Show that, $2\sum_{s=1}^{\infty}\frac{1-\beta(2s+1)}{2s+1}=\ln\left(\frac{\pi}{2}\right)-2+\frac{\pi}{2}$. The Dirichlet beta function is defined as for Re(s)>0
$$\beta(s)=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^s}.$$
Show that,
$$2\sum_{s=1}^{\infty}\frac{1-\beta(2s+1)}{2s+1}=\ln\left(\frac{\pi}{2}\right)-2+\frac{\pi}{2}.$$
Any hints where can we start to prove this identity?
OR any authors would kindly prove it.
 A: $$\begin{align}
2\sum_{m=1}^\infty\frac{1-\beta(2m+1)}{2m+1}&=2\sum_{m=1}^\infty \int_0^1 x^{2m}\,dx\sum_{n=1}\frac{(-1)^{n-1}}{(2n+1)^{2m+1}}\\\\
&=2\sum_{n=1}^\infty \frac{(-1)^{n-1}}{2n+1}\int_0^1\sum_{m=1}^\infty\left(\frac{x^2}{(2n+1)^2}\right)^m\,dx \\\\
&=2\sum_{n=1}^\infty \frac{(-1)^{n-1}}{2n+1}\int_0^1 \frac{x^2}{(2n+1)^2-x^2}\,dx \\\\
&=2\sum_{n=1}^\infty \frac{(-1)^{n-1}}{2n+1}\left((2n+1)\frac12\log\left(\frac{n+1}{n}\right)-1\right)\\\\
&=\sum_{n=1}^\infty(-1)^{n-1}\log\left(\frac{n+1}{n}\right)-2\sum_{n=1}^\infty\frac{(-1)^{n-1}}{2n+1}\\\\
&=\log\left(\frac21\cdot \frac23\cdot \frac43\cdot \frac45\cdots\right)+\frac{\pi}{2}-2\\\\
&=\log\left(\prod_{n=1}^\infty \frac{2n}{2n-1}\frac{2n}{2n+1}\right)+\frac{\pi}{2}-2\\\\
&=\log\left(\frac{\pi}{2}\right)+\frac{\pi}{2}-2
\end{align}$$
as was to be shown!
In the development, we used the series for the arctangent function to evaluate $\arctan(1)=\sum_{n=0}^\infty \frac{(-1)^{n}}{(2n+1)}=\frac{\pi}{4}$ along with the Wallis Product for $\pi$, $\prod_{n=1}^\infty \frac{2n}{2n-1}\frac{2n}{2n+1}=\frac{\pi}{2}$.
