Proving $\int_0^{1/2}\frac{\text{Li}_2(-x)}{1-x}dx=-\text{Li}_3\left(-\frac12\right)-\frac{13}{24}\zeta(3)$

Question

By comparing some results, I found that

$$\int_0^{\frac12}\frac{\text{Li}_2(-x)}{1-x}dx=-\text{Li}_3\left(-\frac12\right)-\frac{13}{24}\zeta(3).\tag{1}$$

I tried to prove it starting with applying IBP:

$$\int_0^{\frac12}\frac{\text{Li}_2(-x)}{1-x}dx=\ln(2)\text{Li}_2\left(-\frac12\right)-\int_0^{\frac12}\frac{\ln(1-x)\ln(1+x)}{x}dx$$

then using the fact that $\ln(1-x)\ln(1+x)=\frac14\ln^2(1-x^2)-\frac14\ln^2\left(\frac{1-x}{1+x}\right)$:

$$\int_0^{\frac12}\frac{\ln(1-x)\ln(1+x)}{x}dx=\frac14\underbrace{\int_0^{\frac12}\frac{\ln^2(1-x^2)}{x}dx}_{1-x^2\to x}-\frac14\underbrace{\int_0^{\frac12}\frac{\ln^2\left(\frac{1-x}{1+x}\right)}{x}dx}_{(1-x)/(1+x)\to x}$$

$$=\frac18\int_{\frac34}^1\frac{\ln^2(x)}{1-x}dx-\frac14\int_{\frac13}^1\frac{\ln^2(x)}{1-x}dx-\frac14\int_{\frac13}^1\frac{\ln^2(x)}{1+x}dx.$$

Using: \begin{gather} \int\frac{\ln^2(x)}{1-x}dx=\sum_{n=1}^\infty\int x^{n-1}\ln^2(x)dx\\\ \overset{\text{IBP}}{=}\sum_{n=1}^\infty\left(\ln^2(x)\frac{x^n}{n}-2\ln(x)\frac{x^n}{n^2}+2\frac{x^n}{n^3}\right)\\ =-\ln^2(x)\ln(1-x)-2\ln(x)\operatorname{Li}_2(x)+2\operatorname{Li}_3(x), \end{gather}

and

\begin{gather} \int\frac{\ln^2(x)}{1+x}dx=\sum_{n=1}^\infty(-1)^{n-1}\int x^{n-1}\ln^2(x)dx\\\ \overset{\text{IBP}}{=}\sum_{n=1}^\infty (-1)^{n-1}\left(\ln^2(x)\frac{x^n}{n}-2\ln(x)\frac{x^n}{n^2}+2\frac{x^n}{n^3}\right)\\\ =\ln^2(x)\ln(1+x)+2\ln(x)\operatorname{Li}_2(-x)-2\operatorname{Li}_3(-x). \end{gather}

we have:

$$\int_{\frac34}^1\frac{\ln^2(x)}{1-x}dx=2\zeta(3)-2\ln(2)\ln^2(3/4)+2\ln(3/4)\text{Li}_2(3/4)-2\text{Li}_3(3/4),$$

$$\int_{\frac13}^1\frac{\ln^2(x)}{1-x}dx=2\zeta(3)+\ln^2(3)\ln(2/3)-2\ln(3)\text{Li}_2(1/3)-2\text{Li}_3(1/3),$$

$$\int_{\frac13}^1\frac{\ln^2(x)}{1+x}dx=\frac32\zeta(3)-\ln^2(3)\ln(4/3)+2\ln(3)\text{Li}_2(-1/3)+2\text{Li}_3(-1/3).$$

Combining the three integrals, we get

$$\int_0^{\frac12}\frac{\ln(1-x)\ln(1+x)}{x}dx=\frac12(\text{Li}_3(1/3)-\text{Li}_3(-1/3))-\frac13\text{Li}_3(3/4)$$ $$+\frac12\ln(3)(\text{Li}_2(1/3)-\text{Li}_2(-1/3))+\frac14\ln(3/4)\text{Li}_2(3/4)$$

$$-\frac14\ln(2)\ln^2(3/4)+\frac14\ln(2)\ln^2(3)-\frac58\zeta(3)$$

and finally

$$\int_0^{\frac12}\frac{\text{Li}_2(-x)}{1-x}dx=\ln(2)\text{Li}_2(-1/2)-\frac12(\text{Li}_3(1/3)-\text{Li}_3(-1/3))+\frac13\text{Li}_3(3/4)$$ $$-\frac12\ln(3)(\text{Li}_2(1/3)-\text{Li}_2(-1/3))-\frac14\ln(3/4)\text{Li}_2(3/4)$$

$$+\frac14\ln(2)\ln^2(3/4)-\frac14\ln(2)\ln^2(3)+\frac58\zeta(3).$$

and I think by using the polylogarithm identities, we can simplify this result into (1).

