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Re: Re: Limit[(x - Log[Cosh ]) SinIntegral , x -> Infinity]

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• Subject: [mg86990] Re: [mg86947] Re: Limit[(x - Log[Cosh ]) SinIntegral , x -> Infinity]
• From: Daniel Lichtblau <danl at wolfram.com>
• Date: Fri, 28 Mar 2008 03:16:11 -0500 (EST)
• References: <20080326082007.401$ci@newsreader.com> <200803271316.IAA19623@smc.vnet.net> <47EBA98B.8000606@wolfram.com> <f831b3d60803271113w428980e1t99d0a046171313f6@mail.gmail.com>

Szabolcs Horvát wrote:
> [...]
> But I am still curious about why Mathematica can compute the two
> limits separately but not the limit of their product. 

Offhand I do not know. I found that both give results in the development 
version, hence had insufficient incentive to look further.


> I could imagine
> that Limit[] has some built-in constraints, e.g. it only tries a
> limited number of transformations on the expression, or it has some
> internal time constraints.   If I understood correctly, Bob Hanlon
> reported that Mathematica does give an answer on his computer---this
> could be explained by time constraints and a faster computer.

Yes, this could be what is behind it. Also there are some leaf count 
constraints used in places. Neither is as big an issue as are 
constraints used in definite integration, but they are indeed present, 
to an extent, in Limit.


> If this is really the case, is there a way to ask Mathematica to work
> "harder" (perhaps try longer), so that it will be able to give an
> answer?  (For example, Simplify has the option TimeConstraint.)
> 
> Szabolcs Horvát

At this time there is no such option. Not even a hidden one. (Or, if 
there is such, it is hidden from me.)


Daniel Lichtblau
Wolfram Research

• References:
  • Re: Limit[(x - Log[Cosh[x]]) SinIntegral[x], x -> Infinity]
    • From: "Szabolcs HorvÃt" <szhorvat@gmail.com>