TITLE: Closed form of $\sum_{m=1}^{\infty} \frac{(-1)^mH_{\frac{3m}{4}}}{3m}$
QUESTION [6 upvotes]: I've been working on an integral, namely: $${\displaystyle \int_0^1 \frac{x^2}{1 + x^3}\ln(1 - x^4)dx}$$
Which I managed to narrow down to the following expression: $$\sum_{m=1}^{\infty} \frac{(-1)^mH_{\frac{3m}{4}}}{3m}$$
Where $H_n$ is the n-th harmonic number. I managed to get here after converting the integrand into a double summation, recognizing the digamma function hidden inside, and summing back to a harmonic number, but couldn't get past that.
Is there another way to solve the integral, or to solve the summation? I've tried breaking the sum down to its parts, but the best I could do was to find a few terms and was left with
$$\sum_{m=1}^{\infty} \frac{(-1)^m\psi(\frac{3m}{4})}{3m}$$
Where $\psi(x)$ is the digamma function, and still couldn't solve that.

REPLY [5 votes]: \begin{align*}J&=\int_0^1 \frac{x^2\ln(1-x^4)}{1+x^3}dx\\
&\overset{y=\frac{1-x}{1+x}}=3\ln^2 2+3\underbrace{\int_0^1 \frac{\ln y}{1+y}dy}_{=-\dfrac{\pi^2}{12}}+\underbrace{\int_0^1 \frac{\ln(1+y^2)}{1+y}dy}_{=\dfrac{3}{4}\ln^2 2-\dfrac{1}{18}\pi^2}-2\ln^2 2-6\ln 2\underbrace{\int_0^1 \frac{y}{1+3y^2}dy}_{\dfrac{1}{3}\ln 2}-\\&2\underbrace{\int_0^1 \frac{y\ln y}{1+3y^2}dy}_{z=y^2}-2\underbrace{\int_0^1 \frac{y\ln(1+y^2)}{1+3y^2}dy}_{z=y^2}+8\int_0^1 \frac{y\ln(1+y)}{1+3y^2}dy\\
&=-\frac{1}{4}\ln^2 2-\frac{5}{48}\pi^2-\frac{1}{2}\underbrace{\int_0^1 \frac{\ln z}{1+3z}dz}_{=A}-\underbrace{\int_0^1 \frac{\ln(1+ z)}{1+3z}dz}_{=B}+8\underbrace{\int_0^1 \frac{y\ln(1+y)}{1+3y^2}dy}_{=C}
\end{align*}
\begin{align*}
A&\overset{\text{IBP}}=-\frac{1}{3}\int_0^1 \frac{\ln(1+3z)}{z}dz\\
&\overset{u=\frac{1}{1+3z}}=\frac{1}{3}\int_{\frac{1}{4}}^1 \frac{\ln u}{u(1-u)}du\\
&=-\frac{2}{3}\ln^2 2+\frac{1}{3}\int_{\frac{1}{4}}^1 \frac{\ln u}{1-u}du\\
&=-\frac{2}{3}\ln^2 2-\frac{\pi^2}{18}-\frac{1}{3}\int_0^{\frac{1}{4}} \frac{\ln u}{1-u}du\\
&\overset{x=4u}=-\frac{2}{3}\ln^2 2-\frac{\pi^2}{18}-\frac{1}{12}\int_0^1 \frac{\ln\left(\frac{1}{4}x\right)}{1-\frac{1}{4}x}dx\\
&=\frac{2}{3}\ln^2 2-\frac{\pi^2}{18}-\frac{2}{3}\ln 2\ln 3-\frac{1}{12}\int_0^1 \frac{\ln x}{1-\frac{1}{4}x}dx\\
&=\boxed{\frac{2}{3}\ln^2 2-\frac{\pi^2}{18}-\frac{2}{3}\ln 2\ln 3+\frac{1}{3}\text{Li}_2\left(\frac{1}{4}\right)}\\
B&\overset{x=\frac{1}{1+3z}}=\frac{1}{3}\int_{\frac{1}{4}}^1 \frac{\ln(1+2x)}{x}dx+\frac{2}{3}\ln^2 2-\frac{2}{3}\ln 2\ln 3\\
&\overset{\text{IBP}}=\frac{1}{3}\Big[\ln x\ln(1+2x)\Big]_{\frac{1}{4}}^1-\frac{2}{3}\int_{\frac{1}{4}}^1 \frac{\ln x}{1+2x}dx+\frac{2}{3}\ln^2 2-\frac{2}{3}\ln 2\ln 3\\
&=-\frac{2}{3}\int_{\frac{1}{4}}^1 \frac{\ln x}{1+2x}dx\\
&=-\frac{2}{3}\underbrace{\int_0^1 \frac{\ln x}{1+2x}dx}_{y=2x}+\frac{2}{3}\underbrace{\int_0^{\frac{1}{4}} \frac{\ln x}{1+2x}dx}_{y=4x}\\
&=\frac{1}{6}\int_0^1 \frac{\ln\left(\frac{y}{4}\right)}{1+\frac{y}{2}}dy-\frac{1}{3}\int_0^2 \frac{\ln\left(\frac{y}{2}\right)}{1+y}dy\\
