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Topic: analysis.. measure (Read 828 times) |
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MonicaMath
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analysis.. measure
« on: Oct 31st, 2009, 11:06pm » |
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Hi,
I'm working on product measure and I have this quation
H(x,y)=exp(-xy) sin(xy) , in [0,r]X[0,s] r,s>0
why we cannot apply Fubini's Theorem for changing integrals in [0, inf]X[0,inf] ?
can we use it if :
(1) r=inf, s<inf
and
(2) r<inf, s=inf ?
thanks for helping in advance
we can't use
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Eigenray
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Re: analysis.. measure
« Reply #1 on: Nov 1st, 2009, 6:27am » |
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Fubini's theorem requires that the integral of the absolute value of the function be finite, in at least one of the three ways of calculating it (but if one is, the other two are).
In this case, we have
 H(x,y) dy = - e-xy(cos xy + sin xy) / (2x),
so the integral from n /x to (n+1)/x is
(-1)n e-n (1+e-)/(2x)
and so
0 |H(x,y)| dy =  e-n(1+e-)/(2x)
= (e+1)/(e-1) * 1/(2x)
Now it follows that the integral of |H| over (x,y)  [0,) x [0,), or even [0, r] x [0,), is infinite. Switching the roles of x and y the same is of course true for [0,) x [0,s].
If we take just the positive part, H+ = max(H,0), then
H+(x,y) dy = (1+e-)/(2x) n even e-n
= e/(e-1) * 1/(2x),
and similarly for H-, so neither is integrable.
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MonicaMath
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Re: analysis.. measure
« Reply #2 on: Nov 1st, 2009, 6:54am » |
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thanks for replay as you always did,,,
can we follow the same argument for the
function G(x,y)=exp(-xy) sin(x)
??!!
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| « Last Edit: Nov 1st, 2009, 7:00am by MonicaMath » |
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Eigenray
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Re: analysis.. measure
« Reply #3 on: Nov 1st, 2009, 11:38am » |
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This one is more interesting because both G(x,y) dx dy and G(x,y) dy dx exist and equal /2, but the integral of |G(x,y)| is infinite, so Fubini's theorem does not apply directly (all integrals being from 0 to ).
Indeed,
G(x,y) dy = sin(x)/x,
and 0M sin(x)/x dx  /2 as M , but 0 |sin(x)/x| dx = .
Doing it the other way,
G(x,y) dx = 1/(1+y2),
whose integral converges just fine to /2, but
|G(x,y)| dx = (ey+1)/(ey-1) * 1/(1+y2),
and the integral of this diverges due to the behavior as y 0.
So for this one, Fubini applies to one of [0, r]x[0, ) and [0, )x[0, s], but not the other.
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