An attempt to prove the generalization of $\sum_{n=1}^\infty \frac{(-1)^nH_n}{n^{2a}}$
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 $(1)$
$$\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^{2a}}=-\left(a+\frac12\right)\eta(2n+1)+\frac1{2(2a-1)!}\int_0^\infty\frac{\ln^{2a-1}(x)\ln(1+x)}{x(1+x)}dx\tag{3}$$
So any idea how to evaluate the integral in $(3)$ in a way that completes my proof?
In the question body in Eq $(3)$, we reached
$$\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^{2a}}=-\left(a+\frac12\right)\eta(2n+1)+\frac1{2(2a-1)!}\int_0^\infty\frac{\ln^{2a-1}(x)\ln(1+x)}{x(1+x)}dx\tag{1}$$
From following the same approach of this solution, we have
\begin{align} I_a=\int_0^\infty\frac{\ln^{2a-1}(x)\ln(1+x)}{x(1+x)}dx&= - \frac{\partial^{2a-1}}{\partial m^{2a-1}} \frac{\partial}{\partial n} \operatorname{B}(m,n-m) \, \Bigg \rvert_{m=0, \, n=1} \\ &= - \frac{\partial^{2a-1}}{\partial m^{2a-1}} \operatorname{\Gamma}(m) \frac{\partial}{\partial n} \frac{\operatorname{\Gamma}(n-m)}{\operatorname{\Gamma}(n)} \, \Bigg \rvert_{m=0,\, n=1} \\ &= - \frac{\partial^{2a-1}}{\partial m^{2a-1}} \operatorname{\Gamma}(m) \operatorname{\Gamma}(1-m) [\operatorname{\psi}^{(0)} (1-m) + \gamma] ~\Bigg \rvert_{m=0} \\ &= - \frac{\partial^{2a-1}}{\partial m^{2a-1}}\frac{\pi}{\sin(\pi m)} [\operatorname{\psi}^{(0)} (1-m) + \gamma] ~\Bigg \rvert_{m=0} \\ \end{align}
Now if we calculate some cases of $I_a$ and lets pick $I_4$ and $I_5$ we notice that
$$I_a=-\color{blue}{\frac1{2a}}\psi^{(2a)}(1)-2(2a-1)!\sum_{j=1}^{a-1}\frac{\psi^{(2a-2j)}(1)}{(2a-2j)!}\ \eta(2j)$$
$$=-\color{blue}{\frac{(2a-1)!}{(2a)!}}\psi^{(2a)}(1)-2(2a-1)!\sum_{j=1}^{a-1}\frac{\psi^{(2a-2j)}(1)}{(2a-2j)!}\ \eta(2j)$$
$$=(2a-1)!\left[-\frac{\psi^{(2a)}(1)}{(2a)!}-2\sum_{j=1}^{a-1}\frac{\psi^{(2a-2j)}(1)}{(2a-2j)!}\ \eta(2j)\right]$$
$$=(2a-1)!\left[\zeta(2a+1)+2\sum_{j=1}^{a-1}\zeta(2a+1-2j)\ \eta(2j)\right]\tag2$$
where we used $\psi^{(a)}(1)=(-1)^{a-1}a!\zeta(a+1)$ which follows from the generalization ,
$$\psi(x)=-\gamma+\sum_{n=0}^\infty\left(\frac1{n+1}-\frac{1}{n+x}\right)$$ $$\Longrightarrow \psi^{(a)}(x)=\sum_{n=0}^\infty\frac{(-1)^{a-1}a!}{(n+x)^{a+1}}\Longrightarrow \psi^{(a)}(1)=(-1)^{a-1}a!\zeta(a+1)$$
Plug $(2)$ in $(1)$ we get
$$\sum_{n=1}^\infty\frac{(-1)^{n}H_n}{n^{2a}}=-\left(a+\frac 12\right)\eta(2a+1)+\frac12\zeta(2a+1)+\sum_{j=1}^{a-1}\eta(2j)\zeta(2a+1-2j)$$
Big thanks to @ComplexYetTrivial for his solution without which I wouldn't have been able to complete my proof.
A better 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
and
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 $q\in N.$