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.