Closed form of $\sum_{m=1}^{\infty} \frac{(-1)^mH_{\frac{3m}{4}}}{3m}$

Question

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.

Answer 1

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Let $\mathcal{I}$ denote the value of the following logarithmic integral:

$$\mathcal{I}:=\int_{0}^{1}\mathrm{d}x\,\frac{3x^{2}}{1+x^{3}}\ln{\left(1-x^{4}\right)}\approx-0.47159.$$

It will be shown that

$$\mathcal{I}=-\frac12\operatorname{Li}_{2}{\left(\frac34\right)}-\frac{5\pi^{2}}{144}+\frac34\ln^{2}{\left(2\right)}.$$

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Making use of the factorizations $\left(1+x^{3}\right)=\left(1+x\right)\left(1-x+x^{2}\right)$ and $\left(1-x^{4}\right)=\left(1-x\right)\left(1+x\right)\left(1+x^{2}\right)$, we can split up the integral into simpler components as

$$\begin{align} \mathcal{I} &=\int_{0}^{1}\mathrm{d}x\,\frac{3x^{2}}{1+x^{3}}\ln{\left(1-x^{4}\right)}\\ &=\int_{0}^{1}\mathrm{d}x\,\left[\frac{1}{1+x}+\frac{2x-1}{1-x+x^{2}}\right]\left[\ln{\left(1-x\right)}+\ln{\left(1+x\right)}+\ln{\left(1+x^{2}\right)}\right]\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{1}{1+x}\left[\ln{\left(1-x\right)}+\ln{\left(1+x\right)}+\ln{\left(1+x^{2}\right)}\right]\\ &~~~~~+\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\left[\ln{\left(1-x\right)}+\ln{\left(1+x\right)}+\ln{\left(1+x^{2}\right)}\right]\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1-x\right)}}{1+x}+\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1+x\right)}}{1+x}+\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1+x^{2}\right)}}{1+x}\\ &~~~~~+\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\ln{\left(1-x\right)}+\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\ln{\left(1+x\right)}\\ &~~~~~+\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\ln{\left(1+x^{2}\right)}\\ &=:\mathcal{I}_{1}+\mathcal{I}_{2}+\mathcal{I}_{3}+\mathcal{I}_{4}+\mathcal{I}_{5}+\mathcal{I}_{6}.\\ \end{align}$$

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Let's focus on the last integral first.

Using the derivatives

$$\frac{d}{dx}\ln{\left(1-x+x^{2}\right)}=\frac{2x-1}{1-x+x^{2}},$$

and

$$\frac{d}{dx}\frac{\ln^{2}{\left(1-x+x^{2}\right)}}{2}=\frac{2x-1}{1-x+x^{2}}\ln{\left(1-x+x^{2}\right)},$$

we show that the following integral vanishes identically:

$$\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\ln{\left(1-x+x^{2}\right)}=\frac{\ln^{2}{\left(1-x+x^{2}\right)}}{2}\bigg{|}_{0}^{1}=0.$$

Then, $\mathcal{I}_{6}$ can be rewritten in the following way:

