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

*To*: mathgroup at smc.vnet.net*Subject*: [mg86950] Re: [mg86932] Limit[(x - Log[Cosh[x]]) SinIntegral[x], x ->*From*: Bob Hanlon <hanlonr at cox.net>*Date*: Thu, 27 Mar 2008 08:16:55 -0500 (EST)*Reply-to*: hanlonr at cox.net

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? >