Prove $\sum_{k=1}^{\infty}\frac{(-1)^k}{(2k+1)^2}(2H_{2k}+H_k)=\frac{\pi^3}{32}-2G\ln2$

Question

How to prove

$$\sum_{k=1}^{\infty}\frac{(-1)^k}{(2k+1)^2}(2H_{2k}+H_k)\stackrel ?=\frac{\pi^3}{32}-2G\ln2,$$ where $G$ is the Catalan's constant.

Attempt

For the first sum, $$\sum_{k=1}^{\infty}\frac{(-1)^k}{(2k+1)^2}H_{2k}=\Re\left\{\sum_{k=1}^{\infty}\frac{i^k}{(k+1)^2}H_{k}\right\},$$ which can be evaluated by using the formula in this post: $$\sum_{n=1}^\infty \frac{H_n}{n^2}\, x^n=\zeta(3)+\frac{\ln(1-x)^2\ln(x)}{2}+\ln(1-x)\operatorname{Li}_2(1-x)+\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x),$$ but we cannot apply the similar approach to the second sum $$\sum_{k=1}^{\infty}\frac{(-1)^k}{(2k+1)^2}H_k.$$ Then, I tried to write the sum as $$\sum_{k=1}^{\infty}\frac{(-1)^k}{(2k+1)^2}\int_0^1\frac{2x^{2k}+x^k-3}{x-1}~\mathrm dx$$ and it become more complicated.

Edit:

Are we able to evaluate the sum directly (avoid calculating integrals and polylogs as much as possible)? The integral given by @Jack D'Aurizio is a bit complicated (see this post).

Answer 1

The series involving $H_k$ and $H_{2k}$ can be studied in a similar way: since $$ \frac{-\log(1-x)}{1-x} = \sum_{n\geq 1} H_n x^{n} $$ we have $ \frac{-\log(1+x^2)}{1+x^2} = \sum_{n\geq 1} H_n(-1)^n x^{2n} $ and $$ \sum_{k\geq 1}\frac{(-1)^k}{(2k+1)^2}H_k = \int_{0}^{1}\frac{\log(1+x^2)\log(x)}{1+x^2}\,dx$$ boils down to $$ \int_{0}^{\pi/4} -2\log(\cos\theta) \log(\tan\theta)\,d\theta $$ which is simple to tackle through well-known Fourier series. It equals

$$ -\frac{\pi^3}{64}-K\log(2)-\frac{\pi}{16}\log^2(2)+2\,\text{Im}\,\text{Li}_3\left(\frac{1+i}{2}\right)\approx -0.07355395672853217. $$

Answer 2

\begin{align} S&=2\sum_{n=0}^\infty\frac{(-1)^nH_{2n}}{(2n+1)^2}+\sum_{n=0}^\infty\frac{(-1)^nH_{n}}{(2n+1)^2}\\ &=2\sum_{n=0}^\infty\frac{(-1)^nH_{2n+1}}{(2n+1)^2}+\sum_{n=0}^\infty\frac{(-1)^nH_{n}}{(2n+1)^2}-2\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^3}\\ &=2\Im\sum_{n=1}^\infty\frac{(i)^nH_{n}}{n^2}+\sum_{n=1}^\infty\frac{(-1)^nH_{n}}{(2n+1)^2}-\frac{\pi^2}{16}\\ &=S_1+S_2-\frac{\pi^3}{16}\tag{1} \end{align} Using the generating function: $$\sum_{n=1}^\infty\frac{x^nH_n}{n^2}=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\frac12\ln x\ln^2(1-x)+\zeta(3)$$ then \begin{align} S_1&=2\Im\left(\operatorname{Li}_3(i)-\operatorname{Li}_3(1-i)+\ln(1-i)\operatorname{Li}_2(1-i)+\frac12\ln(i)\ln^2(1-i)\right)\\ &\boxed{S_1=-2\Im\operatorname{Li}_3(1-i)-G\ln2-\frac{\pi}{8}\ln^22} \end{align}

$$S_2=\sum_{n=1}^\infty\frac{(-1)^nH_n}{(2n+1)^2}=\int_0^1\frac{\ln(1+x^2)\ln x}{1+x^2}\ dx$$ The last integral is evaluated here

