The following problem is proposed by a friend:$$4\int_0^{\pi/4}\left(\int_x^{\pi/4} (x-y)\ln(\tan (x)) \ln\left(\tan \left(y+\frac{\pi }{4}\right)\right) \textrm{d}y\right)\textrm{d}x$$ $$-\sum _{n=1}^{\infty }(-1)^{n-1} \frac{H_n}{(2 n+1)^4}-\sum _{n=1}^{\infty }(-1)^{n-1} \frac{\overline{H_n}}{(2 n+1)^4}$$ $$=\frac{7}{4}\zeta(3)G+\frac{15}{16}\ln(2)\zeta(4)-\frac{1}{768}\ln(2)\psi^{(3)}\left(\frac{1}{4}\right),$$
where $\displaystyle H_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}$ is the $n$th harmonic number and $\displaystyle \overline{H}_n=1-\frac{1}{2}+\cdots+\frac{(-1)^{n-1}}{n}$ represents the $n$th skew-harmonic number.
The integral representations for the two sums are:
$$\sum _{n=1}^{\infty }(-1)^{n} \frac{H_n}{(2 n+1)^4}=\frac16\int_0^1\frac{\ln^3(x)\ln(1+x^2)}{1+x^2}dx;$$
$$\sum _{n=1}^{\infty }(-1)^{n} \frac{\overline{H_n}}{(2 n+1)^4}=-\frac16\int_0^1\frac{\ln^3(x)\ln(1-x^2)}{1+x^2}dx.$$
These two conversions follow from using $\frac{1}{(2n+1)^4}=-\frac16\int_0^1 x^{2n}\ln^3(x)dx$, $\sum_{n=1}^{\infty} H_n (-x^2)^n=-\frac{\ln(1+x^2)}{1+x^2}$, and $\sum_{n=1}^{\infty} \overline{H_n} (-x^2)^n=\frac{\ln(1-x^2)}{1+x^2}.$
The first sum was proposed by @Dr. Wolfgang Hintze and a closed form (involving hypergeometric function) was presented by @pisco. About the second sum, I have no idea how to approach it.
The interesting thing about this problem is that when we combine the double integral and the two sums, the closed form comes out nicely with no hypergeometric representation.
Any idea how to prove this equality? Thanks in advance.
