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>> +++