Well, after full simplification fra1() no longer equals {x}:
sage: ex=fra1(x).full_simplify()
sage: ex
x
It is more interesting to me how Maple finds the correct integral with {x}.
Played with int_a^b f(x,{x}) =? int_a^b f(x,fra1(x))
Whenever Maple could compute the definite integral, it equals
Wolfram alpha integral of f(x,frac(x)) [there are similar other
definitions of {x}]
Is it sound to expect that this approach for integrals of {x} works?
(Possibly $\mod 1$).
This can be generalized to converting sums to definite integrals:
floor(x) = int1(x) = x - fra1(x).
Modulo branches, \sum_{x=a}^b f(x) = \int_a^b f(int1(x)) dx
Maple again computed correctly several infinite complicated sums via
\int.
So far have only one failing? testcase in Maple - can't find the
definite integral, found the indefinite integral but the
limits are "undefined".
--
Georgi
On Sat, Aug 24, 2013 at 03:07:14PM -0700, JamesHDavenport wrote:
> Well, the derivative of the fractional part is indeed 1 where it is
> defined, as
> lim((fra(x+eps)-fra(x))/eps)=lim(eps/eps)=1 unless adding eps crosses a
> boundary,
> which it won't do for eps small enough.
> Maxima (5.29) returns (4 pi log 2 + i log(-1) +pi)/(4 pi).
> Depending on the value of log(-1), this is either log(2) or log(2)+1/2,
> which I must confess I don't understand.
> I conjecture Sage is getting log(2) from taking the "conventional"
> log(-1)=i pi in Maxima's result.
>
> On Saturday, 24 August 2013 12:59:41 UTC+1, Georgi Guninski wrote:
> >
> > Don't claim this is a bug, but don't understand this.
> >
> > Define {x} the fractional part of x by (source mathworld):
> >
> > def fra1(x):
> > """
> > fractional part
> > """
> > return 1/2+I/(2*pi)*log(-exp(-2*pi*I*x))
> >
> >
> > sage: var('x')
> > x
> > sage: ii=integrate(fra1(1/x),x,1/2,1);ii
> > log(2)
> >
> > According to Maple and mathworld this integral equals
> > -1/2 + ln(2)
> >
> > Part of the problem is log() is multivalued, but I suppose
> > for all branches of log(), fra1() should be correct $\mod 1$,
> > yet the result is not correct $\mod 1$.
> >
> > Probably this is related:
> > sage: diff(fra1(x),x)
> > 1
> >
> > Why so?
> >
>
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