Show that $\sum_{n=1}^{\infty} \frac{(-1)^n (\psi(n) - \psi(2n))}{n} = \frac{\pi^2}{16} + \left(\frac{\ln(2)}{2}\right)^2$

Question

Show That

$$\sum_{n=1}^{\infty} \frac{(-1)^n (\psi(n) - \psi(2n))}{n} =\bbox[15px, #B3E0FF, border: 5px groove #0066CC]{\frac{\pi^2}{16} + \left(\frac{\ln(2)}{2}\right)^2}$$

my work

$$\sum_{n=1}^{\infty} \frac{(-1)^n (\psi(n) - \psi(2n))}{n} =\sum_{n=1}^{\infty} \frac{(-1)^n \psi(n)}{n} - \sum_{n=1}^{\infty} \frac{(-1)^n \psi(2n)}{n}$$

I denote

$$\omega_1 =\sum_{n=1}^{\infty} \frac{(-1)^n \psi(n)}{n}$$

$$\omega_2 = \sum_{n=1}^{\infty} \frac{(-1)^n \psi(2n)}{n}$$

$$\omega_1= \sum_{n=1}^{\infty} \frac{(-1)^n \psi(n)}{n} = \sum_{n=1}^{\infty} (-1)^n \left(\frac{H_{n-1} - \gamma}{n}\right) = \sum_{n=1}^{\infty} (-1)^n \frac{H_{n-1}}{n} - \gamma \sum_{n=1}^{\infty} (-1)^n \frac{1}{n}$$

$$=\sum_{n=1}^{\infty} \frac{(-1)^n \left(H_{n} - \frac{1}{n}\right)}{n} + \gamma \log(2) = \sum_{n=1}^{\infty} (-1)^n H_{n} \int_{0}^{1} x^{n-1} \,dx + \gamma \log(2) - \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}$$

$$=-\int_{0}^{1} \frac{\log(1 + x)}{x(1 + x)} \,dx + \gamma \log(2) + \eta(2) = \gamma \log(2) + \frac{1}{2} \zeta(2) - \int_{0}^{1} \frac{\log(1 + x)}{x}dx + \int_{0}^{1} \frac{\log(1 + x)}{x + 1} dx$$

Answer 1

Note that $$ \begin{aligned} \int_{0}^{1} {\frac{x^{2n}}{1+x} \,\mathrm{d}x} & = \sum_{k=0}^{\infty} {(-1)^{k} \int_{0}^{1} {x^{2n+k} \,\mathrm{d}x}} = \sum_{k=0}^{\infty} {\frac{(-1)^{k}}{2n+k+1}} = \sum_{k=2n+1}^{\infty} {\frac{(-1)^{k+1}}{k}}\\ & = \sum_{k=1}^{\infty} {\frac{(-1)^{k+1}}{k}} - \sum_{k=1}^{2n} {\frac{(-1)^{k+1}}{k}} = \ln2 - H_{2n} + H_{n} \end{aligned} $$ so $$ \psi(n)-\psi(2n)=H_{n-1}-H_{2n-1}=H_n-H_{2n}-\frac1{2n}=\int_{0}^{1} {\frac{x^{2n}}{1+x} \,\mathrm{d}x} - \ln2 - \frac1{2n} $$ hence $$ \begin{aligned} &\, \sum_{n=1}^{\infty} \frac{(-1)^n}{n}(\psi(n)-\psi(2n)) \\ =&\, \int_{0}^{1}\left(\frac{1}{1+x}\sum_{n=1}^{\infty} {\frac{(-1)^nx^{2n}}{n}}\right) \mathrm{d}x-\ln2\sum_{n=1}^{\infty} {\frac{(-1)^n}{n}}-\frac1{2}\sum_{n=1}^{\infty} {\frac{(-1)^n}{n^2}}\\ =& -\int_{0}^{1} \frac{\ln(1+x^2)}{1+x}\,\mathrm{d}x + \ln^22 + \frac{\pi^2}{24} = \frac{\pi^2}{16} + \frac{\ln^22}{4} \end{aligned} $$ where $$ \int_{0}^{1} \frac{\ln(1+x^2)}{1+x}\,\mathrm{d}x = -\frac{\pi^2}{48} + \frac{3\ln^22}{4} $$ Supplement: to evaluate the last integral by elementary way, one is to denote $$ F(a)=\int_{0}^{1} \frac{\ln(1+a^2x^2)}{1+x}\,\mathrm{d}x $$ and $$ F'(a) = \int_{0}^{1} \frac{2ax^2}{(1+x)(1+a^2x^2)} \,\mathrm{d}x = -\frac{2\arctan a}{1+a^2} + \frac{2a\ln2}{1+a^2} + \frac{\ln(1+a^2)}{a(1+a^2)} $$ then antiderivative with $F(0)=0$ $$ \begin{aligned} F(a) & = \int_{0}^{a} \left(-\frac{2\arctan u}{1+u^2} + \frac{2u\ln2}{1+u^2} + \frac{\ln(1+u^2)}{u(1+u^2)}\right) \mathrm{d}u \\ & = -(\arctan a)^2 + \ln2 \ln(1+a^2) + \int_{0}^{a} \frac{\ln(1+u^2)}{u} \,\mathrm{d}u - \frac1{4}\ln^2(1+a^2) \end{aligned} $$ hence $$ \begin{aligned} \int_{0}^{1} \frac{\ln(1+x^2)}{1+x}\,\mathrm{d}x = F(1) & = -\frac{\pi^2}{16} + \frac{3\ln^22}{4} + \int_{0}^{1} \frac{\ln(1+u^2)}{u} \,\mathrm{d}u \\ & = -\frac{\pi^2}{16} + \frac{3\ln^22}{4} + \frac1{2} \int_{0}^{1} \frac{\ln(1+u)}{u} \,\mathrm{d}u \\ & = -\frac{\pi^2}{48} + \frac{3\ln^22}{4} \end{aligned} $$ Supplement 2: From this post, recalling that $$ \arctan x \ln (1+x^2) = 2\sum_{n=0}^{\infty} {\frac{(-1)^{n+1}H_{2n}}{2n+1} x^{2n+1}} $$ whose derivative gives you $$ \sum_{n=0}^{\infty} {(-1)^{n}H_{2n}x^{2n}} = -\frac{x\arctan x}{1+x^2} - \frac1{2}\frac{\ln(1+x^2)}{1+x^2} $$ which can help you finish $\omega_2$ from your approach. Yet lots of the terms will be cancelled by subtraction, there is no need to solve it term by term. Moreover, if we denote $$ H_{2n} = \sum_{k=0}^{n-1} \frac1{2n+1} + \frac{H_n}{2} $$ $$ \frac1{2}\sum_{n=0}^{\infty} {(-1)^{n}H_{n}x^{2n}} = -\frac1{2}\frac{\ln(1+x^2)}{1+x^2} $$ you can clearly find $$ \sum_{n=0}^{\infty} {(-1)^{n}\sum_{k=0}^{n-1} \frac{x^{2n}}{2n+1}} = -\frac{x\arctan x}{1+x^2} $$