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Trouble with Limit[I ... k]/k, k-> Infinity]

Thank you all for the various answers and suggestions (sent directly to me
or otherwise). Let me summarize.

> May be it will be a good idea to break the original expression into
> these two subexpressions?

The original expression is very large (LeafCount[] ~40000) and contains
many slightly different instances of Limit[Exp[I A k]/k, k-> Infinity]
among other things. The less I have to do by hand the better.

> My suggestion is to use ComplexExpand prior to asking for the limit. (I
> also dropped the condition that A be nonzero since it seemed
> irrelevant.)
> In[3]:=
> Assuming[Element[A, Reals],
>   Limit[ComplexExpand[Exp[I*A*k]]/k, k -> Infinity]]
> Out[3]=
> 0

This seems to be the best bet. I agree that it is silly to write {a != 0}.
I wonder why it works.

Peter Pein:
> Union[Limit[
>   Exp[I k A]/k, k -> Infinity, Assumptions -> #] & /@ {A > 0, A < 0}]

Clever, I didn't realize that this was a way.

> I think it is slightly tricky for the internal code to figure out that
> there are two possibilities that behave in the same way. If you do
> Limit[Exp[I*k*t]/k, k->Infinity, Assumptions->t>0]
> or
> you get zero. I guess this means you will need to split into cases in
> such situations.

The trouble is that this is not a question of two cases in my actual
example. I am still struggling to understand the structure of the
resulting expression.

I realize that it can be hard for Mathematica for find its way around some
expressions but I am happy to help :-) The trouble is sometimes how.

In my case, I pattern match to replace Exp[I _ k_|-I _ k_]/_ k_^n_ by k^n
and (by the way) ExpIntegral[I (...) k] by I Pi Sign[(...)] as (...) is
real. This seems ugly but, I hope, right in my cases.

> For example: I had some expressions describing a three (ideal) gas
> mixture. Mathematica could not integrate them until I had told it the
> domain of every variable in the integration *and* that the sum of
> every combination of two mol fractions was less than 1.

In my case, I go on to evaluate the result by integration and this I have
not made work yet. Unfortunately, I will have to take care of a great
number of cases.

Thanks a lot of the help. I will post a follow up to ask for more help.


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