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Author Comment/Response
Peter Pein
08/26/12 5:17pm

Hi Amir,

after losing patience I used a simpler approach. y is just a phase shift and has no influence to the value of the integral. 2 Pi is a multiple of the period w.r.t. x and due to the even powers there is a bit more symmetry.

So defining
f[x_, y_, a_] = (1 + a^4 Tan[2 x - 2 y]^2)/(1 + a^2 Tan[2 x - 2 y]^2);

the value at x==k*Pi/4, k integer, y==0 has to be determined by the use of limits. So I added an intermediate point of integration at Pi/4.

the value of the integral is

value = Simplify[
4*Integrate[f[x, 0, a], {x, 0, Pi/4, Pi/2},
Assumptions -> Element[a, Reals]]
]

this evaluates to
Pi*(2 - a + 2*a^2 - Abs[a])
in a few seconds.

For a in {-1, 0, 1} this value is 2*Pi.

maybe the wrong result you got is due to the removable singularities at x = k*Pi/4 + Mod[y, Pi/4]?

Peter

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Subject (listing for 'Bug in Integration')
Author Date Posted
Bug in Integration Amir 08/21/12 09:32am
Re: Bug in Integration Nasser M. Ab... 08/21/12 11:41am
Re: Bug in Integration jf 08/21/12 1:13pm
Re: Re: Bug in Integration Amir 08/24/12 10:31am
Re: Bug in Integration Peter Pein 08/26/12 5:17pm
Re: Re: Bug in Integration Amir 08/31/12 09:41am
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