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After a restart, entering: int(sin(x),x,0,∞), repeatedly, produces the results shown in the image. What is meant by "bounded_function(n)+1", where n increments by 2, after each entry?

[attachment=5792]
Hm, "undef" would be a reasonable answer, as the integral does not converge. Bounded-function is clear, so the number that is increased by 2 is interesting.
Arno
Found bounded-function in an xcas document:
For regular series expansion, order_size is a bounded function, but for non
regular series expansion, it might tend slowly to infinity, for example like a power of ln(x).
So its left over by the tried taylor expansion of sin(x) to get the integral done, then simply returned as the result, but the argument isn't explained.
Arno
It's a divergent integral, limit does not exist, but I didn't grasp the idea of "boundary_function(n)+1", where the n is always even. What does the "n" tell us, and is the +1, just there to represent a constant? Sometimes the way these things are taught varies by country, but I haven't encountered this kind of result anywhere.
I think the one simply is -(-cos(0)), so the antiderivative is first evaluated at its lower bound, n may be an exponent used by taylor, using cosine instead of sine makes it increase by 6, but here Bernard may have an answer.
Arno
I introduced bounded functions in order to solve limits like
limit(sin(x+1/x)-sin(x),x,inf)
What does the function value represent?

bounded_function(2)+1
Exactly that: a bounded function + 1 (the +1 comes from evaluation of cos at 0).
(03-30-2018 06:14 AM)parisse Wrote: [ -> ]Exactly that: a bounded function + 1 (the +1 comes from evaluation of cos at 0).

Does the “2” just enumerate bounded functions?

Nigel (UKj
Interesting ... I shut down the system last night, and this morning when I turned it back on and ran the emulator, all the pre-existing "bounded_function(n)" entries had changed to "undef+1"

Why does the number increment by 2 each time similar integrals are entered?

Further divergent entries like:

int(cos(x),x,0,∞);
int(sin(x),x,0,∞);

still result in bounded_function(2*n) (I am using n to represent the number of divergent integrals entered). Can programmers use that information for some suitable purpose?
No. I'm thinking of replacing bounded_function by undef in the limit output.
It seems like undef would be more meaningful, Parisse, and thank you, very much, for your explanations, and guidance!

-Dale-
Change has been committed.
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