqrt{2}}
\end{eqnarray*}

Now
$J=\ds\int_{-R}^{-\epsilon}+\ds\int_{\Gamma_1}+\ds\int_{\epsilon}^{R}
+\ds\int_{\Gamma_2}=\ds\frac{-\pi^2\sqrt{2}}{8}+\ds\frac{\pi^2
i}{2\sqrt{2}}$

Take limit at $R \to \infty, \epsilon \to 0\ \ \ds\int_{\Gamma_1}$
and $\ds\int_{\Gamma_2} \to 0$



Thus

$\ds\int_{\infty}^0 \undb{\ds\frac{d(xe^{i\pi})\
\log(xe^{i\pi})}{(xe^{i\pi})^4+1}}
+\ds\int_0^{\infty}\undb{\ds\frac{dx\ \log
x}{x^4+1}}=\ds\frac{-\pi^2\sqrt{2}}{8}+\ds\frac{\pi^2
i}{2\sqrt{2}}$

\ \ \ \ setting $z=xe^{i\pi}$\ \ \ \ \ \ \ \ \ \ \ $z=x$

(consistent with cut along $\arg(z)=-\frac{\pi}{2}$)

$-\ds\int_{\infty}^0 \ds\frac{dx\ \log x}{x^4+1}+i\pi
\ds\int_0^{\infty} \ds\frac{dx}{x^4+1}+\ds\int_0^{\infty}
\ds\frac{dx\ \log
x}{x^4+1}=\ds\frac{-\pi^2\sqrt{2}}{8}+\ds\frac{\pi^2
i}{2\sqrt{2}}$

$\Rightarrow 2\ds\int_0^{\infty}\ds\frac{dx\ \log x}{x^4+1}+i\pi
\ds\int_0^{\infty}\undb{\ds\frac{dx}{x^4+1}}=\ds\frac{-\pi^2\sqrt{2}}{8}+\ds\frac{\pi^2
i}{2\sqrt{2}}$

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
$=\ds\frac{\pi}{2\sqrt{2}}$

(poles, simple at $x=e^{i\frac{\pi}{4}},\ x=e^{i\frac{3\pi}{4}}$
and using a $D$-shaped contour)

PICTURE \vspace{1in}

$\Rightarrow \ds\int_0^{\infty} \ds\frac{dx\ \log
x}{x^4+1}=\ds\frac{-\pi^2\sqrt{2}}{16}+\ds\frac{\pi^2
i}{4\sqrt{2}}-\ds\frac{\pi^2
i}{4\sqrt{2}}=\ds\frac{-\pi^2\sqrt{2}}{16}$

\end{document}
