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        riddles >> medium >> analysis.. measure
        
(Message started by: MonicaMath on Oct 31^st, 2009, 11:06pm)

Title: analysis.. measure
Post by MonicaMath on Oct 31^st, 2009, 11:06pm

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

Title: Re: analysis.. measure
Post by Eigenray on Nov 1^st, 2009, 6:27am

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

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif H(x,y) dy = - e^-xy(cos xy + sin xy) / (2x),

so the integral from nhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/x to (n+1)http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/x is

(-1)ⁿ e^{-nhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif} (1+e^{-http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif})/(2x)

and so

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif₀^{http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif} |H(x,y)| dy = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif e^{-nhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif}(1+e^{-http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif})/(2x)
= (e^{http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif}+1)/(e^{http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif}-1) * 1/(2x)

Now it follows that the integral of |H| over (x,y) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif [0,http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif) x [0,http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif), or even [0, r] x [0,http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif), is infinite. Switching the roles of x and y the same is of course true for [0,http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif) x [0,s].

If we take just the positive part, H⁺ = max(H,0), then

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif H⁺(x,y) dy = (1+e^{-http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif})/(2x) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif_{n even} e^{-nhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif}
= e^{http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif}/(e^{http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif}-1) * 1/(2x),

and similarly for H^-, so neither is integrable.

Title: Re: analysis.. measure
Post by MonicaMath on Nov 1^st, 2009, 6:54am

thanks for replay as you always did,,,
can we follow the same argument for the
function G(x,y)=exp(-xy) sin(x)

??!!

Title: Re: analysis.. measure
Post by Eigenray on Nov 1^st, 2009, 11:38am

This one is more interesting because both http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif G(x,y) dx dy and http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif G(x,y) dy dx exist and equal http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/2, but the integral of |G(x,y)| is infinite, so Fubini's theorem does not apply directly (all integrals being from 0 to http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif).

Indeed,
http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif G(x,y) dy = sin(x)/x,
and http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif₀^M sin(x)/x dx http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/2 as M http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif, but http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif₀^{http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif} |sin(x)/x| dx = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif.

Doing it the other way,
http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif G(x,y) dx = 1/(1+y²),
whose integral converges just fine to http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/2, but
http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif |G(x,y)| dx = (e^{http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gify}+1)/(e^{http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gify}-1) * 1/(1+y²),
and the integral of this diverges due to the behavior as y http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif 0.

So for this one, Fubini applies to one of [0, r]x[0, http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif) and [0, http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif)x[0, s], but not the other.