Euler's work VIII

Nice Euler sum involving the Apéry's constant $\zeta(3)$ and the Catalan's constant $\beta(2)$

Today we show the proof of this nice Euler sum proposed by @Ali39342137 $$\sum_{n=0}^{\infty}\frac{(-1)^{n-1}H_{2n}}{2n+1}\left(H_{\frac{n}{2}}-H_{\frac{n-1}{2}}\right) = \frac{\pi^2 \ln\left(2\right)}{8}-\frac{\pi}{2}\beta(2) + \frac{21}{32}\zeta(3)$$ The proof will rely on some properties of the digamma function and the use of Fourier series.

Proof:

Recall the following integral for the difference of digamma functions (proof in Appendix 1): $$\int_{0}^{1} \frac{x^{\mu-1}}{1+x} dx = \frac{1}{2}\left[\psi\left(\frac{\mu+1}{2}\right)-\psi\left(\frac{\mu}{2}\right)\right]$$ Let $n \in \mathbb{N}$. If we put $\mu -1 = n$ $$\int_{0}^{1} \frac{x^{n}}{1+x} dx = \frac{1}{2}\left[\psi\left(\frac{n}{2}+1\right)-\psi\left(\frac{n}{2}+\frac{1}{2}\right)\right]$$ Hence $$\int_{0}^{1} \frac{x^{n}}{1+x} dx = \frac{1}{2}\left[\psi\left(\frac{n}{2}+1\right)+\gamma-\left(\psi\left(\frac{n-1}{2}+1\right)+\gamma\right)\right] = \frac{1}{2}\left[H_{\frac{n}{2}}-H_{\frac{n-1}{2}}\right]$$ $$\Longrightarrow H_{\frac{n}{2}}-H_{\frac{n-1}{2}} = 2\int_{0}^{1} \frac{x^{n}}{1+x} dx$$ Therefore $$\Longrightarrow \frac{(-1)^{n-1}H_{2n}}{2n+1}\left(H_{\frac{n}{2}}-H_{\frac{n-1}{2}}\right) = 2\int_{0}^{1} \frac{1}{1+x} \frac{(-1)^{n-1}H_{2n}}{2n+1}x^n dx$$ Hence $$\begin{align*} S = \sum_{n=0}^{\infty}\frac{(-1)^{n-1}H_{2n}}{2n+1}\left(H_{\frac{n}{2}}-H_{\frac{n-1}{2}}\right) = &\sum_{n=0}^{\infty} 2\int_{0}^{1} \frac{1}{1+x} \frac{(-1)^{n-1}H_{2n}}{2n+1}x^n dx\\ = &2\int_{0}^{1} \frac{1}{1+x} \left(\sum_{n=0}^{\infty} \frac{(-1)^{n-1}H_{2n}}{2n+1}x^n\right) dx \quad \textrm{ (Fubini-Tonelli)} \\ = &4\int_{0}^{1} \frac{1}{1+w^2} \left(\sum_{n=0}^{\infty} \frac{(-1)^{n-1}H_{2n}}{2n+1}w^{2n+1}\right) dw \quad \left(w^2 \mapsto x\right) \end{align*}$$ We proved previously (proof in Appendix 2): $$\sum_{n=0}^{\infty} \frac{(-1)^{n-1}H_{2n}}{2n+1}w^{2n+1} = \frac{1}{2}\arctan(w)\ln(1+w^2)$$ Hence $$S = \sum_{n=0}^{\infty}\frac{(-1)^{n-1}H_{2n}}{2n+1}\left(H_{\frac{n}{2}}-H_{\frac{n-1}{2}}\right) = 2\int_{0}^{1} \frac{\arctan(w)\ln(1+w^2)}{1+w^2} dw$$ We just have to find the integral in the right hand side $$\begin{align*} S = 2\int_{0}^{1} \frac{\arctan(w)\ln(1+w^2)}{1+w^2} dw =& 2\int_{0}^{\frac{\pi}{4}} \theta \ln\left(1+\tan^2(\theta)\right)d\theta \quad \left(\theta = \arctan(w)\right) \\ =& 2\int_{0}^{\frac{\pi}{4}} \theta \ln\left(\frac{\cos^2(\theta)}{\cos^2(\theta)}+\frac{\sin^2(\theta)}{\cos^2(\theta)}\right)d\theta \\ =& 2\int_{0}^{\frac{\pi}{4}} \theta \ln\left(\frac{1}{\cos^2(\theta)}\right)d\theta \\ =& -4\int_{0}^{\frac{\pi}{4}} \theta \ln\left(\cos(\theta)\right)d\theta \\ =& -\int_{0}^{\frac{\pi}{2}} t \ln\left(\cos\left(\frac{t}{2}\right)\right)dt \quad \left( t \mapsto 2\theta\right) \\ =& -\int_{0}^{\frac{\pi}{2}} t\ln\left(\frac{2}{2}\cos\left(\frac{t}{2}\right)\right)dt \\ =& \ln(2)\int_{0}^{\frac{\pi}{2}} t dt - \int_{0}^{\frac{\pi}{2}} t \ln\left(2\cos\left(\frac{t}{2}\right)\right)dt \\ =& \frac{\pi^2 \ln\left(2\right)}{8} -\underbrace{\int_{0}^{\frac{\pi}{2}} t \ln\left(2\cos\left(\frac{t}{2}\right)\right)dt}_{I} \\ \end{align*}$$ We can expand the integrand in $I$ with the following Fourier series: $$\sum_{k=1}^{\infty} \frac{(-1)^{k-1}\cos(kx)}{k} = \ln\left(2\cos\left(\frac{x}{2}\right)\right) \quad -\pi\lt x\lt\pi$$ Therefore $$\begin{align*} I = \int_{0}^{\frac{\pi}{2}} t \ln\left(2\cos\left(\frac{t}{2}\right)\right)dt =& \int_{0}^{\frac{\pi}{2}} t \sum_{k=1}^{\infty} \frac{(-1)^{k-1}\cos(kt)}{k}dt\\ =& \sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k} \int_{0}^{\frac{\pi}{2}} t\cos(kt)dt \quad \textrm{ (Fubini-Tonelli)}\\ =& \sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k^3} \int_{0}^{\frac{k\pi}{2}} s\cos(s)ds \quad (s \mapsto kt) \\ \stackrel{IBP}{= }&\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k^3}\left[ s\sin(s)\Big|_{0}^{\frac{k\pi}{2}} - \int_{0}^{\frac{k\pi}{2}}\sin(s) ds\right] \\ =& \sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k^3}\left[ \left(\frac{k\pi}{2}\right)\sin\left(\frac{k\pi}{2}\right) + \cos\left(\frac{k\pi}{2}\right)-1\right] \\ =& \frac{\pi}{2}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}\sin\left(\frac{k\pi}{2}\right)}{k^2} + \sum_{k=1}^{\infty}\frac{(-1)^{k-1}\cos\left(\frac{k\pi}{2}\right)}{k^3} - \sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k^3} \\ \end{align*}$$ Note that $$\left(\sin\left(\frac{k\pi}{2}\right)\right)_{k} = \left(1,0,-1,0,1,0,-1,...\right)$$ $$\left(\cos\left(\frac{k\pi}{2}\right)\right)_{k} = \left(0,-1,0,1,0,-1,0,...\right)$$ Therefore $$\begin{align*} I =& \frac{\pi}{2}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}\sin\left(\frac{k\pi}{2}\right)}{k^2} + \sum_{k=1}^{\infty}\frac{(-1)^{k-1}\cos\left(\frac{k\pi}{2}\right)}{k^3} - \sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k^3}\\ =& \frac{\pi}{2}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{(2k-1)^2} + \sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{(2k)^3} - \sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k^3}\\ =& \frac{\pi}{2}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{(2k-1)^2} - \frac{7}{8}\sum_{k=1}^{\infty}\frac{(-1)^{k}}{k^3}\\ =& \frac{\pi}{2}\beta(2) - \frac{7}{8}\eta(3) \quad \textrm{ (eta and beta dirichlet functions)}\\ \end{align*}$$ Finally, using the identity: $$\eta(v) = (1-2^{1-v})\zeta(v)$$ $$I = \frac{\pi}{2}\beta(2) - \frac{21}{32}\zeta(3)$$ Therefore $$\boxed{S = \sum_{n=0}^{\infty}\frac{(-1)^{n-1}H_{2n}}{2n+1}\left(H_{\frac{n}{2}}-H_{\frac{n-1}{2}}\right) = \frac{\pi^2 \ln\left(2\right)}{8}-\frac{\pi}{2}\beta(2) + \frac{21}{32}\zeta(3)}$$

