real analysis - Evaluation of $sum_{n=1}^infty frac{(-1)^{n-1}eta(n)}{n} $ without using the Wallis Product

In THIS ANSWER, I showed that

$$2\sum_{s=1}^{\infty}\frac{1-\beta(2s+1)}{2s+1}=\ln\left(\frac{\pi}{2}\right)-2+\frac{\pi}{2}$$

where $\beta(s)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^s}$ is the Dirichlet Beta Function.

In the development, it was noted that

$$\begin{align}
\sum_{n=1}^\infty(-1)^{n-1}\log\left(\frac{n+1}{n}\right)&=\log\left(\frac21\cdot \frac23\cdot \frac43\cdot \frac45\cdots\right)\\\\
&=\log\left(\prod_{n=1}^\infty \frac{2n}{2n-1}\frac{2n}{2n+1}\right)\\\\
&=\log\left(\frac{\pi}{2}\right) \tag 1
\end{align}$$

where I used Wallis's Product for $\pi/2$.

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If instead of that approach, I had used the Taylor series for the logarithm function, then the analysis would have led to

$$\sum_{n=1}^\infty(-1)^{n-1}\log\left(\frac{n+1}{n}\right)=\sum_{n=1}^\infty \frac{(-1)^{n-1}\eta(n)}{n} \tag 2$$

where $\eta(s)=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s}$ is the Dirichlet eta function.

Given the series on the right-hand side of $(2)$ as a starting point, it is evident that we could simply reverse steps and arrive at $(1)$.

But, what are some other distinct ways that one can take to evaluate the right-hand side of $(2)$?

For example, one might try to use the integral representation

$$\eta(s)=\frac{1}{\Gamma(s)}\int_0^\infty \frac{x^{s-1}}{1+e^x}\,dx$$

and arrive at

$$\sum_{n=1}^\infty \frac{(-1)^{n-1}\eta(n)}{n} =\int_0^\infty \frac{1-e^{-x}}{x(1+e^x)}\,dx =\int_1^\infty \frac{x-1}{x^2(x+1)\log(x)}\,dx \tag 3$$

Yet, neither of these integrals is trivial to evaluate (without reversing the preceding steps).

And what are some other ways to handle the integrals in $(3)$?

Answer

Another way to handle $(2)$ is using the identity $$\eta\left(s\right)=\left(1-\frac{1}{2^{s-1}}\right)\zeta\left(s\right)

$$ hence, since $\eta\left(1\right)=\log\left(2\right)
$, $$\sum_{n\geq1}\frac{\left(-1\right)^{n-1}}{n}\eta\left(n\right)=\log\left(2\right)+\sum_{n\geq2}\frac{\left(-1\right)^{n-1}}{n}\eta\left(n\right)
$$ $$=\log\left(2\right)+\sum_{n\geq2}\frac{\left(-1\right)^{n-1}}{n}\zeta\left(n\right)-\sum_{n\geq2}\frac{\zeta\left(n\right)}{n}\left(-\frac{1}{2}\right)^{n-1}
$$ and now we can use the identity $$\sum_{n\geq2}\frac{\zeta\left(n\right)}{n}\left(-x\right)^{n}=x\gamma+\log\left(\Gamma\left(x+1\right)\right),\,-1 $$ which can be proved taking the log of the Weierstrass product of Gamma. So $$\sum_{n\geq1}\frac{\left(-1\right)^{n-1}}{n}\eta\left(n\right)=\log\left(2\right)+2\log\left(\frac{\sqrt{\pi}}{2}\right)=\log\left(\frac{\pi}{2}\right).$$