My Math Forum can't understand the proof of "Interchange of integration"

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November 28th, 2013, 08:18 PM   #1
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can't understand the proof of "Interchange of integration"

<Introduction to Calculus and Analysis>by courant (vol 2 page 81)
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 November 28th, 2013, 08:20 PM #2 Newbie   Joined: Nov 2013 Posts: 9 Thanks: 0 Re: can't understand the proof of "Interchange of integratio if the last equation is right, do it mean u(x,a)=0 ?
November 29th, 2013, 06:33 AM   #3
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Re: can't understand the proof of "Interchange of integratio

Quote:
 Originally Posted by ittechbay if the last equation is right, do it mean u(x,a)=0 ?
No, I think this old guy just meant that:

$u_1(x,y)=\int_{\alpha}^y u_y(x,\eta)d\eta=u(x,y)-u(x,\alpha)$ (by Newton-Leibniz formula), thus $u(x,y)=u(x,\alpha)+u_1(x,y)=u(x,\alpha)+\int_{\alp ha}^y u_y(x,\eta)d\eta$.

November 29th, 2013, 03:20 PM   #4
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Re: can't understand the proof of "Interchange of integratio

Quote:
Originally Posted by stainburg
Quote:
 Originally Posted by ittechbay if the last equation is right, do it mean u(x,a)=0 ?
No, I think this old guy just meant that:

$u_1(x,y)=\int_{\alpha}^y u_y(x,\eta)d\eta=u(x,y)-u(x,\alpha)$ (by Newton-Leibniz formula), thus $u(x,y)=u(x,\alpha)+u_1(x,y)=u(x,\alpha)+\int_{\alp ha}^y u_y(x,\eta)d\eta$.
I know you said, I can't undersand the following equation
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 November 29th, 2013, 05:02 PM #5 Senior Member     Joined: Oct 2010 From: Changchun, China Posts: 492 Thanks: 14 Re: can't understand the proof of "Interchange of integratio The second equality holds because he has already said that: $v(x,y)=\int_{\alpha}^{y}f(x,\eta)d\eta$ , $u(x,y)=\int_{a}^{y} v(\xi,y)d\xi$

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