# A peculiar Euler sum

I would like a hand in the computation of the following Euler sum (Why isn't here a tag for Euler sums?)

$$S=\sum_{m,n\geq 0}\frac{(-1)^{m+n}}{(2m+1)(2n+1)^2(2m+2n+1)} \tag{1}$$ which arises from the computation of $$\int_{0}^{1}\frac{\arcsin\sqrt{x}\arcsin\sqrt{1-x}}{\sqrt{x(1-x)}}\,dx=\frac{\pi^3}{24}$$ via FL-expansions.
By symmetry it equals $$\frac{G^2}{2}+\frac{1}{2}\int_{0}^{1}\left(\sum_{n\geq 0}\frac{x^{2n}(-1)^n}{(2n+1)^2}\right)^2\,dx \tag{2}$$ with $$G$$ being Catalan's constant, and we may also recast $$S$$ as $$\iiint_{(0,1)^3}\frac{-\log y}{(1+x^2 z^2)(1+y^2 z^2)}\,dx\,dy\,dz. \tag{3}$$ An explicit form for $$S$$ is equivalent to an explicit form for $$T= \sum_{n\geq 0}\frac{(-1)^n}{2n+1}\left[\sum_{k>n}\frac{(-1)^k}{2k+1}\right]^2\tag{4}$$ but in any case we have to deal with alternating Euler sums with weight $$3$$, and the theory from Flajolet and Salvy might be helpful.

• I have to say that the computation of $S$ or $T$ is equivalent to the computation of $$\int_{0}^{\pi/2}\frac{\theta}{\sin\theta}\log(\sin\theta)\,d\theta$$ or $$\sum_{k\geq 0}\frac{4^k}{(2k+1)^3 \binom{2k}{k}}.$$ – Jack D'Aurizio Mar 3 '19 at 23:56
• I have managed to reduce it to $\int_{0}^{1}\frac{\log(x)\log(1+x^2)}{1+x^2}\,dx$, which is known, by applying a ton of transformations. It turns out that $S$ and $T$ are polynomials in $\pi, G, \log(2)$ and $\operatorname{Im}\operatorname{Li}_3\left(\tfrac{1+i}{2}\right)$, but I would be happy in seeing a cleaner derivation than mine. – Jack D'Aurizio Mar 4 '19 at 0:24
• Do you mind sharing your solution? – NoName Mar 8 '19 at 7:41
• I will share it soon, I am just a bit overworked at the moment – Jack D'Aurizio Mar 8 '19 at 8:40

Here it is my approach, with some definitions first: $$\mathscr{H}_n$$ is $$\sum_{k=0}^{n}\frac{1}{2k+1}$$; $$D^{1/2}$$ is the semi-derivative operator, i.e. a linear operator which acts on power series (centered at $$0$$ or $$1$$) by mapping $$x^{\alpha}$$ into $$x^{\alpha-1/2}\frac{\Gamma(\alpha+1)}{\Gamma(\alpha+1/2)}$$; $$D^{1/2}_{\perp}$$ is the adjoint operator of $$D^{1/2}$$ with respect to the standard inner product of $$L^2(0,1)$$, which is denoted through $$\langle\cdot,\cdot\rangle$$; $$K$$ is Catalan's constant; $$K(x)$$ and $$E(x)$$ are the complete elliptic integrals of the first and second kind, with the parameter being the elliptic modulus. $$\tau$$ is the involutive and self-adjoint operator bringing $$g(x)$$ into $$g(1-x)$$; $$\text{IBP},\text{SBP}$$ stand for integration/summation by parts; $$\text{FL}$$ stands for Fourier-Legendre expansion(s).

$$\newcommand{Li}{\operatorname{Li}}$$

By semi-integration by parts $$\zeta(2)$$ equals

$$\int_{0}^{1}\frac{\arcsin(\sqrt{x})\arcsin(\sqrt{1-x})}{\sqrt{x(1-x)}}\,dx=\frac{\pi}{2}\int_{0}^{1}D^{1/2}\log(1-x)\cdot D^{1/2}_{\perp}\log(x)\,dx.$$ where the LHS can be written both in terms of power series and Fourier-Legendre expansions, leading to: $$\begin{eqnarray*}\frac{\pi^3}{24}&=&\sum_{m,n\geq 0}\frac{\left[\frac{1}{4^m}\binom{2m}{m}\right]\left[\frac{1}{4^n}\binom{2n}{n}\right]}{(2n+1)(2m+1)(m+n+1)\binom{m+n}{n}}\\ &=& \sum_{n\geq 0}\frac{(-1)^n}{2n+1}\left[\pi(-1)^n+\frac{2}{(2n+1)}-4(-1)^n\sum_{k=0}^{n}\frac{(-1)^k}{2k+1}\right]^2\end{eqnarray*}$$ or, equivalently: $$\begin{eqnarray*} \frac{\pi^3}{24}&=&\sum_{n\geq 0}\frac{(-1)^n}{2n+1}\left[\frac{2}{(2n+1)}+4(-1)^n\sum_{k>n}\frac{(-1)^k}{2k+1}\right]^2\\&=&\frac{\pi^3}{8}+16\sum_{n\geq 0}\frac{1}{(2n+1)^2}\sum_{k>n}\frac{(-1)^k}{2k+1}+16\sum_{n\geq 0}\frac{(-1)^n}{2n+1}\left[\sum_{k>n}\frac{(-1)^k}{2k+1}\right]^2\end{eqnarray*}$$

