On Feb 1, 4:47 pm, Simon <simonjty...@gmail.com> wrote:
> Hi, this is hopefully an easy question:
>
> As a simple exercise, I'm trying to show that \int_0^{2\pi} e^{i (m-n)
> x}dx = 2\pi\delta_{mn} for integer m, n.
> Here's how I did it:
>
> sage: var('m,n'); w = SR.wild(0);
> sage: assume(n, 'integer');assume(m, 'integer')
> sage: int = integrate(e^(i*(m-n)*x),x,0,2*pi)
> sage: print int.limit(m=n)
> sage: print int.subs({e^(w):cosh(w)+sinh(w)}).simplify_trig()
> 2*pi
> 0
>
> The bit I don't like is using the substitution... it should not be
> nesc. The problem, I think, lies at
>
> sage: sin(2*pi*n).simplify_trig() # this works
> sage: e^(i*2*pi*m).simplify_full() # this doesn't work
> 0
> e^(2*I*pi*m)
In Maxima you would use rectform to convert the expression from polar
to rect form:
sage: int._maxima_().rectform()
0
sage: e^(i*2*pi*m)._maxima_().rectform()
e^(2*I*pi*m)
sage: exp(i*2*pi*m)._maxima_().rectform()
1
HTH, Andrej
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