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

  • To: mathgroup at smc.vnet.net
  • Subject: [mg86968] Re: Limit[(x - Log[Cosh[x]]) SinIntegral[x], x -> Infinity]
  • From: "David W.Cantrell" <DWCantrell at sigmaxi.net>
  • Date: Thu, 27 Mar 2008 08:20:17 -0500 (EST)
  • References: <fsd6q7$9hi$1@smc.vnet.net>

Szabolcs_Horvát <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]

I confirm the precise behavior you described. It's surprising that a
restart was required.

> Are there any options (perhaps SystemOptions) that would allow Limit to
> give an answer directly for the product of the terms?

In[4]:= ClearSystemCache[]

In[5]:= Limit[(x - Log[Cosh[x]]) SinIntegral[x], x -> Infinity]

Out[5]= Limit[(x - Log[Cosh[x]]) SinIntegral[x], x -> Infinity]

In[6]:= Limit[FullSimplify[(x - Log[Cosh[x]]) SinIntegral[x], x > 0],
  x -> Infinity]

Out[6]= (1/2)*Pi*Log[2]

Note that using FullSimplify merely changes -Log[Cosh[x]] to +Log[Sech[x]],
but curiously, that seems to help.

David


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