&=\frac{1}{3}\text{Li}_2\left(-\frac{1}{2}\right)+\frac{2}{3}\ln^2 2-\frac{1}{3}\ln 2\ln 3 -\frac{1}{3}\int_0^2 \frac{\ln y}{1+y}dy\\
&=\frac{1}{3}\text{Li}_2\left(-\frac{1}{2}\right)+\frac{2}{3}\ln^2 2-\frac{1}{3}\ln 2\ln 3 -\frac{1}{3}\int_0^1 \frac{\ln y}{1+y}dy-\frac{1}{3}\underbrace{\int_1^2 \frac{\ln y}{1+y}dy}_{z=\frac{1}{y}}\\
&=\frac{1}{3}\text{Li}_2\left(-\frac{1}{2}\right)+\frac{2}{3}\ln^2 2-\frac{1}{3}\ln 2\ln 3 -\frac{1}{3}\int_0^1 \frac{\ln y}{1+y}dy+\frac{1}{3}\int_{\frac{1}{2}}^1 \frac{\ln z}{(1+z)z}dz\\
&=\frac{1}{3}\text{Li}_2\left(-\frac{1}{2}\right)+\frac{1}{2}\ln^2 2-\frac{1}{3}\ln 2\ln 3 -\frac{1}{3}\int_0^1 \frac{\ln y}{1+y}dy-\frac{1}{3}\int_{\frac{1}{2}}^1 \frac{\ln z}{1+z}dz\\
&=\frac{1}{3}\text{Li}_2\left(-\frac{1}{2}\right)+\frac{1}{2}\ln^2 2-\frac{1}{3}\ln 2\ln 3 -\frac{2}{3}\underbrace{\int_0^1 \frac{\ln y}{1+y}dy}_{=-\frac{1}{12}\pi^2}+\frac{1}{3}\underbrace{\int_0^{\frac{1}{2}} \frac{\ln z}{1+z}dz}_{x=2z}\\
&=\boxed{\frac{2}{3}\text{Li}_2\left(-\frac{1}{2}\right)+\frac{5}{6}\ln^2 2-\frac{2}{3}\ln 2\ln 3 +\frac{1}{18}\pi^2}\\
\end{align*}
\begin{align*}
C'&=\int_0^1 \frac{y\ln(1-y)}{1+3y^2}dy\\
C+C'&\overset{z=1-y^2}=\frac{1}{8}\int_0^1\frac{\ln z}{1-\frac{3}{4}z}dz\\
&=-\frac{1}{6}\text{Li}_2\left(\frac{3}{4}\right)\\
C^\prime-C&\overset{z=\frac{1-y}{1+y}}=\int_0^1 \frac{(1-z)\ln z}{2(1+z)(z^2-z+1)}dz\\
&\overset{\text{IBP}}=\left[\left(\frac{1}{3}\ln(1+z)-\frac{1}{6}\ln(1-z+z^2)\right)\ln z\right]_0^1-\frac{1}{3}\int_0^1 \frac{\ln(1+z)}{z}dz+\frac{1}{6}\int_0^1 \frac{\ln\left(\frac{1+z^3}{1+z}\right)}{z}dz\\
&=-\frac{1}{2}\int_0^1 \frac{\ln(1+z)}{z}dz+\frac{1}{6}\underbrace{\int_0^1 \frac{\ln\left(1+z^3\right)}{z}dz}_{u=z^3}\\
&=-\frac{4}{9}\underbrace{\int_0^1 \frac{\ln(1+z)}{z}dz}_{=\frac{1}{12}\pi^2}\\
&=\boxed{-\frac{1}{27}\pi^2}
\end{align*}
Therefore,
\begin{align*}
C&=\boxed{-\frac{1}{12}\text{Li}_2\left(\frac{3}{4}\right)+\frac{1}{54}\pi^2}\\
J&=\boxed{\ln 2\ln 3-\frac{17}{12}\ln^2 2-\frac{1}{6}\text{Li}_2\left(\frac{1}{4}\right)+\frac{7}{432}\pi^2-\frac{2}{3}\text{Li}_2\left(-\frac{1}{2}\right)-\frac{2}{3}\text{Li}_2\left(\frac{3}{4}\right)}
\end{align*}
Since,
\begin{align*}0\leq x<1,\text{Li}_2(x)+\text{Li}_2(1-x)&=\frac{\pi^2}{6}-\ln(1-x)\ln x\\
-1\leq x<1,\text{Li}_2(x)+\text{Li}_2(-x)&=\frac{1}{2}\text{Li}_2(x^2)
\end{align*}
Then,
\begin{align*}
\text{Li}_2\left(\frac{3}{4}\right)&=\frac{1}{6}\pi^2-4\ln^2 2+2\ln2 \ln 3-\text{Li}_2\left(\frac{1}{4}\right)\\
\text{Li}_2\left(-\frac{1}{2}\right)&=\frac{1}{2}\text{Li}_2\left(\frac{1}{4}\right)-\underbrace{\text{Li}_2\left(\frac{1}{2}\right)}_{=\frac{1}{12}\pi^2-\frac{1}{2}\ln^2 2}\\
\end{align*}
Thus,
$\boxed{\displaystyle J=\frac{11}{12}\ln^2 2-\frac{1}{3}\ln 2\ln 3-\frac{17}{432}\pi^2+\frac{1}{6}\text{Li}_2\left(\frac{1}{4}\right)}$
NB: I assume that,
\begin{align*}-1\leq a\leq 1,\text{Li}_2\left(a\right)&=-a\int_0^1 \frac{\ln t}{1-at}dt\\
\int_0^1 \frac{\ln x}{1+x}dx&=-\frac{1}{12}\pi^2\\
\int_0^1 \frac{\ln(1+x)}{x}dx&=\frac{1}{12}\pi^2\\
\end{align*}