Appendix 1

We show the proof of this nice integral representation $$\int_{0}^{1} \frac{x^{\mu-1}}{1+x} dx = \frac{1}{2}\left[\psi\left(\frac{\mu+1}{2}\right)-\psi\left(\frac{\mu}{2}\right)\right]$$

Proof

$$\begin{align*} \int_{0}^{1} \frac{x^{\mu-1}}{1+x} dx =& \int_{0}^{1} \sum_{j=0}^{\infty} (-1)^jx^jx^{\mu-1} dx\\ =& \sum_{j=0}^{\infty} (-1)^j\int_{0}^{1} x^{j+\mu-1} dx\\ =& \sum_{j=0}^{\infty} \frac{(-1)^j}{j+\mu} \\ =& \sum_{j=0}^{\infty}\left[ \frac{1}{2j+\mu} - \frac{1}{2j+1+\mu}\right] \\ =& \frac{1}{2}\sum_{j=0}^{\infty}\left[ \frac{1}{j+\frac{\mu}{2}} - \frac{1}{j+\frac{1+\mu}{2}}\right] \\ =& \frac{1}{4}\sum_{j=0}^{\infty}\frac{1}{\left(j+\frac{\mu}{2}\right)\left(j+\frac{1+\mu}{2}\right)} \end{align*}$$ The difference between two digamma functions may be expressed as $$\psi(x)-\psi(y) = (x-y)\sum_{j=0}^{\infty} \frac{1}{(j+x)(j+y)}$$ If we put $\displaystyle x = \frac{\mu}{2}, y = \frac{\mu+1}{2}$ $$\int_{0}^{1} \frac{x^{\mu-1}}{1+x} dx = \frac{1}{4}\sum_{j=0}^{\infty}\frac{1}{\left(j+\frac{\mu}{2}\right)\left(j+\frac{1+\mu}{2}\right)} = \frac{1}{2}\left[\psi\left(\frac{\mu+1}{2}\right)-\psi\left(\frac{\mu}{2}\right)\right]$$

