An integral of Romanian Mathematical Magazine

Find a closed form: \[ \Omega = \int_{1}^{e^{\pi}} \int_{0}^{1} \frac{\arcsin(x^a) \arccos(x^a)}{x} \, dx \, da \] Proposed by Ankush Kumar Parcha - India 

Solution by Ose Favour - Nigeria \[ \Omega = \int_{0}^{1} \frac{\arccos(x) \arcsin(x)}{x} \, dx = \int_{0}^{1} \frac{\arcsin(x)}{x} \left( \frac{\pi}{2} - \arcsin(x) \right) \, dx = \] \[ = \frac{\pi}{2} \int_{0}^{1} \frac{\arcsin(x)}{x} \, dx - \int_{0}^{1} \frac{(\arcsin(x))^2}{x} \, dx \cong \] \[ = \frac{\pi}{2} \int_{0}^{\frac{\pi}{2}} y \cot(y) \, dy - \int_{0}^{\frac{\pi}{2}} y^2 \cot(y) \, dy = \frac{\pi^2}{4} \ln(2) + 2 \int_{0}^{\frac{\pi}{2}} y \ln(\sin(y)) \, dy = \] \[ = \frac{\pi^2}{4} \ln(2) - \frac{\pi^2}{8} \ln(2) - \sum_{k=1}^{\infty} \frac{1}{k} \int_{0}^{\frac{\pi}{2}} y \cos(2ky) \, dy = \frac{7}{8} \zeta(3). \]