TITLE: Evaluate $\int_0^{\pi/4}{(4\cot x\ln\sec x-x)\ln^2\tan xdx}$
QUESTION [29 upvotes]: Question: How can we prove $$I=\int_0^{\pi/4}{(4\cot x\ln\sec x-x)\ln^2\tan xdx}=\frac5{2304}\pi^4?$$

I confirmed, numerically, that it holds for 1000 decimal places. This integral came up when I was calculating $\sum_{n=1}^\infty\frac{H_{4n}}{n^3}$.
Attempt
Note that $\tan x=i\frac{1-e^{2ix}}{1+e^{2ix}}$, substituing $t=e^{ix}$ gives
$$I=\int_1^{e^{i\pi/4}}\frac it\left(\frac{1+t^2}{1-t^2}\ln\frac{2t}{1+t^2}-\frac1i\ln t\right)\left(\frac{\pi i}2+\ln\frac{1-t^2}{1+t^2}\right)^2dt,$$
then I tried $t^2\mapsto t$,
$$I=\frac12\int_1^{i}\frac it\left(\frac{1+t}{1-t}(\ln2+(\ln t)/2-\ln(1+t))-\frac1{2i}\ln t\right)\left(\frac{\pi i}2+\ln\frac{1-t}{1+t}\right)^2dt,$$
For the next step I substituted $u=\frac{1-t}{1+t}$, but it makes the integral more complex. Moreover, the substitution seem to turn the integral into the original form. I got stuck here.

REPLY [13 votes]: $\DeclareMathOperator{\arctanh}{arctanh}$
We may write 
\begin{align*}
I =&\int_0^{\frac \pi 4} \Big(4\cot x\log (\sec x) -x \Big)\log^2 (\tan x) \ dx\\
 =&  4\underbrace{\int_0^{\frac \pi 4} \cot x\log (\sec x) \log^2 (\tan x) \ dx}_{\tan x \mapsto x} -\underbrace{\int_0^{\frac \pi 4} x \log^2 (\tan x) \ dx}_{=:I_2}\\
=& 2\underbrace{\int_0^1 \frac{\log^2 x\log(1+x^2)}{x(1+x^2)} dx}_{x^2 \to x} - I_2\\
=&\underbrace{\frac 1 4 \int_0^1 \frac{\log^2 x \log(1+x)}{x(1+x)}dx}_{=:I_1} - I_2\\
=&I_1 - I_2.
\end{align*} Using the Maclaurin series of $\frac{\log(1+x)}{1+x} = -\sum_{k=1}^\infty (-1)^{k}H_k x^k,$ we get
\begin{align*}I_1 =&- \frac 1 4 \sum_{k=1}^\infty (-1)^{k}H_k \int_0^1 x^{k-1} \log^2 x\ dx\\
=&\boxed{-\frac 1 2 \sum_{k=1}^\infty \frac{(-1)^{k}H_k}{k^3}}
\end{align*} as can be found in @omegadot's answer. (In fact, it holds that
$$
 \sum_{k=1}^\infty  \frac{(-1)^{k}H_k}{k^3}=2\text{Li}_4(1/2) + \frac {7\ln 2 \zeta(3)} 4  - \frac {\pi^2 \ln^2 2 }{12}  -\frac {11\pi^4}{360} + \frac {\ln^4 2}{12},
$$ though we don't really need this fact for our purpose now.)

