The problem is just a typo: In front of the fraction, the sign should be a minus, rather than a plus.

David

GACK, it's more serious than a typo! I got it by plugging e=f=1 into a broken version of the identity Integrate[t^f*Floor[t]^e, {t, 0, x}] == (1/(1 + f))*(x^(1 + f)*Floor[x]^e + Sum[(-1)^i*Binomial[e, i]*(-HurwitzZeta[-1 - e - f + i, 1 + Floor[x]] + Zeta[-1 - e - f + i]), {i, e}]) == -((1/(1 + f))*((-x^(1 + f))*Floor[x]^e + Floor[x]^(1 + e + f) - Sum[Binomial[1 + f, i]*(-HurwitzZeta[-1 - e - f + i, Floor[x]] + HurwitzZeta[-1 - e - f + i, 0]), {i, 1 + f}])) where e or f is an integer and e,f>=0. Thanks Rich and David for catching this! --rwg

----- Original Message ----- From: <rwg@sdf.lonestar.org> ----- Original Message ----- From: <rcs@xmission.com> To: <math-fun@mailman.xmission.com> Sent: Monday, January 19, 2009 20:29 Subject: Re: [math-fun] integrate(floor(sqrt(x)))

...

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

_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun