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
> 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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