Next, we derive that:
\begin{align*}
\boxed{I_2 = \frac 1 8 \int_0^1 \frac{\log x \log^2\left(\frac {1+x}{1-x}\right)}{x}dx +\frac{\pi^4}{256}.}
\end{align*} 
Then, using the Maclaurin series of $$\displaystyle\frac 1 4
\log^2\left(\frac {1+x}{1-x}\right)=\arctanh^2 x =\sum_{k= 1}^\infty \frac {H_{2k}-\frac 1 2 H_k}{k}x^{2k} $$ we get:
\begin{align*}
\frac 1 8 \int_0^1 \frac{\log x \log^2\left(\frac {1+x}{1-x}\right)}{x}dx =&\sum_{k=1}^\infty \frac{H_{2k}-\frac 1 2 H_k}{2k}\int_0^1 x^{2k-1} \log x\ dx\\
=- &\sum_{k=1}^\infty \frac{H_{2k}-\frac 12  H_k}{8k^3}  \\
=&-\frac 1 2\sum_{k=1}^\infty\frac{(1+(-1)^k)H_k}{k^3} +\frac 1 {16} \sum_{k=1}^\infty\frac{H_k}{k^3}\\
=&-\frac 1 2\sum_{k=1}^\infty\frac{(-1)^kH_k}{k^3} - \frac 7 {16} \sum_{k=1}^\infty\frac{H_k}{k^3}\\
=& I_1 - \frac 7 {16} \sum_{k=1}^\infty\frac{H_k}{k^3}.
\end{align*} 
This implies that
\begin{align*}
I_2 =& I_1 - \frac 7 {16} \sum_{k=1}^\infty\frac{H_k}{k^3} +\frac{\pi^4}{256}\\
\end{align*} hence
$$
\boxed{I = I_1 - I_2 = \frac 7{16}\sum_{k=1}^\infty\frac{H_k}{k^3}-\frac {\pi^4}{256} = \frac 7 {16}\frac {\pi^4} {72} - \frac {\pi^4}{256}= \frac{5\pi^4}{2304}.}
$$ Here, Euler's formula $$ 2\sum_{k=1}^\infty\frac{H_k}{k^n} = (n+2)\zeta(n+1) - \sum_{k=1}^{n-2}\zeta(k+1)\zeta(n-k),\qquad n\ge 2$$ can be used to evaluate $\sum_{k=1}^\infty\frac{H_k}{k^3} = \frac{\pi^4}{72}$. 

Evaluation of $I_2$: We note by the symmetry that
\begin{align*}
\int_0^{\frac \pi 4} \log^2(\tan(x)) dx= \frac 1 2\int_0^{\frac \pi 2} \log^2\left(\tan\left(\frac x 2\right)\right) dx= \frac 1 4 \int_{0}^{\pi} \log^2\left(\tan\left(\frac x 2\right)\right) dx = \frac{\pi^3}{16}.
\end{align*} This follows from $\displaystyle \log\left(\tan\left(\frac x 2\right)\right) = \sum_{k=1}^\infty \frac{(-1)^k -1}{k} \cos(kx)$ and Parseval's identity. Thus it follows
\begin{align*}
I_2 =& \underbrace{\int_0^{\frac\pi 4} \left(x-\frac \pi 4\right) \log^2(\tan(x)) dx }_{\frac \pi 4 -x\mapsto x}+ \frac{\pi^4}{64}\\
=& -\underbrace{\int_0^{\frac \pi 4} x \log^2\left(\frac {1+\tan x}{1-\tan x}\right) dx}_{\tan x = y} + \frac{\pi^4}{64} \\
=& -\underbrace{4\int_0^1 \frac{\arctan y}{1+y^2}\arctanh^2 y\ dy}_{=:J}+ \frac{\pi^4}{64}\\=&-J +\frac{\pi^4}{64}.
\end{align*} 

Now let us define 
$$
f(z) = \frac{\arctan z}{1+z^2}\left( \arctanh^2 z +\frac {\pi^2} {16}\right)
$$ so that $f$ is analytic on $|z|<1$ and continuous on $|z|\le 1$. We will use the symmetric nature of $f$ to evaluate the $J$.
We first notice that
$$
\int_{0}^1 f(z)dz =\frac 1 4 J +\frac{\pi^2}{16}\int_0^1\frac{\arctan z}{1+z^2}dz =\frac 1 4 J + \frac{\pi^4}{512} .
$$ 
On the other hand, we have that
\begin{align*}
\int_{0}^i f(z)dz=&i\int_0^1 f(ix)dx \\
=&i\int_0^1 \frac{\arctan(ix)}{1-x^2}\left(\arctanh^2(ix) + \frac{\pi^2}{16}\right) dx\\
=&  \int_0^1 \frac{\arctanh x}{1-x^2} \left(\arctan^2 x - \frac{\pi^2}{16}\right) dx\\
\underset{\text{IBP}}{=}&-\int_0^1 \arctanh^2 x\ \frac{\arctan x}{1+x^2} dx\\
=&-\frac 1 4 J
\end{align*} where we have used the relations $\arctan(ix) = i\arctanh x$ and $\arctanh(ix) = i \arctan x$.
Combining these, we know that
\begin{align*}
J =& \frac 12 J +\frac 1 2 J\\
=&2\left(\int_{0}^1f(z)dz -\frac{\pi^4}{512}\right) -2 \int_{0}^i f(z)dz\\=&-\underbrace{2 \int_1^i f(z)dz}_{=:K} -\frac{\pi^4}{256}\\
=&-K-\frac{\pi^4}{256}.
\end{align*}

