Beautiful closed form of $\sum_{k=0}^{\infty}\frac{(-1)^{k+1}}{2(k+1)^2}\int_0^1\ln (k+1-k^x)+2\ln (1+k^x)dx$

Question

While evaluating the following the integral which I found here in brilliant

$$\int_0^1\ln(4-3^x)+\ln(1+3^x)dx$$

I conjectured the general variant of the aforementioned integral as

$$\begin{aligned}\sum_{k=0}^{\infty}\frac{(-1)^{k+1}}{2(k+1)^2}\int_0^1\ln (k+1-k^x)+2\ln (1+k^x)dx=\eta(2)\ln\left(\frac{4e^{\gamma}\pi}{A^{12}}\right)\end{aligned},$$

where $\eta(.)$ is Dirichlet eta funtion, $A$ is Glashier-kinkelin constant, $\gamma$ is Euler-Mascheroni constant and $e$ is Euler-number.

Is the closed form obtained correct?

Answer 1

Update: I have managed to explain the simplifications.

We claim $$ \int_0^1\log(k+1-k^x)+2\log(1+k^x)\,dx=2\log(k+1) $$Indeed, if this is the case, we are left with the sum $$ \sum_{k=0}^{\infty}\frac{(-1)^{k+1} \log(k+1)}{(k+1)^2}=\eta'(2)=\eta(2)\left(\gamma+\log\left(\frac{4\pi}{A^{12}}\right)\right); $$this can be shown, for instance, on this MSE post. Now we have to prove the claim. We have $$ \int _0^1 \log(k+1-k^x)\,dx =\log(k+1)+ \frac{\text{Li}_2\left(\frac{1}{k+1}\right)-\text{Li}_2\left(\frac{k}{k+1}\right) }{\log (k)}; $$ $$ 2\int _0^1 \log(1+k^x)\,dx =-\frac{2 \text{Li}_2(-k)+\frac{\pi ^2}{6}}{ \log (k)}; $$Combining, we have $$\int _0^1 \log(k+1-k^x)+2\log(1+x^k)\,dx $$ $$ =\log(k+1)+\underbrace{\frac{\text{Li}_2\left(\frac{1}{k+1}\right)-\text{Li}_2\left(\frac{k}{k+1}\right)-2 \text{Li}_2(-k)-\frac{\pi ^2}{6}}{\log (k)}}_{\color{red}{L(k)}} $$Let's show $L(k)=\log(k+1)$. We will use two functional identities which can be found here. First, we'll use Euler's reflection formula to eliminate $\text{Li}_2\left(\frac{k}{k+1}\right)$: $$ \text{Li}_2(z)=-\text{Li}_2(1-z)-\log (1-z) \log (z)+\frac{\pi ^2}{6}; $$ $$ L(k) = \frac{-2 \text{Li}_2(-k)+2 \text{Li}_2\left(\frac{1}{k+1}\right)+\log \left(\frac{1}{k+1}\right) \log \left(\frac{k}{k+1}\right)-\frac{\pi ^2}{3}}{\log(k)} $$ Next, use an inversion formula, valid for $\Re(z)\le 0$ with $z=-k$: $$\text{Li}_2(z)=\text{Li}_2\left(\frac{1}{1-z}\right)+\frac{1}{2} \log (1-z) \log \left(\frac{1-z}{z^2}\right)-\frac{\pi ^2}{6}; $$ $$ L(k) = \frac{\log \left(\frac{1}{k+1}\right) \log \left(\frac{k}{k+1}\right)-\log (k+1) \log \left(\frac{k+1}{k^2}\right)}{\log(k)} $$ $$ = \frac{ \log (k+1)\log \left(\frac{k+1}{k}\right)+\log (k+1) \log \left(\frac{k^2}{k+1}\right)}{\log(k)} $$ $$ = \frac{ \log (k+1)\log \left(\frac{k+1}{k}\cdot \frac{k^2}{k+1}\right)}{\log(k)} $$ $$ =\log(k+1), $$as was to be shown.