# Interesting Integral

Calculus Level 4

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

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.

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