For $K$, we parametrize $[1,i]$ by $z=e^{i\theta}, 0\le \theta \le \frac \pi 2$. Then exploiting the fact that  $$\arctan(e^{i\theta}) = -\frac i 2 \Big[\log\big(\cot\big(\scriptsize{\frac \theta 2+\frac\pi 4} \normalsize\big)\big)+ \frac {\pi i} 2\Big]$$
$$
\arctanh(e^{i\theta}) = \frac 1 2 \Big[\log\big(\cot\big(\scriptsize \frac \theta 2 \normalsize\big)\big) +\frac{\pi i}2\Big]
$$ we can see that (by Cauchy's integral theorem)
\begin{align*}
K=&2 \int_0^{\frac \pi 2} f(e^{i\theta}) ie^{i\theta} d\theta \\
=&2i\int_0^{\frac \pi 2} \frac{\arctan(e^{i\theta})}{1+e^{i2\theta}} \left(\arctanh^2(e^{i\theta}) + \frac{\pi^2}{16}\right)e^{i\theta}d\theta \\
=&\frac 1 {8} \int_0^{\frac\pi 2} \frac{\log\big(\cot\big(\frac \theta 2 +\frac \pi 4\big)\big)+\frac{\pi i}2}{\cos \theta}\Big[\log^2\big(\cot\big(\scriptsize \frac \theta 2 \normalsize\big)\big)+\pi i \log\big(\cot\big(\scriptsize \frac \theta 2 \normalsize\big)\big)\Big]d\theta.
\end{align*} Since $K$ is real, we can take the real part of the integral and perform the change of variable $\tan(\frac \theta 2) = y$ to get
\begin{align*}
K =&\frac 1 {8}\int_0^{\frac\pi 2} \frac{\log^2\big(\cot\big(\frac \theta 2\big)\big)\log\big(\cot\big(\frac \theta 2 +\frac \pi 4\big)\big) -\frac{\pi^2}2 \log\big(\cot\big(\frac\theta 2\big)\big)}{\cos \theta}d\theta\\
=&\frac 1 {4} \int_0^1 \frac{1}{1-y^2}\left(\log^2 y\log\left(\frac {1-y}{1+y}\right) +\frac{\pi^2}2 \log y\right)dy \\
=&\frac 1 {4} \underbrace{\int_0^1 \frac{\log^2 y\log\left(\frac {1-y}{1+y}\right)}{1-y^2}dy}_{\frac{1-y}{1+y} = x} +\frac{\pi^2}{8} \int_0^1 \frac{\log y}{1-y^2}dy\\
=&\frac 1 {8}\int_0^1 \frac{\log x\log^2\left(\frac {1+x}{1-x}\right)}{x}dx +\frac{\pi^2}{8} \sum_{n=0}^\infty \int_0^1 y^{2n}\log y\ dy \\
=&\frac 1 {8}\int_0^1 \frac{\log x\log^2\left(\frac {1+x}{1-x}\right)}{x}dx -\frac{\pi^2}{8} \underbrace{\sum_{n=0}^\infty\frac 1{(2n+1)^2}}_{=\frac 3 4 \zeta(2) = \frac{\pi^2}{8}} \\
=&\frac 1 {8}\int_0^1 \frac{\log x\log^2\left(\frac {1+x}{1-x}\right)}{x}dx -\frac{\pi^4}{64}.
\end{align*} 

This gives
\begin{align*}
J =& -K -\frac{\pi^4}{256}\\
=&-\frac 1 {8}\int_0^1 \frac{\log x\log^2\left(\frac {1+x}{1-x}\right)}{x}dx +\frac{3\pi^4}{256}
\end{align*} and
\begin{align*}
I_2 =& -J +\frac{\pi^4}{64}\\
=&\boxed{\frac 1 8 \int_0^1 \frac{\log x \log^2\left(\frac {1+x}{1-x}\right)}{x}dx +\frac{\pi^4}{256}}
\end{align*} as wanted.
Thus it holds
$$
I = \frac {5\pi^4}{2304}
$$ by the previous argument.