Re: Computing integer sequences via integrals

Dear Georgi,

On Fri 20 Jan 2023 at 08:20, Georgi Guninski wrote:

I am trying to compute integer sequences via closed
forms of definite integrals, but probably this is impossible.

It depends what you intend by "impossible". The mathematical _expression_

might be possible, but you may not be able to compute it using a well-defined

primitive of the function you are integrating.

The question is: which integer sequences can be computed
by integrals?

Partial results:

Trying to "discretize" the reals, we are using floor()
and the area of the definite integrals.

For a start consider the plot of floor(x).
? ploth(x=1,10,floor(x))

The area is the union of rectangles and for integer
bounds the area is integer.

sagemath computes the following indefinite integrals:
sage: integrate(floor(x),x)
1/2*(2*x - floor(x) - 1)*floor(x)

This is a slight abuse: at integers the resulting function has no derivative,

so it is not really a primitive of floor(x). However this function coincides with

the integral of floor(t) for t from 0 to x.

sage: integrate(floor(x)^2,x)
x*floor(x)^2

sage does not know a primitive function for floor(x)^2. If I did not make

a mistake, I think

(x-floor(x)*2/3-1/2)*floor(x)^2+floor(x)/6

could be a function which coincides with the integral of floor(t) for t from 0 to x,

hence the "primitive" you are looking for.

The area for floor(x) and the sum(i=1,N-1,x) appear to be computed correctly:
sage: I1=integrate(floor(x),x)
sage: N=13;[I1(x=N)-I1(x=1)-N*(N-1)/2]
[0]

The area for floor(x)^2 doesn't work for me, why?

Because nobody cared to insert a "primitive" for floor(x)^2 in sage.

floor() might not be elementary function, but there is closed
form for it using exp() and log(), taking the principal
branch of the logarithm:

? floor1(x)=x - 1/2*I*log(-exp(-2*I*Pi*x))/Pi - 1/2
%12 = (x)->x-1/2*I*log(-exp(-2*I*Pi*x))/Pi-1/2
? floor1(13.1)
%13 = 13.000000000000000000000000000000000000 + 0.E-39*I

The closed form cannot be used to compute a "primitive", because there

are jumps at integers.

Best,

Loïc