Re: Forcing a parameter to be integer when using 'Integrate'
- To: mathgroup at smc.vnet.net
- Subject: [mg67570] Re: Forcing a parameter to be integer when using 'Integrate'
- From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
- Date: Sat, 1 Jul 2006 05:11:50 -0400 (EDT)
- Organization: The Open University, Milton Keynes, UK
- References: <e82nvi$r49$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Ian Linington wrote:
> Hello,
>
> I am a mathematica novice, so please excuse me if this question is a
> bit dumb...
>
> I would like to solve the integral:
>
>> Integrate[E^((n*I)*x)/Sqrt[1 + a^2*Sin[x]^2], {x,0,2*Pi}]
>
> with n integer and a real and positive.
> The problem is that I don't know how to tell Mathematica about these
> conditions on a and n.
>
> If I explicitly assign an integer value for n (i.e. 4 in this example),
> Mathematica solves the integral:
>
>> Integrate[E^((4*I)*x)/Sqrt[1 + a^2*Sin[x]^2], {x,0,2*Pi}]
>
> (with a few conditions) and gives the output
>
>> If[(Re[a^(-1)] != 0 || -Im[a^(-1)] < 0) &&
> (Re[a^(-1)] != 0 || -Im[a^(-1)] >= 0) && ...
> ... (lots more)...,
> (4*(-8*(2 + a^2)*EllipticE[-a^2] +
> (4 + a^2)*(4 + 3*a^2)*EllipticK[-a^2]))/(3*a^4),
> Integrate[E^((4*I)*x)/Sqrt[1 + a^2*Sin[x]^2],
> {x, 0, 2*Pi},
> Assumptions -> !((Re[a^(-1)] != 0 || -Im[a^(-1)] < 0)
> && (Re[a^(-1)] != 0 || -Im[a^(-1)] >= 0) && ...)
> ... (lots more) ...]]
>
> Simlilarly if we pick other integer values for n.
>
> The question is, how to get a generic formula in terms of a and n?
>
> Can anybody help please?
>
> Many thanks,
> Ian
>
Use Assumptions. For instance
Integrate[E^((n*I)*x)/Sqrt[1 + a^(2)*Sin[x]^(2)],
{x, 0, 2*Pi}, Assumptions -> {Element[n, Integers], a > 0}]
HTH,
Jean-Marc