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Re: troublesome integral

  • To: mathgroup at
  • Subject: [mg126032] Re: troublesome integral
  • From: Yi Wang <tririverwangyi at>
  • Date: Fri, 13 Apr 2012 04:51:57 -0400 (EDT)
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  • References: <jm64ib$6la$>

Yes, this shouldn't be a hard integral. On my computer Mathematica cannot do it either. But there is a workaround: first let beta -> beta+alpha. Then the integral becomes =E2=88=AB cos[=CE=B2+=CE=B1] exp[ iz cos=CE=B2 ] d=CE=B2. Then expand cos[=CE=B2+=CE=B1] to cos[=CE=B2]cos[=CE=B1]-sin[=CE=B2]sin[=CE=B1]. Then Mathematica knows how to calculate.

In[17]:= Integrate[
 Cos[\[Alpha]] Cos[\[Beta]] Exp[I z Cos[\[Beta]]], {\[Beta], 0,
  2 \[Pi]}, Assumptions -> z \[Element] Reals]

Out[17]= 2 I \[Pi] BesselJ[1, z] Cos[\[Alpha]]

In[18]:= Integrate[-Sin[\[Alpha]] Sin[\[Beta]] Exp[
   I z Cos[\[Beta]]], {\[Beta], 0, 2 \[Pi]},
 Assumptions -> z \[Element] Reals]

Out[18]= 0

Thus the final result is the same as yours.


On Thursday, April 12, 2012 4:42:51 AM UTC-4, peter lindsay wrote:
> A couple of colleagues wondered about this. I've sent it on to support @ wolfram who are escalating it to the developers. Possibly someone here has an answer though ?
> Integrate[Cos[\[Beta]] Exp[I z Cos[\[Beta]-\[Alpha]]],{\[Beta],0,2 \[Pi]},Assumptions->z\[Element]Reals]
> doesn't seem to run.
> Answer should be
> 2 I \[Pi] BesselJ[1,z] Cos[\[Alpha]]  [ I think ]
> thanks
> Peter Lindsay

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