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.
by
Harsh Shrivastava