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