Whenever someone says something like "for ANY exponent", I can bet that they won't be going to throw an easy problem at any offered code.
So, looking at the zipf distribution, it turns out to be an easy one to deal with, sort of. The zipf pdf is simply
p_zipf(x,p) = -x^(p+1)/zeta(p+1)
This is a discrete distribution in x, over the positive integers.
If we look at the pdf, we recognize that zeta(p+1) is the limit of the sum of powers of x, x^-(p+1), where x ranges from 1 to inf. So zeta(p+1) merely normalizes that pdf to have unit area.
Here the zipf exponent that was referred to would be p, I assume, but there are unresolved questions hidden in the question. I'll quote the line you wrote:
"However, i need to generate Zipf numbers which can be less than 1."
So, IF your question is to generate zipf samples that are non-integer, this makes no sense, since the zipf distribution is a discrete one, that samples over the positive integers.
If you are calling the zipf exponent (p+1), and you wish to sample such that p+1 is less than 1, so p is less than zero, then you will have a great deal of trouble.
If your question is to sample for non-integer values of p that are less than one, but greater than zero, then this is possible, but a serious bit of an effort for many values of p less than 1.
By the way, since the zipf pdf involves zeta(p+1), this would employ the symbolic toolbox, because for small p, the zeta function converges pretty slowly.
Sadly, I cannot proceed further without some resolution of the above questions.
Edit: Though I've not yet gotten a response to my questions, I took a quick look at the randraw code. Randraw defines the zipf pdf such that
p_zipf(x,a) = -x^(-a)/zeta(a)
and then requires only that a>1.
It would appear that you want to generate zipf numbers for a<=1, as randraw defines that distribution. If this is so, then you cannot do so. The sum...
sum(x^-a)
where x is the set of positive integers, is inf for a<=1.
This is why randraw requires that a>1, as otherwise, the zeta function has a problem. Even for a near 1, I would predict that randraw becomes slow.