---

My question is how to prove (1) without going through all this mess if possible?

---

Edit: I also tried the Cauchy product $$\left(\sum_{n=1}^\infty a_n x^n\right)\left(\sum_{n=1}^\infty b_n x^n\right)=\sum_{n=1}^\infty x^{n+1}\left(\sum_{k=1}^n a_k b_{n-k+1}\right)$$ of the integrand:

$$\frac{\text{Li}_2(-x)}{1-x}=\left(\text{Li}_2(-x)\right)\left(\frac1{1-x}\right)=\left(\sum_{n=1}^\infty\frac{(-1)^n x^n}{n^2}\right)\left(\frac1x\sum_{n=1}^\infty x^{n}\right)$$

take $a_n=\frac{(-1)^n}{n^2}$ and $b_n=1=n^0$

$$=\frac1x\sum_{n=1}^\infty x^{n+1}\left(\sum_{k=1}^n\frac{(-1)^k(n-k+1)^0}{k^2}\right)=\sum_{n=1}^\infty x^{n}\left(\sum_{k=1}^n\frac{(-1)^k}{k^2}\right).$$

By using the definition of the $n$th generalized skew harmonic number of order $2$:

$$\overline{H}_n^{(2)}=\sum_{k=1}^n\frac{(-1)^{k-1}}{k^2}$$

we have

$$\frac{\text{Li}_2(-x)}{1-x}=-\sum_{n=1}^\infty x^n \overline{H}_n^{(2)}$$

or in general:

$$\frac{\text{Li}_a(-x)}{1-x}=-\sum_{n=1}^\infty x^n \overline{H}_n^{(a)}.$$

Employing this series expansion, we have

$$\int_0^{\frac12}\frac{\text{Li}_2(-x)}{1-x}dx=-\sum_{n=1}^\infty \overline{H}_n^{(2)}\int_0^{\frac12}x^n dx$$

$$=-\sum_{n=1}^\infty \frac{\overline{H}_n^{(2)}}{(n+1)2^{n+1}}$$

let the index start from zero since $\overline{H}_0^{(a)}=0$

$$=-\sum_{n=0}^\infty \frac{\overline{H}_n^{(2)}}{(n+1)2^{n+1}}$$

shift the index

$$=-\sum_{n=1}^\infty \frac{\overline{H}_{n-1}^{(2)}}{n2^{n}}.$$

Write $\overline{H}_{n-1}^{(2)}=\overline{H}_{n}^{(2)}+\frac{(-1)^n}{n^2}$, we get

$$\int_0^{\frac12}\frac{\text{Li}_2(-x)}{1-x}dx=-\sum_{n=1}^\infty \frac{\overline{H}_{n}^{(2)}}{n2^{n}}-\sum_{n=1}^\infty \frac{(-1)^n}{n^32^{n}}=-\sum_{n=1}^\infty \frac{\overline{H}_{n}^{(2)}}{n2^{n}}-\text{Li}_3\left(-\frac12\right).$$

Now we need to find this sum, which is, by comparing with (1), equal to $\frac{13}{24}\zeta(3).$

Answer 1

Using the fact that

$$\frac{1}{1-x}=\frac{1}{x(1-x)}-\frac1x$$ and writing

$$\frac{\text{Li}_2(-x)}{x}=\int_0^1\frac{\ln(y)}{1+xy}dy$$

we have

$$\int_0^{\frac12}\frac{\text{Li}_2(-x)}{1-x}dx=\int_0^{\frac12}\frac{\text{Li}_2(-x)}{x(1-x)}dx-\int_0^{\frac12}\frac{\text{Li}_2(-x)}{x}dx$$

$$=\int_0^1\ln(y)\left(\int_0^{\frac12}\frac{dx}{(1-x)(1+xy)}\right)dy-\text{Li}_3\left(-\frac12\right)$$

$$=\int_0^1\ln(y)\left(\frac{\ln(2+y)}{1+y}\right)dy-\text{Li}_3\left(-\frac12\right)$$

This integral is calculated here:

$$\int_0^1\frac{\ln(y)\ln(2+y)}{1+y}dy=-\frac{13}{24}\zeta(3)$$

and so

$$\int_0^{\frac12}\frac{\text{Li}_2(-x)}{1-x}dx=-\frac{13}{24}\zeta(3)-\text{Li}_3\left(-\frac12\right).$$

---

Bonus: Since we showed in the question body that

$$\int_0^{\frac12}\frac{\text{Li}_2(-x)}{1-x}dx=-\sum_{n=1}^\infty \frac{\overline{H}_{n}^{(2)}}{n2^{n}}-\text{Li}_3\left(-\frac12\right)$$

we also have $$\sum_{n=1}^\infty \frac{\overline{H}_{n}^{(2)}}{n2^{n}}=\frac{13}{24}\zeta(3).$$