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Re: Same Limit: OK in 5.2, fails in 6.0; Packages gone in 6.0 ???
*To*: mathgroup at smc.vnet.net
*Subject*: [mg77982] Re: [mg77959] Same Limit: OK in 5.2, fails in 6.0; Packages gone in 6.0 ???
*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>
*Date*: Thu, 21 Jun 2007 05:31:34 -0400 (EDT)
*References*: <200706200941.FAA10343@smc.vnet.net>
On 20 Jun 2007, at 18:41, jrc wrote:
> Why?
>
> I have,
>
> $Assumptions = {a > 0, k1 > 0, k2 >0, x e(in) Reals}
> (repeats ok)
>
> i1b = (1/a)* Integral from -a/2 to +a/2 of integrand:
>
> exp(- i k2 x) exp(i k1 x) dx
>
> with the expected result,
>
> i1b = fraction with numerator = 2 Sin[(1/2)a(k1-k2)]
> and denominator = a(k1-k2)
>
> Now I want the limit of this result, as the parameter a goes to
> infinity:
>
> Limit[i1b, a -> Infinity]
>
> Mathematica 5.2 gives correct result, zero;
> (note this is just lim(sin(x)/x, x -> inf), which is obviously
> zero)
>
> Mathematica 6.0 is unable to evaluate the limit.
>
> Ruskeepaa's "Navigator", 2nd ed, (written for v5.2) claims there
> is a package, "Calculus`Limit`" that makes 'Limit' work better
> (p. 395 and p. 430). However, no such package seems to exist in
> v6.0.
>
> How many packages no longer exist in 6.0 ???? Is there a list ????
>
> Can anyone give me a reason for this obvious failure?
>
> jrc
>
Whose failure?
Integrate[(Exp[(-I)*k2*x]*Exp[I*k1*x])/a, {x, -a/2, a/2}]
(2*Sin[(1/2)*a*(k1 - k2)])/(a*k1 - a*k2)
Assuming[Element[k1 | k2, Reals], Limit[%, a -> Infinity]]
0
The result is, of course, not true without the assumption (e.g. put
k1=2I, k2 = I).
Andrzej Kozlowski
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