The first integral seems to have a problem when the upper limit x crosses the square of an integer: The integrand jumps, but the integral is continuous. (No problem yet.) But the RHS expression will have a big jump, as sqrt(x) crosses an integer value, and floor(sqrt(x)) jumps, and the positive product in the RHS jumps. What am I missing? Rich --------------------- Quoting rwg@sdf.lonestar.org:
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
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