I'm trying to seek ideas related to generalizing the following integral,
$$I(q)=\int_0^1\frac{\ln^{2q-1}( x)\ln(1+x^2)}{1+x^2}\ dx=\sum_{n=1}^\infty\frac{(-1)^nH_{n}}{(2n+1)^{2q}}$$
The motivation is to generalize the series is due to the existence of its counterpart,
$$\sum_{n=1}^{\infty} \frac{(-1)^n H_n}{(2n+1)^{2q+1}} = (2q+1)\beta(2q+2) + \frac{\pi}{(2q)!4^{q+1}} \lim_{m \to \frac{1}{2}} \frac{\mathrm{d}^{2q}}{\mathrm{d}m^{2q}} \frac{\psi(1 - m) + \gamma}{\sin(m\pi)}$$
$\int_0^1\frac{\ln^{2q}( x)\ln(1+x^2)}{1+x^2}\ dx$ remains covered.
Not too much work to be shown,
$$\int_0^1 \frac{\ln^{2q - 1} x \ln(1 + x^2)}{1 + x^2} \, dx = \frac{1}{2^{2q + 1}} \int_0^\infty \frac{\ln^{2q - 1} x \ln(1 + x)}{(1 + x)\sqrt{x}} \, dx - \int_0^1 \frac{\ln^{2q} x}{1 + x^2} \, dx$$
The first integral has,
$$\lim_{(m,n)\to\left(\frac12,1\right)}\int_0^\infty \frac{x^{m-1} \ln^{2q-1} x \cdot \ln(1 + x)}{(1 + x)^n} \, dx=-\lim_{(m,n)\to\left(\frac12,1\right)}\frac{\partial}{\partial n \partial m^{2q-1}}{\frac{\Gamma(m)\Gamma(n-m)}{\Gamma(n)}}$$
- Question 1 : Can this above expression be simplified to a nicer extent?
The second integral has,
$$\int_0^1 \frac{\ln^{2q}(x)}{1+x^2}dx= \Gamma(2q+1)\beta(2q+1)$$
- Question 2 : Is there a better way to approach this and eventually generalize $\sum_{n=1}^\infty\frac{(-1)^nH_{n}}{(2n+1)^{2q}}$
I'm hoping this helps continue to solving for $\sum_{n=1}^\infty\frac{(-1)^nH_{\frac{n}{2}}}{(2n+1)^{2q}}$, again as its counterpart $\sum_{n=1}^\infty\frac{(-1)^nH_{\frac{n}{2}}}{(2n+1)^{2q-1}}$ exist for different $q$'s.
