Show that $\int_{0}^{\infty }\frac {\ln x}{x^4+1}\ dx =-\frac{\pi^2 \sqrt{2}}{16}$ I could prove it using the residues but I'm interested to have it in a different way (for example using Gamma/Beta or any other functions) to show that
$$
\int_{0}^{\infty}\frac{\ln\left(x\right)}{x^{4} + 1}\,{\rm d}x
=-\frac{\,\pi^{2}\,\sqrt{\,2\,}\,}{16}.
$$
Thanks in advance.
 A: Another approach, we can split the denominator part as follows
$$
\frac{1}{x^4+1}=\frac{1}{2i}\left(\frac{1}{x^2-i}-\frac{1}{x^2+i}\right).
$$
Consequently, the integral becomes
$$
\int_{0}^{\infty }\frac {\ln x}{x^4+1}\ dx =\frac{1}{2i}\int_{0}^{\infty }\left(\frac{\ln x}{x^2-i}-\frac{\ln x}{x^2+i}\right)\ dx.
$$
Using formula from here,
$$
\int_0^{\infty}\frac{\ln x}{x^2+a^2}\ dx=\frac {\pi \ln a}{2a},
$$
we obtain
$$
\begin{align}
\int_{0}^{\infty }\frac {\ln x}{x^4+1}\ dx&=\frac{1}{2i}\left(\frac {\pi \ln \sqrt{-i}}{2\sqrt{-i}}-\frac {\pi \ln \sqrt{i}}{2\sqrt{i}}\right)\\
&=\frac{\pi}{4i}\left(\frac {\ln i^{\frac{3}{2}}}{i^{\frac{3}{2}}}-\frac {\ln i^{\frac{1}{2}}}{i^{\frac{1}{2}}}\right).
\end{align}
$$
Taking $0\le\theta\le2\pi$, from Euler's formula we have
$$
e^\frac{i\pi}{2}=\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}=i.
$$
Thus
$$
\begin{align}
\int_{0}^{\infty }\frac {\ln x}{x^4+1}\ dx
&=\frac{\pi}{4i}\left(\frac {\ln e^\frac{3i\pi}{4}}{e^\frac{3i\pi}{4}}-\frac {\ln e^\frac{i\pi}{4}}{e^\frac{i\pi}{4}}\right)\\
&=\frac{\pi}{4i}\left(-\frac{i\pi}{2\sqrt{2}}\right)\\
&=\boxed{\color{blue}{-\Large\frac{\pi^2}{16}\sqrt{2}}}
\end{align}
$$
$$\\$$

$$\large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\int_{0}^{\infty}{\ln\pars{x} \over x^{4} + 1}\,\dd x
     =-\,{\pi^2 \root{2} \over 16}:\ {\large ?}}$.

\begin{align}
&\overbrace{\color{#c00000}{\int_{0}^{\infty}{\ln\pars{x} \over x^{4} + 1}\,\dd x}}
^{\ds{x^{4} \mapsto x}}
={1 \over 16}\int_{0}^{\infty}{x^{-3/4}\ln\pars{x} \over x + 1}\,\dd x
={1 \over 16}\lim_{\mu \to -3/4}\partiald{}{\mu}
\int_{0}^{\infty}{x^{\mu} \over x + 1}\,\dd x 
\\[3mm]&={1 \over 16}\lim_{\mu \to -3/4}\partiald{}{\mu}
\int_{0}^{\infty}x^{\mu}\int_{0}^{\infty}\expo{-\pars{x + 1}t}\,\dd t\,\dd x 
\\[3mm]&={1 \over 16}\lim_{\mu \to -3/4}\partiald{}{\mu}
\int_{0}^{\infty}\expo{-t}\int_{0}^{\infty}x^{\mu}\expo{-xt}\,\dd x\,\dd t
\\[3mm]&={1 \over 16}\lim_{\mu \to -3/4}\partiald{}{\mu}
\pars{\int_{0}^{\infty}t^{-\mu - 1}\expo{-t}\,\dd t}
\pars{\int_{0}^{\infty}x^{\mu}\expo{-t}\,\dd x}
\\[3mm]&={1 \over 16}\lim_{\mu \to -3/4}
\partiald{\bracks{\Gamma\pars{-\mu}\Gamma\pars{\mu + 1}}}{\mu} 
\end{align}
  where $\ds{\Gamma\pars{z}}$ is the Gamma Function ${\bf\mbox{6.1.1}}$.

