TITLE: An attempt to prove the generalization of $\sum_{n=1}^\infty \frac{(-1)^nH_n}{n^{2a}}$
QUESTION [13 upvotes]: The following classical generalization

$$\sum_{n=1}^\infty\frac{(-1)^{n}H_n}{n^{2a}}=-\left(a+\frac 12\right)\eta(2a+1)+\frac12\zeta(2a+1)+\sum_{j=1}^{a-1}\eta(2j)\zeta(2a+1-2j)$$
where $\eta(a)=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^a}=(1-2^{1-a})\zeta(a)$ is the Dirichlet eta function.

was proved by G. Bastien here page 7 Eq. 17 and also by Cornel here.

I am trying to prove it in a different way but came across an integral that can be calculated by Beta function but I want it in $\zeta$ if possible to get the right result.
Here is my approach which follows from the same idea of my solution here:
By using $$\frac1{n^{2a}}=-\frac1{(2a-1)!}\int_0^1x^{n-1}\ln^{2a-1}(x)\ dx$$
we can write
$$\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^{2a}}=-\frac1{(2a-1)!}\int_0^1\frac{\ln^{2a-1}(x)}{x}\left(\sum_{n=1}^\infty(-x)^nH_n\right)\ dx$$
$$=\frac1{(2a-1)!}\int_0^1\frac{\ln^{2a-1}(x)\ln(1+x)}{x(1+x)}dx=\frac1{(2a-1)!}I_a\tag1$$

$$I_a=\int_0^1\frac{\ln^{2a-1}(x)\ln(1+x)}{x(1+x)}dx=\int_0^\infty\frac{\ln^{2a-1}(x)\ln(1+x)}{x(1+x)}dx-\underbrace{\int_1^\infty\frac{\ln^{2a-1}(x)\ln(1+x)}{x(1+x)}dx}_{x\mapsto 1/x}$$
$$=\int_0^\infty\frac{\ln^{2a-1}(x)\ln(1+x)}{x(1+x)}dx+\color{blue}{\int_0^1\frac{\ln^{2a-1}(x)\ln(1+x)}{1+x}dx}-\int_0^1\frac{\ln^{2a}(x)}{1+x}dx$$
By adding
$$I_a=\int_0^1\frac{\ln^{2a-1}(x)\ln(1+x)}{x(1+x)}dx=\int_0^1\frac{\ln^{2a-1}(x)\ln(1+x)}{x}dx-\color{blue}{\int_0^1\frac{\ln^{2a-1}(x)\ln(1+x)}{1+x}dx}$$
to both sides, the blue integral nicely cancels out and we get
$$2I_a=\int_0^\infty\frac{\ln^{2a-1}(x)\ln(1+x)}{x(1+x)}dx+\underbrace{\int_0^1\frac{\ln^{2a-1}(x)\ln(1+x)}{x}dx}_{IBP}-\int_0^1\frac{\ln^{2a}(x)}{1+x}dx$$
$$=\int_0^\infty\frac{\ln^{2a-1}(x)\ln(1+x)}{x(1+x)}dx-\frac{1+2a}{2a}\int_0^1\frac{\ln^{2a}(x)}{1+x}dx$$
where
$$\int_0^1\frac{\ln^{2a}(x)}{1+x}dx=\sum_{n=1}^\infty(-1)^{n-1}\int_0^1 x^{n-1}\ln^{2a}(x)dx=(2a)!\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^{2a+1}}=(2a)!\eta(2a+1)$$
so
$$I_a=\frac12\int_0^\infty\frac{\ln^{2a-1}(x)\ln(1+x)}{x(1+x)}dx-\left(a+\frac12\right)(2a-1)!\eta(2a+1)\tag2$$
Plug $(2)$ in $(1)$
$$\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^{2a}}=-\left(a+\frac12\right)\eta(2a+1)+\frac1{2(2a-1)!}\int_0^\infty\frac{\ln^{2a-1}(x)\ln(1+x)}{x(1+x)}dx\tag{3}$$
So any idea how to evaluate the integral in $(3)$ in a way that completes my proof?