$$\begin{align} \mathcal{I}_{6} &=\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\ln{\left(1+x^{2}\right)}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\ln{\left(1+x^{2}\right)}-\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\ln{\left(1-x+x^{2}\right)}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\left[\ln{\left(1+x^{2}\right)}-\ln{\left(1-x+x^{2}\right)}\right]\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\ln{\left(\frac{1+x^{2}}{1-x+x^{2}}\right)}\\ &=\int_{1}^{0}\mathrm{d}y\,\frac{\left(-2\right)}{\left(1+y\right)^{2}}\cdot\frac{\left(1+y\right)\left(1-3y\right)}{\left(1+3y^{2}\right)}\ln{\left(\frac{2+2y^{2}}{1+3y^{2}}\right)};~~~\small{\left[x=\frac{1-y}{1+y}\right]}\\ &=\int_{0}^{1}\mathrm{d}y\,\frac{2}{\left(1+y\right)}\cdot\frac{\left(1-3y\right)}{\left(1+3y^{2}\right)}\ln{\left(\frac{2+2y^{2}}{1+3y^{2}}\right)}\\ &=\int_{0}^{1}\mathrm{d}y\,\left[\frac{2}{\left(1+y\right)}-\frac{6y}{\left(1+3y^{2}\right)}\right]\ln{\left(\frac{2+2y^{2}}{1+3y^{2}}\right)}\\ &=\int_{0}^{1}\mathrm{d}y\,\frac{2}{\left(1+y\right)}\ln{\left(\frac{2+2y^{2}}{1+3y^{2}}\right)}-\int_{0}^{1}\mathrm{d}y\,\frac{6y}{\left(1+3y^{2}\right)}\ln{\left(\frac{2+2y^{2}}{1+3y^{2}}\right)}\\ &=\int_{0}^{1}\mathrm{d}y\,\frac{2}{\left(1+y\right)}\left[\ln{\left(2\right)}+\ln{\left(1+y^{2}\right)}-\ln{\left(1+3y^{2}\right)}\right]\\ &~~~~~-\int_{0}^{1}\mathrm{d}t\,\frac{3}{\left(1+3t\right)}\ln{\left(\frac{2+2t}{1+3t}\right)};~~~\small{\left[y^{2}=t\right]}\\ &=2\ln{\left(2\right)}\int_{0}^{1}\mathrm{d}y\,\frac{1}{\left(1+y\right)}+2\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y^{2}\right)}}{\left(1+y\right)}-2\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+3y^{2}\right)}}{\left(1+y\right)}\\ &~~~~~-\int_{0}^{1}\mathrm{d}u\,\frac{3}{\left(1+3u\right)}\ln{\left(1+u\right)};~~~\small{\left[t=\frac{1-u}{1+3u}\right]}\\ &=2\ln^{2}{\left(2\right)}+2\mathcal{I}_{3}-2\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+3y^{2}\right)}}{\left(1+y\right)}\\ &~~~~~+\operatorname{Li}_{2}{\left(\frac34\right)}-\operatorname{Li}_{2}{\left(\frac12\right)}+\operatorname{Li}_{2}{\left(-1\right)}\\ &=\operatorname{Li}_{2}{\left(\frac34\right)}-\operatorname{Li}_{2}{\left(\frac12\right)}+\operatorname{Li}_{2}{\left(-1\right)}+2\ln^{2}{\left(2\right)}+2\mathcal{I}_{3}\\ &~~~~~-2\int_{1}^{2}\mathrm{d}t\,\frac{\ln{\left(1+3(t-1)^{2}\right)}}{t};~~~\small{\left[1+y=t\right]}\\ &=\operatorname{Li}_{2}{\left(\frac34\right)}-\operatorname{Li}_{2}{\left(\frac12\right)}+\operatorname{Li}_{2}{\left(-1\right)}+2\ln^{2}{\left(2\right)}+2\mathcal{I}_{3}\\ &~~~~~-2\int_{1}^{2}\mathrm{d}t\,\frac{\ln{\left(3t^{2}-6t+4\right)}}{t}\\ &=\operatorname{Li}_{2}{\left(\frac34\right)}-\operatorname{Li}_{2}{\left(\frac12\right)}+\operatorname{Li}_{2}{\left(-1\right)}+2\ln^{2}{\left(2\right)}+2\mathcal{I}_{3}\\ &~~~~~-2\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}u\,\frac{\ln{\left(4u^{2}-4\sqrt{3}\,u+4\right)}}{u};~~~\small{\left[t=\frac{2u}{\sqrt{3}}\right]}\\ &=\operatorname{Li}_{2}{\left(\frac34\right)}-\operatorname{Li}_{2}{\left(\frac12\right)}+\operatorname{Li}_{2}{\left(-1\right)}+2\ln^{2}{\left(2\right)}+2\mathcal{I}_{3}\\ &~~~~~-2\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}u\,\frac{\ln{\left(4\right)}}{u}-2\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}u\,\frac{\ln{\left(u^{2}-\sqrt{3}\,u+1\right)}}{u}\\ &=2\mathcal{I}_{3}-\frac{\pi^{2}}{6}-\frac32\ln^{2}{\left(2\right)}+\operatorname{Li}_{2}{\left(\frac34\right)}-2\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{3}\,x+x^{2}\right)}}{x}.\\ \end{align}$$

