How to evaluate $\int_{0}^{\infty}\frac{(x^2-1)\ln{x}}{1+x^4}dx$? How to evaluate the following integral
$$I=\int_{0}^{\infty}\dfrac{(x^2-1)\ln{x}}{1+x^4}dx=\dfrac{\pi^2}{4\sqrt{2}}$$
without using residue or complex analysis methods?
 A: Noting that
$$ \int_0^1x^n\ln x\,dx=-\frac{1}{(n+1)^2} $$
we have
\begin{eqnarray}
\int_0^\infty\frac{(x^2-1)\ln x}{1+x^4}dx&=&2\int_0^1\frac{(x^2-1)\ln x}{1+x^4}\,dx\\
&=&2\int_0^1\sum_{n=0}^\infty(-1)^n(x^2-1)x^{4n}\ln x\,dx\\
&=&2\sum_{n=0}^\infty\int_0^1(-1)^n(x^{4n+2}-x^{4n})\ln x\,dx\\
&=&2\sum_{n=0}^\infty(-1)^n\left(\frac1{(4n+1)^2}-\frac1{(4n+3)^2}\right)\\
&=&2\sum_{n=-\infty}^\infty(-1)^n\frac1{(4n+1)^2}\\
&=&\frac{1}{32}\left(\sum_{n=-\infty}^\infty\frac1{(n+\frac{1}{8})^2}-\sum_{n=-\infty}^\infty\frac1{(n+\frac{3}{8})^2}\right)\\
&=&\lim_{b\to0}\frac{1}{32}\left(\frac{\pi\sinh2\pi b}{b\left(\cosh2\pi b-\cos2\pi a\right)}\bigg|_{a=-\frac{1}{8}}-\frac{\pi\sinh2\pi b}{b\left(\cosh2\pi b-\cos2\pi a\right)}\bigg|_{a=-\frac{3}{8}}\right)\\
&=&\frac{\pi^2}{4\sqrt2}.
\end{eqnarray}
Here we use this.
A: Consider the contour integral
$$\oint_C dz \frac{(z^2-1) \log^2{z}}{1+z^4} $$
where $C$ is a keyhole contour about the positive real axis having an outer radius $R$ and an inner radius $\epsilon$.  As $R \to \infty$ and $\epsilon \to 0$, the integral may be shown to be equal to
$$-i 4 \pi \int_0^{\infty} dx \frac{(x^2-1) \log{x}}{1+x^4} + 4 \pi^2 \int_0^{\infty} dx \frac{x^2-1}{1+x^4} $$
The contour integral is equal to $i 2 \pi$ times the sum of the residues of the poles of the integrand, which are at $e^{i (2 k+1) \pi/4}$ for $k=0,1,2,3$, or
$$i \frac{\pi}{2} \left [\frac{(i-1) (-\pi^2/16)}{e^{i 3 \pi/4}} - \frac{(i+1) (-9\pi^2/16)}{e^{i  \pi/4}} + \frac{(i-1) (-25 \pi^2/16)}{e^{-i \pi/4}} - \frac{(i+1) (-49\pi^2/16)}{e^{-i 3 \pi/4}}  \right ]$$
which simplifies to $-i (\pi^3/32) 16 \sqrt{2} = -i \pi^3/\sqrt{2}$.  Equating real and imaginary parts, we find that
$$\int_0^{\infty} dx \frac{(x^2-1) \log{x}}{1+x^4} = \frac{\pi^2}{4 \sqrt{2}} $$
A: \begin{align}
&\int_{0}^{\infty}\frac{(x^2-1)\ln{x}}{1+x^4}dx 
=2\int_{0}^{1}\frac{(x^2-1)\ln{x}}{1+x^4}dx \\
=&\ \sqrt2 \int_{0}^{1}\ln x \ d\left( -\tanh^{-1}\frac{\sqrt2x}{1+x^2}\right)
\overset{ibp}={\sqrt2} \int_0^1 \frac1x \tanh^{-1}\frac{\sqrt2x}{1+x^2}dx\\
=& \ {\sqrt2} \int_0^1 \int_0^{\pi/4} 
\frac{2(x^2+1)\cos t}{(x^2+1)^2-4x^2\sin^2t}dt\ dx\\ 
= & \ {\sqrt2} \int_0^{\pi/4} \tan^{-1}\frac{2x\cos t}{1-x^2}\bigg|_{x=0}^{x=1}\ dt=  {\sqrt2} \int_0^{\pi/4} \frac\pi2 dt
=\frac{\pi^2}{4\sqrt{2}}
\end{align}
A: We have a well-known formula below
$$J(a,b)=\int_0^\infty\frac{x^{\large a-1}}{1+x^b}\mathrm dx=\frac{\pi}{b}\csc\left(\frac{a\pi}{b}\right)\tag{1}$$
Differentiating $(1)$ with respect to $a$ once, we have
$$J'(a,b)=\int_0^\infty\dfrac{x^{\large a-1}\ln x}{1+x^b}\mathrm dx=-\frac{\pi^2}{b^2}\csc\left(\frac{a\pi}{b}\right)\cot\left(\frac{a\pi}{b}\right)\tag{2}$$
then, by using $(2)$, we can obtain the result of our integral as follows
\begin{align}
I&=\int_{0}^{\infty}\frac{(x^2-1)\ln{x}}{1+x^4}\mathrm dx\\[10pt]
&=\int_{0}^{\infty}\frac{x^2\;\ln{x}}{1+x^4}\mathrm dx-\int_{0}^{\infty}\frac{\ln{x}}{1+x^4}\mathrm dx\\[10pt]
&=J'(3,4)-J'(1,4)\\[10pt]
&=\frac{\pi^2}{8\sqrt{2}}+\frac{\pi^2}{8\sqrt{2}}\\[10pt]
