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Re: nasty integrals that Mathematica can't do

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  • Subject: [mg10819] Re: nasty integrals that Mathematica can't do
  • From: Paul Abbott <>
  • Date: Tue, 10 Feb 1998 21:01:33 -0500
  • Organization: University of Western Australia
  • References: <6baoqa$>

Burn Microsoft...Burn! wrote:

> In my research, i've come up with the following integral:
>     Integrate[ CosIntegral[a x] Cos[b x], {x,0,Infinity} ]
> Gradshteyn and Ryzhik have this integral, but apparently, Mathematica
> 3.0.1 can't do it.

Mathematica can do the indefinite integral.
> My question: Is there a way to "teach" Mathematica how to do a
> particular type of integral?  Can I somehow plug G+R's solution in
> Mathematica?

Sort of ...

In[1]:= Unprotect[Integrate];

In[2]:= Integrate[Cos[p_ x] CosIntegral[q_ x], {x, 0, Infinity}] := 
	Which[p^2 > q^2, -(Pi/(2 p)), p^2 == q^2, -(Pi/(4 p)), p^2 < q^2, 0]

In[3]:= Protect[Integrate];

In[4]:=Integrate[Cos[a x] CosIntegral[b x], {x, 0, Infinity}] Out[4]=
       2    2    Pi     2     2    Pi     2    2 Which[a  > b , -(---),
a  == b , -(---), a  < b , 0]
                 2 a               4 a

In[5]:=Integrate[Cos[3 x] CosIntegral[2 x], {x, 0, Infinity}] Out[5]=

In[6]:=Integrate[Cos[2 x] CosIntegral[3 x], {x, 0, Infinity}] Out[6]= 0

However, if you have an integrand which includes other terms there is no
guarantee that the above pattern will be recognized before the
integrator tries to evaluate the whole expression.

Note that Mathematica can do 

	Integrate[ Exp[-c x] Cos[a x]/a Cos[b x], {x,0,Infinity} ]

and that integrating over a gives you an equivalent integral (and a very
simple form).  You can then take take the limit as c->0.


Paul Abbott                                   Phone: +61-8-9380-2734
Department of Physics                           Fax: +61-8-9380-1014
The University of Western Australia            Nedlands WA  6907  AUSTRALIA                   

            God IS a weakly left-handed dice player

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