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MathGroup Archive 2007

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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



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