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MathGroup Archive 2001

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Re: integral function

  • To: mathgroup at smc.vnet.net
  • Subject: [mg31713] Re: integral function
  • From: pasquale.nardone at ulb.ac.be (Pasquale Nardone)
  • Date: Wed, 28 Nov 2001 01:29:36 -0500 (EST)
  • References: <9tp4ab$ja8$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

In article <9tp4ab$ja8$1 at smc.vnet.net>, "Higinio Ramos" <higra at usal.es> wrote:

> I have a function expressed in integral form, where the integral cannot 
> be made,
> y[t _ ]: = Integrate[Exp[Cos[s]]*Sin[t - s ], { s, 0, t } ]. 
> Nevertheless, if I indicate Mathematica that draws, it does. 
> Plot[y[t],{t,0,1}].
> What happens, does it make the numerical integration ?. How does it make 
> it, with what precision, what method?
> On the other hand, the function Derivative does not make it correctly, 
> it simply derives the part from the function integrating that depends on 
> the variable t although by hand it can be obtained without difficulty, 
> (and even with Mathematica, expanding previously the function Sin[t-s]). 
> Why?
> Thanks
> 
> Higinio Ramos

First the derivative is correct because Sin[t-s] when s->t is 0, so it
remains the derivative of Sin[t-s] which gives the Cos[t-s] (if you try
with a Cos[t-s] you will see the Exp[Cos[t]] appears correctly)

Yes i suppose that Plot generate an evaluation. 
By the way you can evaluate approximatively the integral by changing
variable u=Cos[t]
then integrate, you will end with an integral of Exp[u] u (1-u^2)^(-1)
or you can first expand Sin[t-s]=Sin[t]Cos[s]-Cos[t]Sin[s] then integrate
what you can then expand Exp[Cos[s]]=Cos[s]^n then evaluate again.


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