Evaluating $\int_0^1 \frac{\log x \log \left(1-x^4 \right)}{1+x^2}dx$ I am trying to prove that
\begin{equation}
\int_{0}^{1}\frac{\log\left(x\right)
\log\left(\,{1 - x^{4}}\,\right)}{1 + x^{2}}
\,\mathrm{d}x = \frac{\pi^{3}}{16} - 3\mathrm{G}\log\left(2\right)
\tag{1}
\end{equation}
where $\mathrm{G}$ is Catalan's Constant.
I was able to express it in terms of Euler Sums but it does not seem to be of any use.
\begin{align}
&\int_{0}^{1}\frac{\log\left(x\right)
\log\left(\,{1 - x^{4}}\,\right)}{1 + x^{2}}
\,\mathrm{d}x
\\[3mm] = &\
\frac{1}{16}\sum_{n = 1}^{\infty}
\frac{\psi_{1}\left(1/4 + n\right) -
\psi_{1}\left(3/4 + n\right)}{n} \tag{2}
\end{align}
Here $\psi_{n}\left(z\right)$ denotes the polygamma function.
Can you help me solve this problem $?$.
 A: This is a partial solution.
Let us put, for $0\leq t\leq 1$,
$$F(t) = \int_0^1 \frac{\log x \log(1-tx^4)}{1+x^2} dx$$
Then
$$F'(t) = -\int_0^1 \frac{x^4\log x}{(1+x^2)(1-tx^4)} dx =  -\int_0^1 \frac{x^4\log x}{1+x^2} \sum_{n=0}^\infty t^nx^{4n} dx$$
$$=-\sum_{n=0}^\infty t^{n} C_{4(n+1)}$$
where $$C_m = \int_0^1 \frac{x^{m}\log x}{1+x^2} dx.$$
One has $C_0 = -G$. Multiplying both sides of the identity
$$x^m = \frac{x^m}{1+x^2} + \frac{x^{m+2}}{1+x^2}$$
by $\log x$ and integrating from $0$ to $1$, one finds the recurrence formula
$$C_m + C_{m+2} = \frac{-1}{(1+m)^2}$$
and therefore
$$C_{m+4} - C_m = \frac{-1}{(3+m)^2} + \frac{1}{(1+m)^2}.$$
Therefore, 
$$C_0 = -G$$
$$C_4 = -G +1 - \frac{1}{3^2}$$
$$C_8 = -G + 1 - \frac{1}{3^2} + \frac{1}{5^2} - \frac{1}{7^2}.$$
and so on. (Remark that $C_{4n} \to 0$ by definition of $G$.) Now, remark that $F(0) = 0$, so your integral is
$$F(1) = \int_0^1 F'(t) dt = -\sum_{n=0}^\infty \frac{C_{4(n+1)}}{n+1} = -\sum_{n=1}^\infty \frac{C_{4n}}{n}.$$
Now, it should be a matter of partial summation to transform the sum $-\sum_{n=1}^\infty \frac{C_{4n}}{n}$ into $\pi^3/16 -3G\log 2$ (in a manner similar to this), but I don't see it right away. I'll think about it a bit more later.
