Problem - S. Stewart (Australia)
Prove
\[\sum_{n=1}^\infty\frac{\bar{H}_{2n}}{n^2}=\frac{3\zeta(3)}{4}\]where
\[\bar{H}_n=\sum_{k=1}^n\frac{(-1)^{k-1}}{k}\]is the nth skew-harmonic number.
Solution
Before we evaluate the sum in question we will first evaluate several useful integrals and series.
Define
\[L=\int_0^1\frac{1}{x}\ln^2\left(\frac{1-x}{1+x}\right)\,\mathrm dx.\]Upon making the substitution $x=\frac{1-t}{1+t}$ we find that
\[\begin{align*} L &= 2\int_0^1\frac{1+t}{1-t}\cdot\frac{1}{(1+t)^2}\ln^2(t)\,\mathrm dt \\ &= 2\int_0^1\frac{\ln^2(t)}{1-t^2}\,\mathrm dt \\ &= 2\int_0^1\ln^2(t)\sum_{n=0}^\infty t^{2n} dt, \\ &\qquad\text{by monotone convergence} \\ &= 2\sum_{n=0}^\infty\int_0^1\ln^2(t)t^{2n}\,\mathrm dt \\ &= 4\sum_{n=0}^\infty\frac{1}{(2n+1)^3} \\ &= \frac{7\zeta(3)}{2}.\end{align*}\]In the above, the integral $\int_0^1\ln^2(t)t^\alpha\,\mathrm dt$ may be evaluated by differentiating $\int_0^1 t^\alpha\,\mathrm dt=\frac{1}{\alpha+1}$ under the integral sign twice. And the sum $\sum_{n=0}^\infty\frac{1}{(2n+1)^3}$ can be evaluated by considering $\sum_{n=1}^\infty\frac{1}{n^3}=\sum_{n=0}^\infty\frac{1}{(2n+1)^3}+\sum_{n=1}^\infty\frac{1}{(2n)^3}$.
Let
\[K=\int_0^1\frac{\ln^2(1-x^2)}{x}\,\mathrm dx.\]If we make the change of variables $x=\sqrt{1-t}$ so that $\mathrm dx=\frac{-1}{2\sqrt{1-t}}\,\mathrm dt$ we get
\[\begin{align*} K &= \frac{1}{2}\int_0^1\frac{\ln^2(t)}{1-t}\,\mathrm dt \\ &= \frac{1}{2}\int_0^1\ln^2(t)\sum_{n=0}^\infty t^n\,\mathrm dt, \\ &\qquad\text{by monotone convergence}\\ &= \sum_{n=0}^\infty\frac{1}{(n+1)^3} \\ &= \zeta(3).\end{align*}\]A curious integral that pops up is
\[I=\int_0^1\frac{\ln(1-x)\ln(1+x)}{x}\,\mathrm dx\]and it may be evaluated using a sneaky trick,
\[\begin{align*} I &= \frac{1}{4}\int_0^1\frac{(\ln(1-x)+\ln(1+x))^2-(\ln(1-x)-\ln(1+x))^2}{x}\,\mathrm dx \\ &= \frac{1}{4}\left(\int_0^1\frac{(\ln(1-x)+\ln(1+x))^2}{x}\,\mathrm dx-\int_0^1\frac{(\ln(1-x)-\ln(1+x))^2}{x}\,\mathrm dx\right) \\ &= \frac{1}{4}(K-L) \\ &= -\frac{5\zeta(3)}{8}.\end{align*}\]$I$ also admits a nice representation as an alternating Euler sum,