The integral $\mathcal{I}_{3}$ can in turn be reduced to

$$\begin{align} \mathcal{I}_{3} &=\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1+x^{2}\right)}}{1+x}\\ &=\int_{1}^{2}\mathrm{d}y\,\frac{\ln{\left(1+(y-1)^{2}\right)}}{y};~~~\small{\left[1+x=y\right]}\\ &=\int_{1}^{2}\mathrm{d}y\,\frac{\ln{\left(2-2y+y^{2}\right)}}{y}\\ &=\int_{\frac{1}{\sqrt{2}}}^{\sqrt{2}}\mathrm{d}t\,\frac{\ln{\left(2-2t\sqrt{2}+2t^{2}\right)}}{t};~~~\small{\left[y=t\sqrt{2}\right]}\\ &=\int_{\frac{1}{\sqrt{2}}}^{\sqrt{2}}\mathrm{d}t\,\frac{\ln{\left(2\right)}}{t}+\int_{\frac{1}{\sqrt{2}}}^{\sqrt{2}}\mathrm{d}t\,\frac{\ln{\left(1-\sqrt{2}\,t+t^{2}\right)}}{t}\\ &=\ln^{2}{\left(2\right)}+\int_{\frac{1}{\sqrt{2}}}^{\sqrt{2}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{2}\,x+x^{2}\right)}}{x}.\\ \end{align}$$

Next up, $\mathcal{I}_{5}$ can be reduced to a similar expression involving the same undetermined integral in the expression for $\mathcal{I}_{6}$:

$$\begin{align} \mathcal{I}_{5} &=\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\ln{\left(1+x\right)}\\ &=-\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1-x+x^{2}\right)}}{1+x};~~~\small{I.B.P.s}\\ &=-\int_{1}^{2}\mathrm{d}y\,\frac{\ln{\left(3-3y+y^{2}\right)}}{y};~~~\small{\left[x=y-1\right]}\\ &=-\int_{\frac{1}{\sqrt{3}}}^{\frac{2}{\sqrt{3}}}\mathrm{d}t\,\frac{\ln{\left(3-3t\sqrt{3}+3t^{2}\right)}}{t};~~~\small{\left[y=t\sqrt{3}\right]}\\ &=-\int_{\frac{1}{\sqrt{3}}}^{\frac{2}{\sqrt{3}}}\mathrm{d}t\,\frac{\ln{\left(3\right)}}{t}-\int_{\frac{1}{\sqrt{3}}}^{\frac{2}{\sqrt{3}}}\mathrm{d}t\,\frac{\ln{\left(1-\sqrt{3}\,t+t^{2}\right)}}{t}\\ &=-\ln{\left(2\right)}\ln{\left(3\right)}-\int_{\frac{1}{\sqrt{3}}}^{\frac{2}{\sqrt{3}}}\mathrm{d}t\,\frac{\ln{\left(1-\sqrt{3}\,t+t^{2}\right)}}{t}\\ &=-\ln{\left(2\right)}\ln{\left(3\right)}-\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}u\,\frac{\ln{\left(\frac{1-\sqrt{3}\,u+u^{2}}{u^{2}}\right)}}{u};~~~\small{\left[t=u^{-1}\right]}\\ &=-\ln{\left(2\right)}\ln{\left(3\right)}+\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}u\,\frac{2\ln{\left(u\right)}-\ln{\left(1-\sqrt{3}\,u+u^{2}\right)}}{u}\\ &=-\ln{\left(2\right)}\ln{\left(3\right)}+\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}u\,\frac{2\ln{\left(u\right)}}{u}-\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}u\,\frac{\ln{\left(1-\sqrt{3}\,u+u^{2}\right)}}{u}\\ &=-\ln^{2}{\left(2\right)}-\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{3}\,x+x^{2}\right)}}{x}.\\ \end{align}$$

