Prove $- \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \sum_{k=0}^{\infty} \frac{1}{(n+k)^2} = \frac{\zeta(3)}{4}-\frac{\pi^2}{4} \log (2)$

Question

How can I prove that $$- \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \sum_{k=0}^{\infty} \frac{1}{(n+k)^2} = \frac{\zeta(3)}{4}-\frac{\pi^2}{4} \log (2)$$

The problem comes from this MSE post, where the OP wanted to solve integrals of this form

$$f(x) = \int_{0}^{1} \frac{\ln(t)\ln(1 + t^x)}{t(1-t)}$$

Using the MZIntegrate Mathematica Package, the OP arrived at the result $f(1) = \frac{\zeta(3)}{4}-\frac{\pi^2}{4} \log (2)$. I wanted to mathematically arrive at this result, my attempts are shown below.

Substituting in the taylor series for $\ln(x+1)$, and pulling all constant factors out of the integral, we see that $$\int_{0}^{1} \frac{\ln(t)\ln(1 + t)}{t(1-t)} dt = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\int_{0}^{1} \frac{\ln(t)t^n}{t(1-t)} dt = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\int_{0}^{1} \frac{\ln(t)t^{n-1}}{(1-t)} dt$$

Substituting in the geometric series for $\frac{1}{(1-t)}$ $$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\int_{0}^{1} \frac{\ln(t)t^{n-1}}{(1-t)} dt = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\int_{0}^1 \ln(t) t^{n-1} \sum_{k=0}^{\infty}t^k dt$$

Rearranging the sums and integrals, we see that

$$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\int_{0}^1 \ln(t) t^{n-1} \sum_{k=0}^{\infty}t^k dt = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n} \sum_{k=0}^{\infty}\int_{0}^1 \ln(t) t^{n+k-1} dt$$

The inner integral evaluates to $-\frac{1}{(n+k)^2}$, which is a standard result.

$$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n} \sum_{k=0}^{\infty}\int_{0}^1 \ln(t) t^{n+k-1} dt = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n} \sum_{k=0}^{\infty}- \frac{1}{(n+k)^2}$$

Hence, $$\int_{0}^{1} \frac{\ln(t)\ln(1 + t)}{t(1-t)} = - \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n} \sum_{k=0}^{\infty} \frac{1}{(n+k)^2}$$ But how can I continue to show that this sum equals $\frac{\zeta(3)}{4}-\frac{\pi^2}{4} \log (2)$? I am unsure where the denominator of 4 or where $\zeta(3)$ comes from. The inner sum reminds me of the Hurwitz zeta function.

Thanks.

Answer 1

I am sure this integral is quite known and has appeared several times on this site. Regardless, here is an approach: $$\mathcal{J}=\int _0^1\frac{\log \left(x\right)\log \left(1+x\right)}{x\left(1-x\right)}\:\mathrm{d}x=\int _0^1\frac{\log \left(x\right)}{1-x}\int _0^1\frac{1}{1+xt}\:\mathrm{d}t\:\mathrm{d}x$$ $$=\int _0^1\int _0^1\frac{\log \left(x\right)}{\left(1-x\right)\left(1+xt\right)}\:\mathrm{d}x\:\mathrm{d}t$$ $$=\int _0^1\frac{t}{1+t}\int _0^1\frac{\log \left(x\right)}{1+tx}\:\mathrm{d}x\:\mathrm{d}t+\int _0^1\frac{1}{1+t}\int _0^1\frac{\log \left(x\right)}{1-x}\:\mathrm{d}x\:\mathrm{d}t$$ Note $\int _0^1\frac{\log \left(x\right)}{1+tx}\:\mathrm{d}x=\frac{\operatorname{Li}_2\left(-t\right)}{t}$ $$\mathcal{J}=\int _0^1\frac{\operatorname{Li}_2\left(-t\right)}{1+t}\:\mathrm{d}t-\zeta (2)\int _0^1\frac{1}{1+t}\:\mathrm{d}t$$ Integrate by parts to obtain $$\mathcal{J}=-\frac{1}{2}\log \left(2\right)\zeta (2)+\int _0^1\frac{\log ^2\left(1+t\right)}{t}\:\mathrm{d}t-\log \left(2\right)\zeta (2)$$ $$=\frac{1}{4}\zeta (3)-\frac{3}{2}\log \left(2\right)\zeta (2)$$ Where I used that $\int _0^1\frac{\log ^2\left(1+t\right)}{t}\:\mathrm{d}t=\int _0^1\frac{\log ^2\left(t\right)}{1-t^2}\:\mathrm{d}t-\frac{3}{4}\int _0^1\frac{\log ^2\left(t\right)}{1-t}\:\mathrm{d}t=\frac{1}{8}\int _0^1\frac{\log ^2\left(t\right)}{1-t}\:\mathrm{d}t=\frac{1}{4}\zeta (3)$