AiryAi
- To: mathgroup at smc.vnet.net
- Subject: [mg75565] AiryAi
- From: dimitris <dimmechan at yahoo.com>
- Date: Sat, 5 May 2007 06:03:35 -0400 (EDT)
Hello.
Consider the integral
In[678]:=
f = HoldForm[Integrate[AiryAi[o], {o, a, b}]]
Then
In[679]:=
{ReleaseHold[f /. {a -> 0, b -> Infinity}], (Rationalize[#1, 10^(-12)]
& )[
ReleaseHold[f /. Integrate -> NIntegrate /. {a -> 0, b ->
Infinity}]]}
Out[679]=
{1/3, 1/3}
Hence, the symbolic and the numerical result agree.
Next,
In[682]:=
ReleaseHold[f /. {a -> -Infinity, b -> 0}]
N@%
Out[682]=
2/3
Out[683]=
0.6666666666666666
I think I can trust the last analytic result.
However, no matter what options I set for NIntegrate, I could get a
satisfactory
result from its application for the integral in (-infinity,0].
I really appreciate any help.
Thanks
Dimitris