probability function over naturals

4.3: 12.13: news.math/probability function over naturals:
askamathematician.com 2010:
. can we define a probability distribution on the set of integers
(rather than the real numbers between 0 and 1)
such that they each occur with equal probability
(i.e. does a uniform distribution on the integers exist)?
askamathematician states:
". it can not be the case that
each of the infinity of integers
has the same probability
and the probability density function integrates to 1"

. but intuitively you can see
that the value is non-zero:
it is the infinitesimal defined by 1/infinity .
. by using  L'Hopital's Rules for Improper Integration
we can get a definite integral from
an infinity of infinitesimals .
[12.13: I doubt that applies to a  constant function over
naturals or integers, but I have another argument ...]

. I think it should be obvious from
the definition of probability function and infinity
that integration(0..infinity) 1/infinity = 1
or even integration(-infinity...infinity) 1/infinity = 1
simply because that is the mathematical way of saying
the fact that the chance of picking
the same item from an infinite set
is infinitesimally unlikely but not impossible:
the probability = 1/infinity = infinitesimal .
. integration(0..infinity) 1/infinity
= infinity * 1/infinity = 1;
but, integration(-infinity...infinity) 1/infinity = 1?
. the basic integration rules have this to say:
integration(-infinity..infinity) f(x)
= integration(-infinity..c) f(x)
+ integration(c..infinity) f(x);
so, integration(-infinity..infinity)
= integration(negatives)  + integration(naturals).
= infinity * 1/infinity + infinity * 1/infinity
= 1+1 =2, which is not what we want
for a probability function .
. if the cardinality of naturals is infinity
then cardinality of integers is 2*infinity
and cardinality of reals is 2*infinity*infinity,
because there an infinity of integers to choose from
and then an infinity of fractions also .