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Interesting Integral

01(lnx)21xdx=2a1xxa+1dx\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 aa is a positive integer constant satisfying the equation above, find aa.

Bonus: Prove that a1xxa+1dx=ζ(a) \displaystyle a \int_1^\infty \dfrac{\lfloor x \rfloor }{x^{a+1}} \, dx = \zeta(a) .

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


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