The remaining three integrals $\mathcal{I}_{1,2,4}$ each have nice exact values in terms of well-known constants: we have

$$\mathcal{I}_{1}=\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1-x\right)}}{1+x}=-\operatorname{Li}_{2}{\left(\frac12\right)}=\frac12\ln^{2}{\left(2\right)}-\frac{\pi^{2}}{12},$$

$$\mathcal{I}_{2}=\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1+x\right)}}{1+x}=\frac12\ln^{2}{\left(2\right)},$$

and

$$\begin{align} \mathcal{I}_{4} &=\int_{0}^{1}\mathrm{d}x\,\frac{2x-1}{1-x+x^{2}}\ln{\left(1-x\right)}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1-x+x^{2}\right)}}{1-x};~~~\small{I.B.P.s}\\ &=\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1-y+y^{2}\right)}}{y};~~~\small{\left[x=1-y\right]}\\ &=\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y^{3}\right)}}{y}-\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y\right)}}{y}\\ &=\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1+t\right)}}{3t};~~~\small{\left[y^{3}=t\right]}\\ &~~~~~-\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y\right)}}{y}\\ &=-\frac23\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1+t\right)}}{t}\\ &=\frac23\operatorname{Li}_{2}{\left(-1\right)}\\ &=-\frac{\pi^{2}}{18}.\\ \end{align}$$

Then,

$$\begin{align} \mathcal{I} &=\mathcal{I}_{1}+\mathcal{I}_{2}+\mathcal{I}_{3}+\mathcal{I}_{4}+\mathcal{I}_{5}+\mathcal{I}_{6}\\ &=\left[\frac12\ln^{2}{\left(2\right)}-\frac{\pi^{2}}{12}\right]+\frac12\ln^{2}{\left(2\right)}+\mathcal{I}_{3}-\frac{\pi^{2}}{18}\\ &~~~~~-\ln^{2}{\left(2\right)}-\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{3}\,x+x^{2}\right)}}{x}\\ &~~~~~+2\mathcal{I}_{3}-\frac{\pi^{2}}{6}-\frac32\ln^{2}{\left(2\right)}+\operatorname{Li}_{2}{\left(\frac34\right)}-2\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{3}\,x+x^{2}\right)}}{x}\\ &=\operatorname{Li}_{2}{\left(\frac34\right)}-\frac{11\pi^{2}}{36}-\frac32\ln^{2}{\left(2\right)}+3\mathcal{I}_{3}\\ &~~~~~-3\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{3}\,x+x^{2}\right)}}{x}\\ &=\operatorname{Li}_{2}{\left(\frac34\right)}-\frac{11\pi^{2}}{36}-\frac32\ln^{2}{\left(2\right)}\\ &~~~~~+3\left[\ln^{2}{\left(2\right)}+\int_{\frac{1}{\sqrt{2}}}^{\sqrt{2}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{2}\,x+x^{2}\right)}}{x}\right]\\ &~~~~~-3\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{3}\,x+x^{2}\right)}}{x}\\ &=\operatorname{Li}_{2}{\left(\frac34\right)}-\frac{11\pi^{2}}{36}+\frac32\ln^{2}{\left(2\right)}\\ &~~~~~+3\int_{\frac{1}{\sqrt{2}}}^{\sqrt{2}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{2}\,x+x^{2}\right)}}{x}-3\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{3}\,x+x^{2}\right)}}{x}.\\ \end{align}$$

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At this point, it will be helpful to introduce a two-variable variant of the dilogarithm function:

$$\operatorname{Li}_{2}{\left(r,\theta\right)}:=-\frac12\int_{0}^{r}\mathrm{d}x\,\frac{\ln{\left(1-2x\cos{\left(\theta\right)}+x^{2}\right)}}{x};~~~\small{\left(r,\theta\right)\in\mathbb{R}^{2}}.$$

This function can be shown to satisfy the following pair of formulas useful to the problem at hand:

$$\operatorname{Li}_{2}{\left(\cos{\left(\theta\right)},\theta\right)}=\frac12\left(\frac{\pi}{2}-\theta\right)^{2}+\frac14\operatorname{Li}_{2}{\left(\cos^{2}{\left(\theta\right)}\right)};~~~\small{\theta\in\left[0,\pi\right]},$$

$$\operatorname{Li}_{2}{\left(2\cos{\left(\theta\right)},\theta\right)}=\left(\frac{\pi}{2}-\theta\right)^{2};~~~\small{\theta\in\left[0,\pi\right]}.$$