$$-\frac{\pi^3}{192}=-\sum_{n\geq 0}\frac{\mathscr{H}_n(2)(-1)^n}{(2n+3)}+\sum_{n\geq 0}\frac{(-1)^n}{2n+1}\left[\sum_{k>n}\frac{(-1)^k}{2k+1}\right]^2$$

$$\begin{eqnarray*}&& \sum_{n\geq 0}\frac{(-1)^n}{2n+1}\left[\sum_{k>n}\frac{(-1)^k}{2k+1}\right]^2\\ &=& \frac{5\pi^3}{192}-\frac{\pi K}{4}+K-\frac{\pi^2}{16}+\frac{\pi\log(2)}{4}-\int_{0}^{1}\sum_{m\geq 0}\frac{x^{2m}(-1)^m}{(2m+1)}\sum_{n\geq 0}\frac{x^{2n}(-1)^n}{(2n+1)^2}\,dx\end{eqnarray*}$$

$$\begin{eqnarray*} \sum_{m,n\geq 0}\frac{(-1)^{m+n}}{(2n+1)^2 (2m+1) (2m+2n+1)}&\stackrel{\text{sym}}{=}&\sum_{m,n\geq 0}\frac{(-1)^{m+n}(m+n+1)}{(2m+1)^2(2n+1)^2(2m+2n+1)}\\ &=&\frac{K^2}{2}+\frac{1}{2}\int_{0}^{1}\left(\sum_{n\geq 0}\frac{x^{2n}(-1)^n}{(2n+1)^2}\right)^2\,dx \end{eqnarray*}$$ via $$\frac{\arcsin(\sqrt{x})}{\sqrt{x}}=\sum_{n\geq 0}P_n(2x-1)\left[\pi(-1)^n+\frac{2}{(2n+1)}-4(-1)^n\sum_{k=0}^{n}\frac{(-1)^k}{2k+1}\right].$$ By summation by arts and Fourier-Legendre expansions we have the following relations: $$\sum_{n\geq 0}\frac{\mathscr{H}_n(2)(-1)^n}{(2n+1)}\stackrel{\text{SBP}}{\longleftrightarrow}\sum_{n\geq 0}\frac{\sum_{k=0}^{n}\frac{(-1)^k}{2k+1}}{(2n+1)^2} \stackrel{\text{FL}}{\longleftrightarrow}\int_{0}^{1}\frac{\arcsin(\sqrt{x})}{\sqrt{x}}K(1-x)\,dx=\left\langle D^{-1/2}\frac{\text{arctanh}{\sqrt{x}}}{x\sqrt{\pi}},K(1-x)\right\rangle$$ where the RHS equals $$\begin{eqnarray*} \left\langle \frac{\text{arctanh}{\sqrt{x}}}{x\sqrt{\pi}},(\tau D^{-1/2})K(x)\right\rangle &=& \int_{0}^{1}\frac{\text{arctanh}{\sqrt{x}}\arcsin(\sqrt{1-x})}{x}\,dx\\ &=& 2\int_{0}^{\pi/2}\theta\tan(\theta)\text{arctanh}{\cos\theta}\,d\theta\end{eqnarray*}$$ and $$\int_{0}^{\pi/2}\theta \tan(\theta)\left(\cos\theta\right)^{2k+1}\,d\theta \stackrel{\text{IBP}}{=}\frac{4^k}{(2k+1)^2 \binom{2k}{k}}$$ allows us to state $$\begin{eqnarray*}\int_{0}^{1}\frac{\arcsin(\sqrt{x})}{\sqrt{x}}K(1-x)\,dx&=&2\sum_{k\geq 0}\frac{4^k}{(2k+1)^3 \binom{2k}{k}}=-2\int_{0}^{1}\frac{\arcsin(x)\log(x)}{x\sqrt{1-x^2}}\,dx\\&=&2\int_{0}^{\pi/2}\frac{-\theta\log(\sin\theta)}{\sin(\theta)}\,d\theta.\end{eqnarray*}$$ We may notice that $$-\log(\sin \theta) = \log(2)+\sum_{n\geq 1}\frac{\cos(2n\theta)}{n},$$ $$\int_{0}^{\pi/2}\frac{\theta}{\sin\theta}\cos(2n\theta)\,d\theta = 2K-2\sum_{k=0}^{n-1}\frac{(-1)^k}{(2k+1)^2},$$ hence the integral $$\int_{0}^{1}\frac{\arcsin(\sqrt{x})}{\sqrt{x}}K(1-x)\,dx$$ boils down to the Euler sum $$\sum_{n\geq 1}\frac{1}{n}\sum_{k\geq n}\frac{(-1)^k}{(2k+1)^2}\stackrel{\text{SBP}}{=}\sum_{n\geq 1}\frac{(-1)^n H_n}{(2n+1)^2}=\int_{0}^{1}\frac{\log(x)\log(1+x^2)}{1+x^2}\,dx$$ getting rid of some alternating Stirling numbers. Additionally the last integral is known (at least) since De Doelder and Flajolet: $$\int_{0}^{1}\frac{\log(x)\log(1+x^2)}{1+x^2}\,dx = -\frac{\pi^3}{64}-K\log(2)-\frac{\pi}{16}\log^2(2)+2\operatorname{Im}\Li_3\left(\tfrac{1+i}{2}\right).$$ Now we just have to re-combine all the pieces to find a closed form for $$\int_{0}^{1}\operatorname{Ti}_2(x)^2\,\frac{dx}{x^2}$$ and the monstrosity $$\sum_{n\geq 0}\frac{(-1)^n}{2n+1}\left(\sum_{k=0}^{n}\frac{(-1)^k}{(2k+1)}\right)^2$$.

