Evaluate $\sum_{n=1}^{\infty} (-1)^{n+1} H_n \left( \frac{1}{n+1} - \frac{1}{n+3} + \frac{1}{n+5} - \ldots \right)$

Question

$$\sum_{n=1}^{\infty} (-1)^{n+1} H_n \left( \frac{1}{n+1} - \frac{1}{n+3} + \frac{1}{n+5} - \ldots \right) = \frac{\pi}{16} \cdot \log(2) + \frac{3}{16} \cdot \log(2) - \frac{\pi^2}{192}$$

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

We know that the generating function of harmonic number is given by $$\sum_{n=1}^{\infty} H_n x^n = -\frac{\log(1 - x)}{1 - x}$$ Now replacing $x$ by and by $-x$ by (1) we obtain

\begin{align*} &\sum_{n=1}^{\infty} (-1)^{n+1} H_{n} \int_{0}^{1} \frac{x^n}{1 + x^2} \,dx \\ &= \int_{0}^{1} \frac{\log(1 + x)}{(1 + x)(1 + x^2)} \,dx \\ &= \frac{1}{2} \int_{0}^{1} \frac{\log(1 + x)}{1 + x} \,dx - \frac{1}{2} \int_{0}^{1} \frac{x \log(1 + x)}{1 + x^2} \,dx + \frac{1}{2} \int_{0}^{1} \frac{\log(1 + x)}{1 + x^2} \,dx \end{align*}

Answer 1

It is not too difficult to show (or discover): \begin{align} J_{1} &= \int \frac{\ln(1+x)}{1+x} \, dx = \frac{1}{2} \, \ln^{2}(1+x) \\ J_{2} &= \int \frac{x \, \ln(1+x)}{1+x^2} \, dx \\ &= \frac{1}{2} \, \left( \text{Li}_{2}\left( \frac{(1-i)(1+x)}{2} \right) + \text{Li}_{2}\left( \frac{(1+i)(1+x)}{2} \right) + \ln(1+x) \, \ln(-i \, (1+x^2)) \right) \\ J_{3} &= \int \frac{\ln(1+x)}{1+x^2} \, dx \\ &= \frac{2}{i} \, \left( \text{Li}_{2}\left( \frac{(1+i)(1+x)}{2} \right) - \text{Li}_{2}\left( \frac{(1-i)(1+x)}{2} \right) + \ln(1+x) \, \ln\left(- \frac{1- i\, x}{1+i \, x}\right) \right). \end{align} Now, by using \begin{align} \text{Li}_{2}\left(\frac{1+i}{2}\right) &= - i G + \frac{\zeta(2)}{8} - \frac{(2 \, \ln 2 - \pi i)^2}{32} \\ \text{Li}_{2}\left(\frac{1-i}{2}\right) &= i G + \frac{\zeta(2)}{8} - \frac{(2 \, \ln 2 + \pi i)^2}{32} \\ \text{Li}_{2}\left(\frac{1+i}{2}\right) + \text{Li}_{2}\left(\frac{1+i}{2}\right) &= \frac{5 \, \zeta(2)}{8} - \frac{\ln^{2}(2)}{4} \\ \text{Li}_{2}\left(\frac{1+i}{2}\right) - \text{Li}_{2}\left(\frac{1+i}{2}\right) &= \frac{i}{4} \, (8 \, G - \pi \, \ln 2) \\ \text{Li}_{2}(1+i) &= i G + \frac{9 \, \zeta(2)}{8} + \frac{\pi i}{2} \, \left( \frac{\ln 2}{2} + \frac{\pi i}{4} \right) \\ \text{Li}_{2}(1-i) &= -i G + \frac{9 \, \zeta(2)}{8} - \frac{\pi i}{2} \, \left( \frac{\ln 2}{2} - \frac{\pi i}{4} \right) \\ \text{Li}_{2}(1+i) + \text{Li}_{2}(1-i) &= \frac{3 \, \zeta(2)}{4} \\ \text{Li}_{2}(1+i) - \text{Li}_{2}(1-i) &= \frac{i}{4} \, (4 \, G + \pi \, \ln 2) \end{align} then \begin{align} I_{1} &= \int_{0}^{1} \frac{\ln(1+x)}{1+x} \, dx = \frac{\ln^{2}(2)}{2} \\ I_{2} &= \int \frac{x \, \ln(1+x)}{1+x^2} \, dx = \frac{\zeta(2) + 2 \, \ln^{2}(2)}{16} \\ I_{3} &= \int_{0}^{1} \frac{\ln(1+x)}{1+x^2} \, dx = \frac{\pi \ln 2}{8}. \end{align}

All of this leads to: \begin{align} S &= \sum_{n=1}^{\infty} (-1)^{n-1} \, H_{n} \, \left( \sum_{k=0}^{\infty} \frac{(-1)^k}{2 k + n + 1} \right) \\ &= \frac{I_{1} - I_{2} + I_{3}}{2} \\ &= \frac{6 \, \ln^{2}(2) + 2 \, \pi \, \ln 2 - \zeta(2)}{32}. \end{align}

Answer 2

$$I= \int \frac{\log(1 + x)}{(1 + x)(1 + x^2)} \,dx $$ $$\frac{1}{(1 + x)(1 + x^2)}=\frac{1}{(x+1)(x+i)(x-i)}$$ Using partial fraction decomposition $$\frac{1}{(1 + x)(1 + x^2)}=\frac{1}{2 (x+1)}-\frac{1+i}{4(x-i)}-\frac{1-i}{4(x+i)}$$ Using one integration by parts $$J(a)=\int \frac{\log(1+x)}{x+a}\,dx=\text{Li}_2\left(-\frac{x+1}{a-1}\right)+\log (x+1) \log \left(\frac{a+x}{a-1}\right)$$

So the three integrals are $$\int \frac{\log (x+1)}{ x+1}\,dx=\frac{1}{2} \log ^2(x+1)$$ $$\int \frac{\log (x+1)}{ x-i}\,dx=\text{Li}_2\left(\left(\frac{1-i}{2}\right) (x+1)\right)+\log \left(\left(-\frac{1-i}{2}\right) (x-i)\right) \log (x+1)$$ $$\int \frac{\log (x+1)}{ x+i}\,dx=\text{Li}_2\left(\left(\frac{1+i}{2}\right) (x+1)\right)+\log \left(\left(-\frac{1+i}{2}\right) (x+i)\right) \log (x+1)$$

Use the bounds for the corresponding definite integrals, recombine and get the desired result.