Whith Euler Reflection Formula
${\bf\mbox{6.1.17}}$:
\begin{align}
&\color{#c00000}{\int_{0}^{\infty}{\ln\pars{x} \over x^{4} + 1}\,\dd x}
={1 \over 16}\,\left.
\partiald{\bracks{\pi\csc\pars{-\pi\mu}}}{\mu}\right\vert_{\mu\ =\ -3/4}
={1 \over 16}\,
\bracks{\pi^{2}\cot\pars{3\pi \over 4}\csc\pars{3\pi \over 4}}
\end{align}

$$\color{#00f}{\large%
\int_{0}^{\infty}{\ln\pars{x} \over x^{4} + 1}\,\dd x
=-\,{\root{2} \over 16}\,\pi^2}
$$

A: Substitute $t=1/(1+x^4)$ then we get
$$\int_{0}^{\infty }\frac {\ln x}{x^4+1}\ dx =\frac{1}{16}\int_0^1 \ln\left(\frac{1-t}{t}\right)(1-t)^{-3/4}t^{-1/4}dt.$$
And $\mathrm{B}(x,y)=\Gamma(x)\Gamma(y)/\Gamma(x+y)$, we get
$$\frac{\partial}{\partial x}\mathrm{B}(x,y)=\mathrm{B}(x,y)[\psi(x)-\psi(x+y)]$$
where $\psi$ is digamma function. And by Euler integral of the first kind we get
$$\frac{\partial}{\partial x}\mathrm{B}(x,y)=\int_0^1 \ln t\cdot t^{x-1}(1-t)^{y-1}dt.$$
So
$$
\begin{array}{lcl}
&&\frac{1}{16}\int_0^1 \ln\left(\frac{1-t}{t}\right)(1-t)^{-3/4}t^{-1/4}dt  \\ &=&\frac{1}{16}\int_0^1 \ln(1-t) (1-t)^{-3/4} t^{-1/4}dt-\frac{1}{16}\int_0^1 \ln(t)\cdot (1-t)^{-3/4} t^{-1/4}dt\\
&=& \frac{1}{16}\int_0^1 \ln (t)\cdot t^{-3/4} (1-t)^{-1/4}dt -\frac{1}{16}\int_0^1 \ln(t)\cdot (1-t)^{-3/4} t^{-1/4}dt\\
&=&\frac{1}{16}\mathrm{B}\left(\frac{1}{4},\frac{3}{4}\right)\left[\psi\left(\frac{1}{4}\right)-\psi(1)\right]-\frac{1}{16}\mathrm{B}\left(\frac{1}{4},\frac{3}{4}\right)\left[\psi\left(\frac{3}{4}\right)-\psi(1)\right] \\
&=&\frac{1}{16}\mathrm{B}\left(\frac{1}{4},\frac{3}{4}\right)\left[\psi\left(\frac{1}{4}\right)-\psi\left(\frac{3}{4}\right) \right]
\end{array}
$$
And $\mathrm{B}\left(\frac{1}{4},\frac{3}{4}\right)\left[\psi\left(\frac{1}{4}\right)-\psi\left(\frac{3}{4}\right) \right]=-\pi^2\sqrt{2}$. (It can easily be derived from reflection formula of gamma and digamma function.)
A: And yet another way for you to enjoy. Define
$$f(z):=\frac{\text{Log}\,z}{z^4+1}\;,\;\;C_R:=[-R,R]\cup\gamma_R:=\{z\in\Bbb C\;;\;z=Re^{it}\,,\,\,0<t<\pi\}\;,\;\;1<R\in\Bbb R$$
Now, the only poles within the region determined by $\,C_R\,$ are the simple (why? And note that $\,z=0\,$ is a pole but with residue equal to zero...) ones 
$$z_1=e^{\frac{\pi i}4}\;,\;\;z_2=e^{\frac{3\pi i}4}\implies$$
$$\begin{align*}\text{Res}_{z=z_1}(f)&=\lim_{z\to z_1}(z-z_1)f(z)\stackrel{\text{l'Hospital}}=\lim_{z\to z_1}\frac{\text{Log}\,z}{4z^3}&=\frac{\pi i}{16e^{\frac{3\pi i}4}}&=\frac{\pi }{16\sqrt2}\left(1-i\right)=\\
\text{Res}_{z=z_2}(f)&=\lim_{z\to z_2}(z-z_2)f(z)\stackrel{\text{l'Hospital}}=\lim_{z\to z_2}\frac{\text{Log}\,z}{4z^3}&=\frac{3\pi i}{16e^{\frac{\pi i}4}}&=\frac{3\pi }{16\sqrt2}\left(1+i\right)\end{align*}$$
So by Cauchy Theorem we get:
$$2\pi i\left(\frac{\pi}{16\sqrt2}\left(1-i\right)+\frac{3\pi}{16\sqrt2}\left(1+i\right)\right)=-\frac{\pi^2}{4\sqrt2}=\oint\limits_{C_R}f(z)\,dz=\int\limits_{-R}^R\frac{dx}{x^4+1}+\int\limits_{\gamma_R}f(z)dz$$
And since
$$\left|\;\int\limits_{\gamma_R}f(z)dz\;\right|\le\frac{\pi R}{R^4-1}\xrightarrow[R\to\infty]{}0$$
we get
$$2\int\limits_0^\infty\frac{dx}{x^4+1}\stackrel{\text{why?}}=\int\limits_{-\infty}^\infty\frac1{x^4+1}=\lim_{R\to\infty}\int\limits_{C_R}f(z)\,dz=-\frac{\pi^2}{4\sqrt2}$$
Note: To do the above we had to choose a branch cut for the complex logarithm function, yet we didn't choose the usual one (i.e., the non-positive reals) but rather the negative purely imaginary axis, so...why can we do that?, and what happened with zero, the great nemesis of the complex logarithm?
A: We can use the following way to solve. It is very simple. In fact
\begin{eqnarray}
I&=&\int_0^\infty \frac{\ln x}{1+x^4}dx=\int_0^1 \frac{\ln x}{1+x^4}dx+\int_1^\infty \frac{\ln x}{1+x^4}dx\\
&=&\int_0^1 \frac{\ln x}{1+x^4}dx-\int_0^1 \frac{x^2\ln x}{1+x^4}dx\tag{1}\\
&=&\int_0^1\sum_{n=0}^\infty(-1)^n(x^{4n}-x^{4n+2})\ln xdx\tag{2}\\
&=&\sum_{n=0}^\infty(-1)^{n+1}\left(\frac{1}{(4n+1)^2}-\frac{1}{(4n+3)^2}\right)\\
&=&\sum_{n=-\infty}^\infty(-1)^{n+1}\frac{1}{(4n+1)^2}\\
&=&-\frac{\sqrt{2}\pi}{16}\tag{3}.
\end{eqnarray}
Here, for (1), (2), and (3), we used the substitute $x\to\frac{1}{x}$, $\int_0^1x^n\ln xdx=-\frac1{(n+1)^2}$, and
$$ \sum_{n=-\infty}^\infty(-1)^{n}\frac{1}{(an+b)^2}=\frac{\pi^2a^2\cos\frac{b\pi}{a}}{\sin^2\frac{b\pi}{a}} $$
respectively.
A: One possible way is to introduce
$$ I(s)=\frac{1}{16}\int_0^{\infty}\frac{y^{s-\frac34}dy}{1+y}.\tag{1}$$
The integral you are looking for is obtained as $I'(0)$ after the change of variables $y=x^4$.
Let us make in (1) another change of variables: $\displaystyle t=\frac{y}{1+y}\Longleftrightarrow y=\frac{t}{1-t},dy=\frac{dt}{(1-t)^2}$. This gives
\begin{align}
I(s)&=\frac{1}{16}\int_0^1t\cdot\left(\frac{t}{1-t}\right)^{s-\frac74}\cdot \frac{dt}{(1-t)^2}=\\
&=\frac{1}{16}\int_0^1t^{s-\frac34}(1-t)^{-s-\frac{1}{4}}dt=\\&
=\frac{1}{16}B\left(s+\frac14,-s+\frac34\right)=\\&
=\frac{1}{16}\Gamma\left(s+\frac14\right)\Gamma\left(-s+\frac34\right)=\\
&=\frac{\pi}{16\sin\pi\left(s+\frac14\right)}.
\end{align}
Differentiating this with respect to $s$, we indeed get
$$I'(0)=-\frac{\pi^2\cos\frac{\pi}{4}}{16\sin^2\frac{\pi}{4}}=-\frac{\pi^2\sqrt{2}}{16}.$$
A: \begin{align}
\int_{0}^{\infty }\frac {\ln x}{x^4+1}\ dx
=&\int_0^1 \frac {(1-x^2)\ln x}{x^4+1}\ dx\\
\overset{ibp}=&-\frac1{2\sqrt2}\int_0^1\frac1x 
\ln\frac{x^2+\sqrt2 x+1}{x^2-\sqrt2 x+1}dx\\
=&-\frac1{2\sqrt2}\int_0^1\int_{-\pi/4}^{\pi/4} 
\frac{2\cos y}{x^2+2x\sin y+1}dy \ dx\\
 =&-\frac1{\sqrt2}\int_{-\pi/4}^{\pi/4} \left(\frac\pi4-\frac y2\right) dy=-\frac{\pi^2 }{8\sqrt2}
\end{align}
