Interesting Integral

Calculus

$\large \int_0^1 \dfrac{(\ln x)^2}{1-x} \, dx = 2a \int_1^\infty \dfrac{\lfloor x \rfloor }{x^{a+1}} \, dx$

Given that $a$ is a positive integer constant satisfying the equation above, find $a$ .

Bonus: Prove that $\displaystyle a \int_1^\infty \dfrac{\lfloor x \rfloor }{x^{a+1}} \, dx = \zeta(a)$ .

Notation: $\zeta(\cdot)$ denotes the Riemann zeta function.