$$\boxed{\int_0^1\frac{\ln(1+x^2)\ln x}{1+x^2}\ dx=\frac3{32}\pi^3+\frac{\pi}8\ln^22-G\ln2+2\text{Im}\operatorname{Li_3}(1-i)=S_2}$$ Plugging $S_1$ and $S_2$ in $(1)$, we get $$\color{blue}{S=\frac{\pi^3}{32}-2G\ln2}$$

note that we used: $$\ln(i)=\frac{\pi}{2}i$$ $$\ln(1-i)=\frac12\ln2-\frac{\pi}{4}i$$ $$\operatorname{Li_2}(1-i)=\frac{\pi^2}{16}-\left(\frac{\pi}{4}\ln2+G\right)i$$ which give us: $$\ln(i)\ln^2(1-i)=\frac{\pi^2}{8}\ln2-\left(\frac{\pi^3}{32}-\frac{\pi}{8}\ln^22\right)i$$ $$\ln(1-i)\operatorname{Li_2}(1-i) =-\frac{\pi}{4}G-\frac{\pi^2}{32}\ln2-\left(\frac12\ln2G+\frac{\pi^3}{64}+\frac{\pi}{8}\ln^22\right)i$$

Answer 3

this summation also has some relationship with integral identity, which asked almost the same time, see this, rewrite as

$$\frac3{2} \int_{0}^{1} {\frac{\arctan x \ln(1+x^2)}{x} \mathrm{d}x} = \int_{0}^{1} {\frac{\arctan x \ln(1+x)}{x} \mathrm{d}x}$$

given some famous series

$$\arctan x = \sum_{n=0}^{\infty} {\frac{(-1)^n x^{2n+1}}{2n+1}}$$

$$\frac1{2} \arctan x \ln (1+x^2) = \sum_{n=0}^{\infty} {\frac{(-1)^{n+1}H_{2n}}{2n+1} x^{2n+1}}$$

on the left

$$\frac3{2} \int_{0}^{1} {\frac{\arctan x \ln(1+x^2)}{x} \mathrm{d}x} = 3\int_{0}^{1} {\sum_{n=0}^{\infty} {\frac{(-1)^{n+1}H_{2n}}{2n+1} x^{2n}} \>\mathrm{d}x} = 3\sum_{n=0}^{\infty} {\frac{(-1)^{n+1}H_{2n}}{(2n+1)^2}}$$

on the right

$$\begin{aligned} \int_{0}^{1} {\frac{\arctan x \ln(1+x)}{x} \mathrm{d}x} & = \int_{0}^{1} {\sum_{n=0}^{\infty} {\frac{(-1)^n}{2n+1}} x^{2n}\ln(1+x) \>\mathrm{d}x}\\ & = 2\ln2\sum_{n=0}^{\infty} {\frac{(-1)^n}{(2n+1)^2}} - \sum_{n=0}^{\infty} {\frac{(-1)^n}{(2n+1)^2} \int_{0}^{1} {\frac{1+x^{2n+1}}{1+x}}\mathrm{d}x}\\ & = 2G\ln2 - \sum_{n=0}^{\infty} {\frac{(-1)^n}{(2n+1)^2} \int_{0}^{1} {\sum_{k=0}^{2n} {(-x)^k}}\mathrm{d}x}\\ & = 2G\ln2 + \sum_{n=0}^{\infty} {\frac{(-1)^n}{(2n+1)^2} \sum_{k=1}^{2n+1} {\frac{(-1)^k}{k}}}\\ & = 2G\ln2 + \sum_{n=0}^{\infty} {\frac{(-1)^n}{(2n+1)^2}(H_{n}-H_{2n+1})} \end{aligned}$$

unscramble this identity

$$\sum_{n=0}^{\infty} {\frac{(-1)^n}{(2n+1)^2}(H_{n}+3H_{2n}-H_{2n+1})} = -2G\ln2$$

or

$$\begin{aligned} \sum_{n=0}^{\infty} {\frac{(-1)^n}{(2n+1)^2}(H_{n}+2H_{2n})} & = \sum_{n=0}^{\infty} {\frac{(-1)^n}{(2n+1)^2}(H_{2n+1}-H_{2n})} - 2G\ln2\\ & = \sum_{n=0}^{\infty} {\frac{(-1)^n}{(2n+1)^3}} - 2G\ln2 = \frac{\pi^3}{32} - 2G\ln2 \end{aligned}$$

where you may need some particular values of Dirichlet Beta Function.