A: $$I=\int_0^1 \frac{\ln x \log \left(1-x^4 \right)}{1+x^2}dx$$
Let,
\begin{align*}
\displaystyle A&=\int_0^1 \dfrac{x\arctan x\ln x}{1+x^2}dx\\
\displaystyle B&=\int_0^1 \dfrac{\ln x \ln(1+x^2)}{1+x^2}dx\\
\displaystyle C&=\int_0^1 \dfrac{\arctan x\ln x}{1+x}dx\\
\end{align*}
From Evaluating $\int_0^{\pi/4} \ln(\tan x)\ln(\cos x-\sin x)dx=\frac{G\ln 2}{2}$ ,
it follows that,
$$\displaystyle \int_0^1\dfrac{\ln x\ln(1+x)}{1+x^2}dx=A-\dfrac{1}{2}B-C-2G\ln 2+\beta(3)$$
and,
$$(1)\boxed{A=\dfrac{1}{64}\pi^3-B-G\ln 2}$$
with,
$$\displaystyle \beta(3)=\sum_{n=1}^{\infty} \dfrac{(-1)^n}{(2n+1)^3}$$
Thus,
$$(2)\boxed{\displaystyle \int_0^1\dfrac{\ln x\ln(1+x^2)}{1+x^2}dx=\dfrac{\pi^3}{64}-\dfrac{3}{2}B-C-3G\ln 2+\beta(3)}$$
$G$, being the Catalan constant,
From Evaluating $\int_0^{\pi/4} \ln(\tan x)\ln(\cos x-\sin x)dx=\frac{G\ln 2}{2}$ ,
it follows that,
$$\displaystyle\int_0^{1}\dfrac{\ln x\ln(1-x)}{1+x^2}dx=A+\dfrac{1}{2}B+\beta(3)+\int_0^1\dfrac{\ln x\arctan x }{1-x}dx$$
and,
$$\displaystyle \int_0^1\dfrac{\ln x\arctan x}{1-x}dx=B-C+2G\ln 2-\dfrac{1}{16}\pi^3$$
Thus, using (1),
$$(3)\boxed{\displaystyle\int_0^{1}\dfrac{\ln x\ln(1-x)}{1+x^2}dx=\dfrac{1}{2}B-C+G\ln 2-\dfrac{3\pi^3}{64}+\beta(3)}$$
Therefore,
\begin{align}
\int_0^1 \frac{\ln x \log \left(1-x^4 \right)}{1+x^2}dx&=\int_0^1\dfrac{\ln x\ln(1+x^2)}{1+x^2}dx+\int_0^1\dfrac{\ln x\ln(1+x)}{1+x^2}dx+\int_0^1\dfrac{\ln x\ln(1-x)}{1+x^2}dx\\
&=B+\dfrac{\pi^3}{64}-\dfrac{3}{2}B-C-3G\ln 2+\beta(3)+\dfrac{1}{2}B-C+G\ln 2-\dfrac{3\pi^3}{64}+\beta(3)\\
&=2\beta(3)-\dfrac{\pi^3}{32}-2G\ln 2-2C
\end{align}
$C$ have been already evaluated (see Evaluating $\int_0^1 \frac{\arctan x \log x}{1+x}dx$ )
\begin{equation}
\boxed{\displaystyle C=\dfrac{G\ln 2}{2}-\dfrac{\pi^3}{64}}
\end{equation}
and, knowing that,
\begin{equation}
 \beta(3)=\dfrac{\pi^3}{32}
\end{equation}
it follows that,
$$\boxed{I=\dfrac{\pi^3}{16}-3G\ln 2}$$
A: We may use a simple strategy for a similar type of integral that appears in (Almost) Impossible Integrals, Sums, and Series, page $80$, integral $J_1$.
So, we have
$$\int_0^1 \frac{\log (x) \log(1-x^4)}{1+x^2} \textrm{d}x=\int_0^1 \frac{ (1-x^2)\log (x)\log(1-x^4)}{1-x^4} \textrm{d}x$$
$$=\frac{1}{16}\underbrace{\int_0^1 \frac{\log (x)\log (1-x) }{x^{3/4}(1-x) }\textrm{d}x}_{\displaystyle \text{Beta function}}-\frac{1}{16}\underbrace{\int_0^1 \frac{\log (x)\log (1-x) }{x^{1/4}(1-x)} \textrm{d}x}_{\displaystyle \text{Beta function}}=\frac{\pi^3}{16}-3\log(2)G.$$
A: Presented below is a self-contained evaluation. With
$\int_0^1 \frac{\ln t}{1+t^2}dt =-G$
\begin{align*}
I & = \int_0^1 \frac{\ln x \ln (1-x^4 )}{1+x^2}dx \\
& = \int_0^1 \ln (1-x^4 ) d\left(\int_1^x \frac{\ln t}{1+t^2}dt \right) \overset{IBP}=\int_0^1 \frac{ 4x^3}{1-x^4} \underset{t=xs }{\left(\int_0^x \frac{\ln t}{1+t^2}dt +G \right) } dx \\