Appendix 2

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

Proof

One of the special cases of the Gauss hypergometric is the following: $${}_{2}F_{2} \left(a,\frac{1}{2}+a;\frac{3}{2};z^{2}\right)=\frac{1}{(2-4a)z}\left((1+z)^{1-2a}-(1-z)^{1-2a}\right)$$ If we put $\displaystyle a = \frac{t}{2} , z = ix$: $${}_{2}F_{2} \left(\frac{t}{2},\frac{1}{2}+\frac{t}{2};\frac{3}{2};-x\right)=\frac{1}{2(1-t)ix}\left((1+ix)^{1-t}-(1-ix)^{1-t}\right)$$ From the definition of $_{2}F_{1}$: $$\begin{align*} {}_{2}F_{1} \left(\frac{t}{2},\frac{t+1}{2};\frac{3}{2}, -x\right) =& \sum_{n=0}^{\infty} \frac{\left(\frac{t}{2}\right)_{n}\left(\frac{1+t}{2}\right)_{n}}{\left(\frac{3}{2}\right)_{n}} \frac{(-1)^nx^n}{n!} \\ =& \sum_{n=0}^{\infty} \frac{4^n\left(\frac{t}{2}\right)_{n}\left(\frac{1+t}{2}\right)_{n}}{4^n(1)_{n}\left(\frac{3}{2}\right)_{n}} (-1)^nx^n \\ =& \sum_{n=0}^{\infty} \frac{(t)_{2n}}{(2)_{2n}} (-1)^nx^n \end{align*}$$ Therefore $$\sum_{n=0}^{\infty} \frac{(t)_{2n}}{(2)_{2n}} (-1)^nx^n = \frac{1}{2(1-t)ix}\left((1+ix)^{1-t}-(1-ix)^{1-t}\right)$$ Differentiating with respect to $t$: $$\frac{d}{dt}\Big|_{t=1} \sum_{n=0}^{\infty} \frac{(t)_{2n}}{(2)_{2n}} (-1)^nx^n = \sum_{n=0}^{\infty} \frac{(1)_{2n}\left[\psi(2n+1)+\gamma\right]}{(2)_{2n}} (-1)^n= \sum_{n=0}^{\infty} \frac{(-1)^{n}H_{2n}}{2n+1}x^{2n}$$ Hence $$\begin{align*} \sum_{n=0}^{\infty} \frac{(-1)^{n}H_{2n}}{2n+1}x^{2n} =& \frac{d}{dt}\Big|_{t=1} \left[\frac{1}{2(1-t)ix}\left((1+ix)^{1-t}-(1-ix)^{1-t}\right)\right]\\ =& \lim_{t \to 1} \left[\underbrace{-\frac{i\left((1+ix)^{1-t}-(1-ix)^{1-t}\right)}{2(1-t)^2x}}_{=0} -\frac{i\left((1-ix)^{1-t}\ln(1-ix)-(1+ix)^{1-t}\ln(1+ix)\right)}{2(1-t)x}\right]\\ =& \lim_{t \to 1} -\frac{i\left((1-ix)^{1-t}\ln(1-ix)-(1+ix)^{1-t}\ln(1+ix)\right)}{2(1-t)x}\\ =& \frac{i\left(\ln^2(1+ix)-\ln^2(1-ix)\right)}{4x} \quad \textrm{ (L'Hôpital's rule) } \\ =& \frac{i\left(\ln(1+ix)-\ln(1-ix)\right)\left(\ln(1+ix)+\ln(1-ix)\right)}{4x}\\ =& \frac{i\left(\ln(1+ix)-\ln(1-ix)\right)\left(\ln(1+ix)+\ln(1-ix)\right)}{4x}\\ =& \frac{1}{2x}\left[-\frac{i}{2}\ln\left(\frac{1-ix}{1+ix}\right)\ln(1+x^2)\right]\\ =& \frac{1}{2x}\arctan(x)\ln(1+x^2) \end{align*}$$ Therefore we can conclude $$\sum_{n=0}^{\infty} \frac{(-1)^{n}H_{2n}}{2n+1}x^{2n+1} = \frac{1}{2}\arctan(x)\ln(1+x^2)$$