Forgot password? New user? Sign up
Existing user? Log in
∫01(lnx)21−x dx=2a∫1∞⌊x⌋xa+1 dx\large \int_0^1 \dfrac{(\ln x)^2}{1-x} \, dx = 2a \int_1^\infty \dfrac{\lfloor x \rfloor }{x^{a+1}} \, dx ∫011−x(lnx)2dx=2a∫1∞xa+1⌊x⌋dx
Given that aaa is a positive integer constant satisfying the equation above, find aaa.
Bonus: Prove that a∫1∞⌊x⌋xa+1 dx=ζ(a) \displaystyle a \int_1^\infty \dfrac{\lfloor x \rfloor }{x^{a+1}} \, dx = \zeta(a) a∫1∞xa+1⌊x⌋dx=ζ(a).
Notation: ζ(⋅) \zeta(\cdot) ζ(⋅) denotes the Riemann zeta function.
Problem Loading...
Note Loading...
Set Loading...