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