[math-fun] integrate(floor(sqrt(x)))

It's very easy to "shew that" / x [ I floor(sqrt(t)) dt ] / 0 (floor(sqrt(x)) + 1) (2 floor(sqrt(x)) + 1) = floor(sqrt(x)) (x + -------------------------------------------), 6 but a little surprising when you look at it. / x [ 3/2 I sqrt(floor(t)) dt = floor (x) + x sqrt(floor(x)) ] / 0 1 1 + hurwitz_zeta(- -, floor(x)) - zeta(- -). 2 2 --rwg