Then, for $\theta\in\left(0,\frac{\pi}{2}\right)$, we have

$$\begin{align} \int_{\cos{\left(\theta\right)}}^{2\cos{\left(\theta\right)}}\mathrm{d}x\,\frac{\ln{\left(1-2x\cos{\left(\theta\right)}+x^{2}\right)}}{x} &=\int_{0}^{2\cos{\left(\theta\right)}}\mathrm{d}x\,\frac{\ln{\left(1-2x\cos{\left(\theta\right)}+x^{2}\right)}}{x}\\ &~~~~~-\int_{0}^{\cos{\left(\theta\right)}}\mathrm{d}x\,\frac{\ln{\left(1-2x\cos{\left(\theta\right)}+x^{2}\right)}}{x}\\ &=-2\operatorname{Li}_{2}{\left(2\cos{\left(\theta\right)},\theta\right)}+2\operatorname{Li}_{2}{\left(\cos{\left(\theta\right)},\theta\right)}\\ &=-2\left(\frac{\pi}{2}-\theta\right)^{2}+2\left[\frac12\left(\frac{\pi}{2}-\theta\right)^{2}+\frac14\operatorname{Li}_{2}{\left(\cos^{2}{\left(\theta\right)}\right)}\right]\\ &=\frac12\operatorname{Li}_{2}{\left(\cos^{2}{\left(\theta\right)}\right)}-\left(\frac{\pi}{2}-\theta\right)^{2}.\\ \end{align}$$

We can now complete our calculation for $\mathcal{I}$:

$$\begin{align} \mathcal{I} &=\operatorname{Li}_{2}{\left(\frac34\right)}-\frac{11\pi^{2}}{36}+\frac32\ln^{2}{\left(2\right)}\\ &~~~~~+3\int_{\frac{1}{\sqrt{2}}}^{\sqrt{2}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{2}\,x+x^{2}\right)}}{x}-3\int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}}\mathrm{d}x\,\frac{\ln{\left(1-\sqrt{3}\,x+x^{2}\right)}}{x}\\ &=\operatorname{Li}_{2}{\left(\frac34\right)}-\frac{11\pi^{2}}{36}+\frac32\ln^{2}{\left(2\right)}\\ &~~~~~+3\int_{\cos{\left(\frac{\pi}{4}\right)}}^{2\cos{\left(\frac{\pi}{4}\right)}}\mathrm{d}x\,\frac{\ln{\left(1-2x\cos{\left(\frac{\pi}{4}\right)}+x^{2}\right)}}{x}\\ &~~~~~-3\int_{\cos{\left(\frac{\pi}{6}\right)}}^{2\cos{\left(\frac{\pi}{6}\right)}}\mathrm{d}x\,\frac{\ln{\left(1-2x\cos{\left(\frac{\pi}{6}\right)}+x^{2}\right)}}{x}\\ &=\operatorname{Li}_{2}{\left(\frac34\right)}-\frac{11\pi^{2}}{36}+\frac32\ln^{2}{\left(2\right)}\\ &~~~~~+3\left[\frac12\operatorname{Li}_{2}{\left(\cos^{2}{\left(\frac{\pi}{4}\right)}\right)}-\left(\frac{\pi}{2}-\frac{\pi}{4}\right)^{2}\right]\\ &~~~~~-3\left[\frac12\operatorname{Li}_{2}{\left(\cos^{2}{\left(\frac{\pi}{6}\right)}\right)}-\left(\frac{\pi}{2}-\frac{\pi}{6}\right)^{2}\right]\\ &=\operatorname{Li}_{2}{\left(\frac34\right)}-\frac{11\pi^{2}}{36}+\frac32\ln^{2}{\left(2\right)}\\ &~~~~~+\frac32\operatorname{Li}_{2}{\left(\frac12\right)}-\frac{3\pi^{2}}{16}\\ &~~~~~-\frac32\operatorname{Li}_{2}{\left(\frac34\right)}+\frac{\pi^{2}}{3}\\ &=-\frac12\operatorname{Li}_{2}{\left(\frac34\right)}+\frac{\pi^{2}}{36}+\frac32\ln^{2}{\left(2\right)}\\ &~~~~~+\frac32\left[\frac{\pi^{2}}{12}-\frac12\ln^{2}{\left(2\right)}\right]-\frac{3\pi^{2}}{16}\\ &=-\frac12\operatorname{Li}_{2}{\left(\frac34\right)}-\frac{5\pi^{2}}{144}+\frac34\ln^{2}{\left(2\right)}.\blacksquare\\ \end{align}$$

