On a relation between Euler sums

Prove that:

$$\sum_{m=1}^{\infty} \frac{(-1)^m \mathcal{H}_{2m} }{2m+1} - \frac{1}{2} \sum_{m=1}^{\infty} \frac{(-1)^m \mathcal{H}_m}{2m+1} = \frac{\pi \log 2}{8} - \frac{\mathcal{G}}{2}$$

Solution

We will use the very known fact

$$\frac{x \arctan x}{x^2+1} = \sum_{m=1}^{\infty} (-1)^m \left ( \mathcal{H}_{2m} - \frac{1}{2} \mathcal{H}_m \right ) x^{2m} \tag{1}$$

Thus, for the initial problem we have that:

\begin{align*}
\sum_{m=1}^{\infty} \frac{(-1)^m \mathcal{H}_{2m} }{2m+1} - \frac{1}{2} \sum_{m=1}^{\infty} \frac{(-1)^m \mathcal{H}_m}{2m+1} &= \sum_{m=1}^{\infty} (-1)^m \left ( \mathcal{H}_{2m} - \frac{1}{2} \mathcal{H}_m \right ) \frac{1}{2m+1}  \\
 &= \sum_{m=1}^{\infty} (-1)^m \left ( \mathcal{H}_{2m} - \frac{1}{2} \mathcal{H}_m \right ) \int_{0}^{1}x^{2m} \, {\rm d}x \\
 &= \int_{0}^{1} \sum_{m=1}^{\infty} (-1)^m \left ( \mathcal{H}_{2m} - \frac{1}{2} \mathcal{H}_m \right )x^{2m} \, {\rm d}x\\
 &\overset{(1)}{=} \int_{0}^{1} \frac{x \arctan x}{x^2+1} \, {\rm d}x \\
 &=\left [ \frac{x \arctan^2 x}{2} \right ]_0^1 - \frac{1}{2}\int_{0}^{1} \arctan^2 x \, {\rm d}x \\
 &=\frac{\pi^2}{32} - \frac{1}{2}\int_{0}^{1} \arctan^2 x \, {\rm d}x \\
 &=\frac{\mathcal{G}}{2} + \frac{\pi^2}{32} - \frac{\pi^2}{32} - \frac{\pi \log 2}{8} \\
 &=\frac{\mathcal{G}}{2} - \frac{\pi \log 2}{8}
\end{align*}