REPLY [2 votes]: A different proof with a big bonus:
By the definition of the skew harmonic number we have
\begin{gather}
\sum_{n=1}^\infty\frac{(-1)^n \overline{H}_n}{n^{2q}}=\sum_{n=1}^\infty \frac{(-1)^n}{n^{2q}}\left(\ln(2)-\int_0^1 \frac{(-x)^n}{1+x}\mathrm{d}x\right)\\
=\ln(2)\sum_{n=1}^\infty \frac{(-1)^n}{n^{2q}}-\sum_{n=1}^\infty\frac{1}{n^{2q}}\int_0^1\frac{x^n}{1+x}\mathrm{d}x\\
=-\ln(2)\eta(2q)-\sum_{n=1}^\infty\frac{1}{n^{2q}}(\ln(2)+H_{\frac n2}-H_n)\\
=-\ln(2)\eta(2q)-\ln(2)\zeta(2q)-\sum_{n=1}^\infty\frac{H_{\frac n2}}{n^{2q}}+\sum_{n=1}^\infty\frac{H_n}{n^{2q}}.\label{fbi}
\end{gather}
To get the first sum, set $p=2$ and replace $q$ by $2q$ in (1),
\begin{gather}
\sum_{n=1}^\infty\frac{H_{\frac n2}}{n^{2q}}=2 \sum_{n=1}^\infty\frac{H_{2n}}{(2n)^{2q}}-\sum_{j=1}^{2q-2}(-2)^{-j}\zeta(2q-j)\zeta(j+1)\\
=\sum_{n=1}^\infty\frac{H_{n}}{n^{2q}}+\sum_{n=1}^\infty\frac{(-1)^nH_{n}}{n^{2q}}-\sum_{j=1}^{2q-2}(-2)^{-j}\zeta(2q-j)\zeta(j+1).
\end{gather}
Thus we have
\begin{gather}
\sum_{n=1}^\infty\frac{(-1)^n \overline{H}_n}{n^{2q}}+\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^{2q}}=\sum_{j=1}^{2q-2}(-2)^{-j}\zeta(2q-j)\zeta(j+1)\\
-\ln(2)(\eta(2q)+\zeta(2q)).\tag{a}
\end{gather}
To establish another relation, let $a_n=\frac{\overline{H}_{n}}{n^{2q}}$ in
$$\sum_{n=1}^\infty a_{2n}=\frac12\sum_{n=1}^\infty a_n+\frac12\sum_{n=1}^\infty (-1)^n a_n,$$
we obtain
\begin{gather*}
\sum_{n=1}^\infty \frac{(-1)^n\overline{H}_{n}}{n^{2q}}+\sum_{n=1}^\infty \frac{\overline{H}_{n}}{n^{2q}}=2\sum_{n=1}^\infty \frac{\overline{H}_{2n}}{(2n)^{2q}}\\
\{\text{substitute $\overline{H}_{2n}=H_{2n}-H_n$}\}\\
=2\sum_{n=1}^\infty \frac{H_{2n}}{(2n)^{2q}}-2\sum_{n=1}^\infty \frac{H_{n}}{(2n)^{2q}}\\
=\sum_{n=1}^\infty \frac{(-1)^nH_{n}}{n^{2q}}+\sum_{n=1}^\infty \frac{H_{n}}{n^{2q}}-2\sum_{n=1}^\infty \frac{H_{n}}{(2n)^{2q}}\\
=\sum_{n=1}^\infty \frac{(-1)^nH_{n}}{n^{2q}}+(1-2^{1-2q})\sum_{n=1}^\infty \frac{H_{n}}{n^{2q}}. 
\end{gather*}
Rearrange the terms,
\begin{equation}
\sum_{n=1}^\infty \frac{(-1)^n\overline{H}_{n}}{n^{2q}}-\sum_{n=1}^\infty \frac{(-1)^nH_{n}}{n^{2q}}=(1-2^{1-2q})\sum_{n=1}^\infty \frac{H_{n}}{n^{2q}}-\sum_{n=1}^\infty \frac{\overline{H}_{n}}{n^{2q}}.\tag{b}
\end{equation}
Taking the difference of $(a)$ and $(b)$ then dividing by $2$ gives
\begin{gather*}
\sum_{n=1}^\infty \frac{(-1)^nH_{n}}{n^{2q}}=-\frac12\ln(2)(\zeta(2q)+\eta(2q))+\frac12\sum_{j=1}^{2q-2}(-2)^{-j}\zeta(2q-j)\zeta(j+1)\\
+\frac12\sum_{n=1}^\infty \frac{\overline{H}_{n}}{n^{2q}}-\frac{1-2^{1-2q}}{2}\sum_{n=1}^\infty \frac{H_{n}}{n^{2q}}
\end{gather*}
and combining $(a)$ and $(b)$ then dividing by $2$ gives
\begin{gather*}
\sum_{n=1}^\infty \frac{(-1)^n\overline{H}_{n}}{n^{2q}}=-\frac12\ln(2)(\zeta(2q)+\eta(2q))+\frac12\sum_{j=1}^{2q-2}(-2)^{-j}\zeta(2q-j)\zeta(j+1)\\
-\frac12\sum_{n=1}^\infty \frac{\overline{H}_{n}}{n^{2q}}+\frac{1-2^{1-2q}}{2}\sum_{n=1}^\infty \frac{H_{n}}{n^{2q}}.
\end{gather*}
Substitute
$$\sum_{m=1}^\infty\frac{H_m}{m^q}
=\frac{q+2}{2}\zeta(q+1)-\frac12\sum_{j=1}^{q-2}\zeta(q-j)\zeta(j+1)$$
and
$$\sum_{m=1}^\infty\frac{\overline{H}_m}{m^q}=\left(1-2^{-q}-\frac{q}{2}\right)\zeta(q+1)+(2-2^{1-q})\ln(2)\zeta(q)$$
$$+\frac12\sum_{j=1}^{q-2}(1-2^{1-q+j})(1-2^{-j})\zeta(q-j)\zeta(j+1)$$
we get
$$
\sum_{n=1}^\infty\frac{(-1)^{n}H_n}{n^{2q}}=-\,\frac{2^{2q+1}q-2q-1}{2^{2q+1}}\zeta(2q+1)$$
$$+\frac14\sum_{j=1}^{2q-2}\left[2-2^{-j}-(-2)^{1-j}-2^{j-2q+1}\right]\zeta(2q-j)\zeta(j+1)\tag{c}$$
and
$$\sum_{n=1}^\infty\frac{(-1)^n \overline{H}_n}{n^{2q}}=(2^{1-2q}-2)\ln(2)\zeta(2q)+\frac{2^{2q+1}q-2q-1}{2^{2q+1}}\zeta(2q+1)$$
$$+\frac14\sum_{j=1}^{2q-2}\left[2^{j-2q+1}-(-2)^{1-j}+2^{-j}-2\right]\zeta(2q-j)\zeta(j+1)\tag{d}.$$

Note: we can see that the the form of $(c)$ is different than the form in the question body but for sure they are equivalent as they both give the right results for any $q\in N.$