MathGroup Archive 2008

[Date Index] [Thread Index] [Author Index]

Search the Archive

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


Starting with a fresh kernel.

$Version

6.0 for Mac OS X x86 (64-bit) (March 13, 2008)

Limit[#, x -> Infinity] & /@ ((x - Log[Cosh[x]])*SinIntegral[x])

(1/2)*Pi*Log[2]

I also get this result after calculating the individual intermediate result=
s in either order (fresh kernel in each case).


Bob Hanlon

---- "Szabolcs Horv=C3=A1t" <szhorvat at gmail.com> wrote:
> I expected that Limit would try to compute the limit of terms in a
> product separately.  Perhaps this is not feasible because of the large
> number of possible combinations?
>
> In the following example it can give result for the terms separately,
> but not for the product:
>
> In[1]:= Limit[(x - Log[Cosh[x]])*SinIntegral[x], x -> Infinity]
> Out[1]= Limit[(x - Log[Cosh[x]])*SinIntegral[x], x -> Infinity]
>
> In[2]:= Limit[x - Log[Cosh[x]], x -> Infinity]
> Out[2]= Log[2]
>
> In[3]:= Limit[SinIntegral[x], x -> Infinity]
> Out[3]= Pi/2
>
> In[4]:= Limit[(x - Log[Cosh[x]])*SinIntegral[x], x -> Infinity]
> Out[4]= Limit[(x - Log[Cosh[x]])*SinIntegral[x], x -> Infinity]
>
> (Note that In[4] still didn't give a result.)
>
> However, if the evaluations are done in a different order (after a
> kernel restart), it is able to compute the result:
>
> In[1]:= Limit[x - Log[Cosh[x]], x -> Infinity]
> Out[1]= Log[2]
>
> In[2]:= Limit[SinIntegral[x], x -> Infinity]
> Out[2]= Pi/2
>
> In[3]:= Limit[(x - Log[Cosh[x]])*SinIntegral[x], x -> Infinity]
> Out[3]= (1/2)*Pi*Log[2]
>
>
> Are there any options (perhaps SystemOptions) that would allow Limit to=

> give an answer directly for the product of the terms?
>



  • Prev by Date: Re: Re: Tagged list processing
  • Next by Date: Re: Counting nonzeros
  • Previous by thread: Re: Equi-sized tick labels
  • Next by thread: Mathematica SIG (Washington DC Area)