$$\sum_{k\geq 0}\frac{4^k}{(2k+1)^3 \binom{2k}{k}} =\phantom{}_4 F_3\left(\tfrac{1}{2},\tfrac{1}{2},1,1;\tfrac{3}{2},\tfrac{3}{2},\tfrac{3}{2};1\right) = -\frac{\pi^3}{32}-\frac{\pi}{8}\log^2(2)+4\operatorname{Im}\Li_3\left(\tfrac{1+i}{2}\right)$$

$$\sum_{n\geq 0}\frac{1}{(2n+1)^2}\left[\pi+\frac{2(-1)^n}{(2n+1)}-4\sum_{k=0}^{n}\frac{(-1)^k}{2k+1}\right]=-\frac{\pi^3}{32}-\frac{\pi}{8}\log^2(2)+4\operatorname{Im}\Li_3\left(\tfrac{1+i}{2}\right)$$

$$\sum_{n\geq 0}\frac{1}{(2n+1)^2}\sum_{k=0}^{n}\frac{(-1)^k}{2k+1}=\frac{7\pi^3}{128}+\frac{\pi}{32}\log^2(2)-\operatorname{Im}\Li_3\left(\tfrac{1+i}{2}\right)$$

$$\sum_{n\geq 0}\frac{\mathscr{H}_n(2)(-1)^n}{2n+3} = \frac{3\pi^3}{128}+\frac{\pi}{32}\log^2(2)-\operatorname{Im}\Li_3\left(\tfrac{1+i}{2}\right)$$

$$\sum_{n\geq 0}\frac{(-1)^n}{2n+1}\left[\sum_{k>n}\frac{(-1)^k}{2k+1}\right]^2=\frac{7\pi^3}{384}+\frac{\pi}{32}\log^2(2)-\operatorname{Im}\Li_3\left(\tfrac{1+i}{2}\right)$$

$$\sum_{m,n\geq 0}\frac{(-1)^{m+n}}{(2n+1)^2 (2m+1) (2m+2n+1)}=\frac{\pi^3}{128}-\frac{\pi K}{4}+K-\frac{\pi^2}{16}+\frac{\pi\log(2)}{4}-\frac{\pi\log^2(2)}{32}+\operatorname{Im}\Li_3\left(\tfrac{1+i}{2}\right)$$

$$\int_{0}^{1}\left(\sum_{n\geq 0}\frac{x^{2n}(-1)^n}{(2n+1)^2}\right)^2\,dx =\frac{\pi^3}{64}-\frac{\pi K}{2}+2K-K^2-\frac{\pi^2}{8}+\frac{\pi\log(2)}{2}-\frac{\pi\log^2(2)}{16}+2\operatorname{Im}\Li_3\left(\tfrac{1+i}{2}\right).$$

• This is truly incredible! – NoName Mar 8 '19 at 13:20
• "summation by arts"... I think that's not a typo, that was truly art. – nospoon Mar 8 '19 at 17:26