& =4\int_0^1 \left( \int_0^1 \frac{x^4 \ln x+x^4\ln s }{(1-x^4 )(1+x^2s^2)}ds +\frac{Gx^3}{1-x^4} \right) dx\\
& =4\int_0^1 \int_0^1 \frac{x^4\ln x}{(1-x^4)(1+x^2s^2)}dsdx -4 \int_0^1\int_0^1 \frac{\ln s}{1+x^2s^2}dx ds \\
& \>\>\>\>\>+ 4 \int_0^1 \left(\int_0^1 \frac{\ln s }{(1-x^4 )(1+x^2s^2)}ds +\frac{Gx ^3}{1-x^4} \right) dx\\
\end{align*}
Integrate the 2nd integral
\begin{align*}
& \int_0^1\int_0^1 \frac{\ln s}{1+x^2s^2}dx ds =\int_0^1 \frac{\ln s\tan^{-1}s}sds
\overset{IBP}=-\frac12 \int_0^1 \frac{\ln^2s}{1+s^2}ds=-\frac{\pi^3}{32}
\end{align*}
and apply the decomposition below in the 3rd integral
$$\frac{1 }{(1-x^4 )(1+x^2s^2)}
= \frac{-s^4}{(1-s^4)(1+x^2s^2)} +\frac1{2(1-s^2)(1+x^2)}+ \frac1{2(1+s^2)(1-x^2)}
$$
Then, the 1st integral cancels and
\begin{align*}
I =& -4\left(-\frac{\pi^3}{32}\right)
+2\int_0^1 \int_0^1 \frac{\ln s }{(1-s^2 )(1+x^2)}dx ds \\
&\>\>\> + 2\int_0^1 \left( \int_0^1 \frac{\ln s }{(1+s^2 )(1-x^2)}ds  +\frac{2Gx ^3}{1-x^4}\right) dx\\
= & \frac{\pi^3}8+ 2\int_0^1 \frac{\ln s ds }{1-s^2}\int_0^1\frac{dx }{1+x^2}
 -2G \int_0^1 \left( \frac{1}{1-x^2} -\frac{2x^3}{1-x^4}\right) dx\\
 = & \frac{\pi^3}8+ 2\left(-\frac{\pi^2}{8}\right) \frac\pi4
 -2G \int_0^1 \left( \frac{x}{1+x^2} +\frac{1}{1+x}\right) dx\\
 = & \frac{\pi^3}{16}
 -3G\ln2\\
\end{align*}
A: I tried substitutions and the differentiation w.r.t a paramater trick like the other posters. Another partial result, or a trail of breadcrumbs to follow, is the following. We try a series expansion,
$$
\frac{\log\left(1-x^4\right)}{1+x^2} = \displaystyle \sum_{k=1}^{\infty} x^{4k}\left(x^{2} -1\right)H_k,
$$
where $H_k$ are the Harmonic numbers. Then 
\begin{align}
\int_0^1 \frac{\log x \log \left(1-x^4 \right)}{1+x^2}\ \mathrm{d}x &=\displaystyle \sum_{k=1}^{\infty}\, H_k\int_0^1 x^{4k}\left(x^{2} -1\right)\log x \ \mathrm{d}x  \\
&=\displaystyle \sum_{k=1}^{\infty} \, \frac{H_k}{(4k+1)^2}-\displaystyle \sum_{k=1}^{\infty} \, \frac{H_k}{(4k+3)^2}.
\end{align}
These sums look very similar to the ones evaluated in this post, in which they are transformed into alternating sums. Using the same techniques, or perhaps working back from the answers, we can hopefully show that
$$
\displaystyle \sum_{k=1}^{\infty} \, \frac{H_k}{(4k+1)^2} = -G\left(\frac{\pi}{4}+\frac{\log 8}{2} \right) +\frac{7}{4}\zeta(3) +\frac{\pi^3}{32} - \frac{\pi^2}{16}\log 8,
$$
$$
\displaystyle \sum_{k=1}^{\infty} \, \frac{H_k}{(4k+3)^2} = -G\left(\frac{\pi}{4}-\frac{\log 8}{2} \right) +\frac{7}{4}\zeta(3) -\frac{\pi^3}{32} - \frac{\pi^2}{16}\log 8,
$$
Subtracting the second from the first gives us 
$$
\frac{\pi^3}{16}-G\log 8.
$$
A: The following is a proof of the formula $$S= \sum_{k=1}^{\infty} \frac{H_{k}}{ (k+a)^{2}}= \left(\gamma + \psi(a) \right) \psi_{1}(a) - \frac{\psi_{2}(a)}{2} \, , \quad a >0.$$ 
This formula is mentioned in a comment under Bennett Gardiner's answer. 