\[\begin{align*} \sum_{n=1}^\infty\frac{(-1)^n}{n^2}H_n &= \sum_{n=1}^\infty\frac{(-1)^n}{n^2}\int_0^1\frac{1-x^n}{1-x}\,\mathrm dx \\ &= \int_0^1\frac{-\eta(2)-\text{Li}_2(-x)}{1-x}\,\mathrm dx,\;\text{by uniform convergence} \\ &= \left[(\eta(2)+\text{Li}_2(-x))\ln(1-x)\right]|_0^1+I, \\ &\qquad \text{ by parts, letting }u=-\eta(2)-\text{Li}_2(-x),\;\mathrm dv=\frac{\mathrm dx}{1-x} \\ &= I.\end{align*}\]The last major integral that needs to be evaluated is
\[J=\int_0^1\frac{\ln^2(1+x)}{x}\,\mathrm dx.\]We attack this integral by expressing the logarithms through the Mercator series and applying the Cauchy product,
\[\begin{align*} J &=\int_0^1\frac{1}{x}\left(\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}x^n\right)^2\,\mathrm dx \\ &= \int_0^1 x\left(\sum_{n=0}^\infty\frac{(-1)^n}{n+1}x^n\right)^2\,\mathrm dx \\ &= \int_0^1 x\sum_{n=0}^\infty\left(\sum_{k=0}^n\frac{(-1)^k(-1)^{n-k}}{(k+1)(n-k+1)}\right)x^n\,\mathrm dx \\ &= \sum_{n=0}^\infty(-1)^n\left(\sum_{k=0}^n\frac{1}{(k+1)(n-k+1)}\right)\int_0^1 x^{n+1}\,\mathrm dx, \\ &\qquad\text{by uniform convergence} \\ &= \sum_{n=0}^\infty\frac{(-1)^n}{n+2}\sum_{k=0}^n\frac{1}{(k+1)(n-k+1)} \\ &= \sum_{n=0}^\infty\frac{(-1)^n}{n+2}\sum_{k=0}^n\left(\frac{1}{(n+2)(k+1)}+\frac{1}{(n+2)(n-k+1)}\right) \\ &= 2\sum_{n=0}^\infty\frac{(-1)^n}{(n+2)^2}\sum_{k=0}^n\frac{1}{k+1},\;\text{by symmetry} \\ &= 2\sum_{n=0}^\infty\frac{(-1)^n}{(n+2)^2}H_{n+1} \\ &= 2\sum_{n=0}^\infty\frac{(-1)^{n+2}}{(n+2)^2}H_{n+2}-2\sum_{n=0}^\infty\frac{(-1)^{n+2}}{(n+2)^3} \\ &= 2\sum_{n=1}^\infty\frac{(-1)^n}{n^2}H_n-2\sum_{n=1}^\infty\frac{(-1)^n}{n^3} \\ &= 2I+2\eta(3) \\ &= \frac{\zeta(3)}{4}.\end{align*}\]We may now evaluate the original sum stated in the problem by expressing $\bar{H}_n$ through an integral formula similar to that for $H_n$,
\[\begin{align*} \sum_{n=1}^\infty\frac{\bar{H}_{2n}}{n^2} &= \sum_{n=1}^\infty\frac{1}{n^2}\int_0^1\frac{1-x^{2n}}{1+x}\,\mathrm dx \\ &= \int_0^1\frac{\zeta(2)-\text{Li}_2(x^2)}{1+x}\,\mathrm dx,\;\text{by uniform convergence} \\ &= \left[(\zeta(2)-\text{Li}_2(x^2))\ln(1+x)\right]|_0^1-2\int_0^1\frac{\ln(1-x^2)\ln(1+x)}{x}\,\mathrm dx, \\ &\qquad\text{by parts, letting } u=\zeta(2)-\text{Li}_2(x^2),\;\mathrm dv=\frac{\mathrm dx}{1+x} \\ &= -2\int_0^1\frac{\ln((1-x)(1+x))\ln(1+x)}{x}\,\mathrm dx \\ &= -2\left(\int_0^1\frac{\ln(1-x)\ln(1+x)}{x}\,\mathrm dx+\int_0^1\frac{\ln^2(1+x)}{x}\,\mathrm dx\right) \\ &= -2(I+J) \\ &= \frac{3\zeta(3)}{4}.\end{align*}\]