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MathGroup Archive 2000

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Re: Difficult integral: correction

  • To: mathgroup at smc.vnet.net
  • Subject: [mg23457] Re: Difficult integral: correction
  • From: Roland Franzius <Roland.Franzius at uos.de>
  • Date: Thu, 11 May 2000 00:54:08 -0400 (EDT)
  • References: <8f559t$49c@smc.vnet.net> <8fb0th$hid@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

You are right, simplification time tends to infinity. But the integral
is tivial.

Standard way: Put Sin[x]-> (Exp[I x]-Exp[-I x])/(2 I)  and use 
Integrate[Exp[I n x] Exp[I m x], {x,0,2Pi}]
especially if n and \rho are integers.

Brute force: Calculate the indefinite integral (don't forget ";"!)
 
f[x_]=Integrate[(Sin[x])^24 Exp[I n x], x];

f[2Pi]-f[0]//Together
-((620448401733239439360000*I*(-1 + E^(2*I*n*Pi)))/
   ((-24 + n)*(-22 + n)*(-20 + n)*(-18 + n)*(-16 + n)*
    (-14 + n)*(-12 + n)*(-10 + n)*(-8 + n)*(-6 + n)*
    (-4 + n)*(-2 + n)*n*(2 + n)*(4 + n)*(6 + n)*(8 + n)*
    (10 + n)*(12 + n)*(14 + n)*(16 + n)*(18 + n)*(20 + n)*
    (22 + n)*(24 + n)))

Marco de Innocentis wrote:
> 
> I am sorry, I made a silly mistake: rho was supposed to be part
> of the exponential, so
> 
> > Integrate [(Sin[x])^[2 rho] Exp[I n x],{x,0,2 Pi}]
> 
> Thanks,
> 
> Marco
> 
> In article <8f559t$49c at smc.vnet.net>,
>   Marco de Innocentis <mdi11 at hotmail.com> wrote:
> 
> > I need to calculate the following integral
> >
> > Integrate [(Sin[x])^2 rho Exp[I n x],{x,0,2 Pi}]
> >
> > as a function of the two parameters rho and n. I have tried it
> > for rho = 1, 4, 8, 10, and it works, but it doesn't work at all
> > for rho = 12. I left the computer on all night but only got the
> > same expression as a result.
> > Any suggestions?
> > Thanks
> >
> > Marco
> >
> > Sent via Deja.com http://www.deja.com/
> > Before you buy.
> >
> >
> 
> --
> http://www.thehungersite.com
> 
> Sent via Deja.com http://www.deja.com/
> Before you buy.

-- 
Roland Franzius

  +++ exactly <<n>> lines of this message have value <<FALSE>> +++


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