&=\bbox[3pt,border:3px #FF69B4 solid]{\color{red}{\large\frac{\pi^2}{4\sqrt{2}}}}
\end{align}
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{{\displaystyle #1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,}
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\begin{align}
{\tt I} & \equiv \color{#44f}{\int_{0}^{\infty}{\pars{x^{2} - 1}\ln\pars{x} \over 1 + x^{4}}\,\dd x}
\\[5mm] \sr{x^{4}\ \mapsto\ x}{=} &
{1 \over 16}\int_{0}^{\infty}{\pars{x^{-1/4} - x^{-3/4}}\ln\pars{x} \over 1 + x}\,\dd x
\\[5mm] & =
\left.{1 \over 16}\partiald{}{\nu}\int_{0}^{\infty}{x^{\pars{\color{red}{\nu\ +\ 3/4}}\ -\ 1}\,\,\, -\,\,\, x^{\pars{\color{red}{\nu\ +\ 1/4}}\ -\ 1}\,\,\, \over 1 + x}\,\dd x\right\vert_{\,\nu\ =\ 0}
\tag{1}\label{1}
\end{align}
Note that $\ds{{1 \over 1 + x} = \sum_{n = 0}^{\infty}
\color{red}{\Gamma\pars{1 + n}}{\pars{-x}^{n} \over n!}}$ such that - see (\ref{1}) -
\begin{align}
& \int_{0}^{\infty}{x^{\pars{\color{red}{\nu\ +\ 3/4}}\ -\ 1}\,\,\, -\,\,\, x^{\pars{\color{red}{\nu\ +\ 1/4}}\ -\ 1}\,\,\, \over 1 + x}\,\dd x
\\[5mm] = & \
\overbrace{\Gamma\pars{\nu + {3 \over 4}}
\Gamma\pars{1 - \bracks{\nu + {3 \over 4}}} -
\Gamma\pars{\nu + {1 \over 4}}
\Gamma\pars{1 - \bracks{\nu + {1 \over 4}}}}
^{\ds{Ramanujan's\ Master\ Theorem}}
\\[5mm] = &
\pi\csc\pars{\pi\bracks{\nu + {3 \over 4}}} -
\pi\csc\pars{\pi\bracks{\nu + {1 \over 4}}}
\tag{2}\label{2}
\end{align}
(\ref{1}) and (\ref{2}) $\ds{\implies}$
\begin{align}
{\tt I} & \equiv \color{#44f}{\int_{0}^{\infty}{\pars{x^{2} - 1}\ln\pars{x} \over 1 + x^{4}}\,\dd x}
\\[5mm] & =
{\pi \over 16}\partiald{}{\nu}\bracks{%
\csc\pars{\pi\bracks{\nu + {3 \over 4}}} -
\csc\pars{\pi\bracks{\nu + {1 \over 4}}}}_{\,\,\,\nu\ =\ 0}
\\[5mm] & = \bbx{\color{#44f}{{\root{2} \over 8}\,\pi^{2}}} \approx 1.7447 \\ &
\end{align}
A: \begin{align}J&=\int_0^\infty \frac{(x^2-1)\ln x}{1+x^4}dx\\
&=\underbrace{\int_0^\infty \frac{x^2\ln x}{1+x^4}dx}_{u=\frac{1}{x}}-\int_0^\infty \frac{\ln x}{1+x^4}dx\\
&=-2\int_0^\infty \frac{\ln x}{1+x^4}dx\\
K&=\int_0^\infty \frac{\ln x}{1+x^4}dx\\
R&=\int_0^\infty\int_0^\infty \frac{\ln(xy)}{(1+x^4)(1+y^4)}dxdy\\
&\overset{u(x)=xy}=\int_0^\infty\int_0^\infty\frac{y^3\ln u}{(u^4+y^4)(1+y^4)}dudy=\frac{1}{4}\int_0^\infty\frac{\ln u}{u^4-1}\left[\ln\left(\frac{1+y^4}{u^4+y^4}\right)\right]_{y=0}^{y=\infty}du\\
&=\int_0^\infty\frac{\ln^2 u}{u^4-1}du=\frac{1}{2}\underbrace{\int_0^\infty\frac{\ln^2 u}{u^2-1}du}_{=0}-\frac{1}{2}\underbrace{\int_0^\infty\frac{\ln^2 u}{u^2+1}du}_{=\frac{\pi^3}{8}}=\boxed{-\frac{\pi^3}{16}}
\end{align}
On the other hand,
\begin{align}R&=2K\int_0^\infty \frac{1}{1+x^4}dx\\
S&=\int_0^\infty \frac{1}{1+x^4}dx\overset{u=\frac{1}{x}}=\int_0^\infty \frac{u^2}{1+u^4}du\\
2S&=\int_0^\infty \frac{1+\frac{1}{x^2}}{\left(x-\frac{1}{x}\right)^2+2}dx\overset{z=x-\frac{1}{x}}=\int_{-\infty}^{+\infty}\frac{1}{z^2+2}dz=\frac{1}{\sqrt{2}}\left[\arctan\left(\frac{x}{\sqrt{2}}\right)\right]_{-\infty}^{+\infty}\\
S&=\boxed{\frac{\pi}{2\sqrt{2}}}
\end{align}
Therefore,
\begin{align}K&=\frac{R}{2S}=\frac{-\frac{\pi^3}{16}}{2\times\frac{\pi}{2\sqrt{2}}}=-\frac{\pi^2}{8\sqrt{2}}\\
J&=-2K=\boxed{\frac{\pi^2}{4\sqrt{2}}}
\end{align}
NB:
I assume that,
\begin{align}\int_0^\infty \frac{\ln^2 x}{1+x^2}dx=\frac{\pi^3}{8}\end{align}