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Answer 2

Maple gives a rather unpleasant expression involving complex dilogarithm functions:

$$-\frac{19 \pi^{2}}{144}+\frac{7 \ln \! \left(2\right)^{2}}{12}-\frac{\ln \! \left(2-\sqrt{3}\right)^{2}}{6}+\frac{\mathit{dilog}\! \left(\frac{1}{2}+\frac{i \sqrt{3}}{6}\right)}{3}-\frac{\mathit{dilog}\! \left(\frac{3}{2}-\frac{i}{2}+\left(-\frac{1}{2}+\frac{i}{2}\right) \sqrt{3}\right)}{3}-\frac{\mathit{dilog}\! \left(\frac{3}{2}-\frac{i}{2}+\left(\frac{1}{2}-\frac{i}{2}\right) \sqrt{3}\right)}{3}-\frac{\mathit{dilog}\! \left(\frac{3}{2}+\frac{i}{2}+\left(\frac{1}{2}+\frac{i}{2}\right) \sqrt{3}\right)}{3}-\frac{\mathit{dilog}\! \left(1-\frac{i \sqrt{3}}{3}\right)}{3}+\frac{\mathit{dilog}\! \left(\frac{1}{2}-\frac{i \sqrt{3}}{6}\right)}{3}-\frac{\mathit{dilog}\! \left(1+\frac{i \sqrt{3}}{3}\right)}{3}-\frac{\mathit{dilog}\! \left(\frac{3}{2}+\frac{i}{2}-\left(\frac{1}{2}+\frac{i}{2}\right) \sqrt{3}\right)}{3}+\frac{\mathit{dilog}\! \left(\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)}{3}+\frac{\mathit{dilog}\! \left(\frac{1}{2}-\frac{i \sqrt{3}}{2}\right)}{3} $$

I suspect this comes from writing the rational function $x^2/(1+x^3)$ in partial fractions and also writing $\ln(1-x^4) = \ln(1-x) + \ln(1+x) + \ln(1-ix) + \ln(1+ix)$.

Answer 3

Too long for comments

I do not know how you arrived at $$S_p=\sum _{m=1}^p \frac{(-1)^m }{3 m}\psi \left(\frac{3 m}{4}\right)$$ with $p\to \infty$ because this does not converge; for example $$S_{10000}\sim 0.341858 \qquad \text{and} \qquad S_{10001}\sim 0.341561$$ and I do not see how this relates to the value of the integral which is $-0.157197$.

Assuming that you wrote $$\frac{x^2}{1+x^3}=\sum _{n=0}^\infty (-1)^n x^{3 n+2}$$ you should have for the integral $${\displaystyle \int_0^1 \frac{x^2}{1 + x^3}\log(1 - x^4)\,dx}=\sum _{n=0}^\infty \frac{(-1)^{n+1}}{3(n+1) } H_{\frac{3 (n+1)}{4}}$$ which converges to the correct value.

Consider numerically $$T_k=\sum _{n=0}^{10^k} \frac{(-1)^{n+1}}{3(n+1) } H_{\frac{3 (n+1)}{4}}\qquad \text{and} \qquad U_k=\sum _{n=0}^{10^k+1} \frac{(-1)^{n+1}}{3(n+1) } H_{\frac{3 (n+1)}{4}}$$ $$\left( \begin{array}{cccc} k & T_k & U_k & T_k - U_k\\ 1 & -0.197575 & -0.118992 & -0.0785825 \\ 2 & -0.165269 & -0.149187 & -0.0160818 \\ 3 & -0.158395 & -0.155999 & -0.0023952 \\ 4 & -0.157355 & -0.157038 & -0.0003166 \\ 5 & -0.157216 & -0.157177 & -0.0000393 \\ 6 & -0.157199 & -0.157194 & -0.0000047 \\ 7 & -0.157197 & -0.157196 & -0.0000005 \end{array} \right)$$

Answer 4

\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*}