(For $a=0$, the right side of the equation should be interpreted as a limit).
$$ \begin{align}  S &= \sum_{k=1}^{\infty} \frac{H_{k}}{(k+a)^{2}} \\ &= \sum_{k=1}^{\infty} \frac{1}{(k+a)^{2}} \sum_{n=1}^{k} \frac{1}{n}  \\& = \sum_{n=1}^{\infty} \frac{1}{n} \sum_{k=n}^{\infty}  \frac{1}{(k+a)^2} \\ &= \sum_{n=1}^{\infty} \frac{\psi_{1}(a+n)}{n}  \\ &= - \sum_{n=1}^{\infty} \frac{1}{n} \int_{0}^{1} \frac{x^{a+n-1} \ln x}{1-x} \, dx \tag{1} \\ &= - \int_{0}^{1}  \frac{x^{a-1} \ln x}{1-x} \sum_{n=1}^{\infty} \frac{x^{n}}{n} \, dx \\ &= \int_{0}^{1} \frac{x^{a-1} \ln x \ln(1-x)}{1-x} \, dx \\ &= \lim_{b \to 0^{+}} \frac{\partial }{\partial a \, \partial b} B(a,b) \\ &= \small  \lim_{b \to 0^{+}} \frac{\Gamma(a) \Gamma(b)}{\Gamma(a+b)} \left( \psi(a) \psi(b) - \psi(a)\psi(a+b) - \psi(b) \psi(a+b) + \psi^{2}(a+b) - \psi_{1}(a+b) \right) \tag{2} \\ &=  \lim_{b \to 0^{+}} \frac{\Gamma(a)}{\Gamma(a+b)} \left( \frac{1}{b} - \gamma + \mathcal{O}(b) \right)\left( \left( \gamma \psi_{1}(a) + \psi(a) \psi_{1} (a)  - \frac{\psi_{2}(a)}{2} \right)b + \mathcal{O}(b^{2}) \right) \\ &= \left(\gamma + \psi(a) \right) \psi_{1}(a) -  \frac{\psi_{2}(a)}{2} \end{align}$$

$(1)$ https://en.wikipedia.org/wiki/Trigamma_function#Calculation 
$(2)$ http://mathworld.wolfram.com/BetaFunction.html (26)
A: I have several pieces of this, but can't quite put them together.  Perhaps someone else can pick up from here.
$\int_0^1 \frac{logxlog(1-x^4)}{(1+x^2)}dx$ 
We are going to let $u = (1-x^4)$ giving du = $-4x^3dx$.
Rewriting the first integral we have 
$\int_0^1 \frac{-4x^3logxlog(1-x^4)}{-4x^3(1+x^2)}dx$ =
$\int_0^1 \frac{log(1-u)^{1/4}log(u)}{-4(1-u)^{3/4}(1+u^{1/2})}du$ = 
(-1/16)$\int_0^1 \frac{log(1-u)log(u)}{(1-u)^{3/4}(1+(1-u)^{1/2})}du$  
Let v = 1-u so dv = -du  which gets us to
(1/16)$\int_0^1 \frac{log(v)log2(v^{1/2}(v^{-1/2}-v^{1/2})/2}{(v)(v^{-1/4}+v^{1/4})}dv \hspace{50px}$   The fact that the 1/16 shows up is encouraging.
Now let w = log v so that v = $e^w$  and dw = (1/v)dv.  So now we have
(1/16)$\int_{-\infty}^0 \frac{w[ log 2 + w/2 + log(-sinh(w))}{2(cosh(v/2)}dw$
Having gotten this far the next step is the Catalan constant which can be defined as $\sum_{n = 0}^{\infty}\frac{(-1)^n}{(2n+1)^2} = 1/1^2 - 1/3^2 + 1/5^2 ... $ 
The log (-sinh(w)) can be expanded in a Taylor's series, with the idea of integrating term by term, and there is some reason to hope that it will produce something helpful. The source of this hope is
Evaluating $\int_0^{\large\frac{\pi}{4}} \log\left( \cos x\right) \, \mathrm{d}x $
What I can't see right now is what would happen with all the other stuff in the integral, but one could hope things might work out. 
